Calculus 1 Exam 3 Review Sheet

Problem 1

Show that (x^5)-5x-1 has exactly 3 real roots.   How many nonreal roots does it have?

Solution

Let f(x)=(x^5)-5x-1.  Then f'(x)=5(x^4)-5.  Setting this equal to zero gives us 5(x^4)-5=0, or (x^4)-1=0.  This factors as (x-1)(x+1)((x^2)+1)=0, which has the roots x=1,-1, i,-i.  So f'(x) has 2 distinct real roots.  The corollaries to Rolle’s Theorem tell us that means f(x) has at most 3 distinct real roots. Making a table of values for f(x)=(x^5)-5x-1 gives us

TableForm[ Table[ Map[Style[ToString[#1], FontFamily-> "TimesNewRoman"]&,{x,x^5-5x-1}],{x,-4,4}], TableHeadings-> {None,{TraditionalForm[Style[x, FontFamily-> "TimesNewRoman"]], TraditionalForm[ Style[x^5-5x-1, FontFamily-> "TimesNewRoman"]]}},, TableAlignments-> Center]

As f(x)=(x^5)-5x-1 is continuous the sign switches in the table and the Intermediate Value Theorem implies that f(x) must have a zero between -2 and -1, between -1 and 0, and between -1 and 2.  So f(x) has at least 3 distinct real roots.

As f(x)=(x^5)-5x-1 has at most 3 distinct real roots and at least 3 distinct real roots, it has exactly 3.  As since by the Fundamental Theorem of Algebra it has 5 roots, it must have 2 non-real roots.

Note: In problems like this you won’t always get an exact match between the lower bound you get from the Intermediate Value Theorem and the upper bound from Rolle’s Theorem and its corollaries.  In that case the best you get is a range for the number of roots (like “between 2 and 5 real roots”).

Problem 2

What does Rolle’s Theorem and its corollaries say about the number of real roots of (x^5)+10(x^4)+50(x^3)+(x^2)-3x+1?

Solution

Let f(x)=(x^5)+10(x^4)+10(x^3)+(x^2)-3x+1.  Then f'(x)=5(x^4)+40(x^3)+150(x^2)+2x-3; this is too complicated for us to try to set equal to 0 and solve.  So we move down another derivative to get f''(x)=20(x^3)+120(x^2)+300x+2.  This is likely too hard for most people to set equal to 0 and solve (see the note at the end of the problem), so we pass down to yet another derivative to get f'''(x)=60(x^2)+240x+300.  As this is quadratic we can find the number of real roots by looking at the quadratic discriminant Δ=(b^2)-4a c=(240^2)-4 60 300=-14400.  As the discriminant is negative f'''(x) has 0 real roots.  As we took 3 derivatives that means the original function  f(x)=(x^5)+10(x^4)+10(x^3)+(x^2)-3x+1 has at most 0+3=3 real roots.

If you use the Intermediate Value Theorem to try to find a lower bound for the real roots you will only be able to find 1 sign change - so you know the polynomial has between 1 and 3 real roots.  In fact it has exactly 1 real root but this is harder to show.

Note:  If you took some advanced algebra (like the Classical Algebra & Geometry summer course) you could tell that f''(x)=20(x^3)+120(x^2)+300x+2 has exactly 1 real root, which would tell you the original has 1+2=3 real roots.

Problem 3

Find those points x=c predicted by the Mean Value Theorem for the function (x^3)-x on the interval [-2,2].

Solution

Let f(x)=(x^3)-x.  Then f(2)=6 and f(-2)=-6.  The MVT states that if f(x) is differentiable on (a,b) and continuous on [a,b] (which our function is), then for at least one point c in (a,b), f'(c)=((f(b)-f(a))/(b-a)).  Applying this with a=-2 and b=2, f'(c)=((f(2)-f(-2))/(2-(-2)))=((6-(-6))/4)=3.  f'(x)=3(x^2)-1. Solving 3(x^2)-1=3 gives (x^2)=(4/3), or x=±(Sqrt[(4/3)]).  Both of these values are in [-2,2].

Problem 4

Show that if a and b are in the interval (-(π/2),(π/2)) with then|tan(b)-tan(a)|≥|b-a|.

Solution

Since the statement |tan(b)-tan(a)|≥|b-a| is the same if you switch a and b, I will assume a<b.  f(x)=tan(x) will be continuous on [a,b] and differentiable on (a,b), so the Mean Value Theorem applies to it on that interval.  Therefore for some point x=c in (a,b), f'(c)=((f(b)-f(a))/(b-a)).  As the derivative of tan(x) is (sec^2)(x), this becomes (sec^2)(c)=((tan(b)-tan(a))/(b-a)) or tan(b)-tan(a)=(sec^2)(c)(b-a).  Taking absolute values gives |tan(b)-tan(a)|=|(sec^2)(c)(b-a)|=|(sec^2)(c)| |b-a|.  |sec(x)|≥1 so |(sec^2)(c)|≥ 1 and we have |tan(b)-tan(a)|=|(sec^2)(c)| |b-a|≥ 1 |b-a|=|b-a|.

Problem 5

Suppose that L(x) is a function which is defined for x>0, L'(x)=(1/x), and L(1)=0.  Show that for any positive number a, L(a x)=L(a)+L(x).

Solution

Using the chain rule (d/(dx))L(a x)=L'(a x) (d/(dx))(a x)=(1/(a x))a =(1/x)=(d/(dx))L(x).  Since L(x) is differentiable for x>0 and has the same derivative as L(a x), L(a x) and L(x) must differ by some constant c; that is L(a x)=L(x)+c as long as x>0.  Plugging in x=1 gives L(a)=0+c, or c=L(a).  So L(a x)=L(x)+L(a) as desired.

Note: L(x) is really the function ln(x), which you may have seen in pre-Calculus. Many of the other properites of logarithms can be proven the same way.

Problem 6

Suppose that s(x) and c(x) are functions with s'(x)=c(x), c'(x)=-s(x), s(0)=0, c(0)=1.  What is the function ((s(x))^2)+((c(x))^2)?

Solution

Let f(x)=((s(x))^2)+((c(x))^2).  Then f(0)=((s(0))^2)+((c(0))^2)=(0^2)+(1^2)=1.  In addition on taking the derivative of f(x)  we see that:

({{({{f'(x)=(d/(dx))(((s(x))^2)+((c(x))^2))=(d/(dx))(((s(x))^2))+(d/(dx))(((c(x))^2))=2s(x)s'(x)+2c(x)c'(x)=}, {2s(x)c(x)-2s(x)c(x)=0}})}})

As f'(x) is always 0, f(x) must be a constant function.  Since f(0)=1, that constant must be 1. So f(x)=((s(x))^2)+((c(x))^2)=1.

Note: Obviously s(x) is meant to be sin(x) and c(x) is cos(x).  This shows you the identity (sin^2)(x)+(cos^2)(x)=1 can be deduced from Calculus as long as you have the derivative rules and some basic values.  In fact it is possible to skip trigonometry altogether and define sine and cosine strictly in terms of their derivatives.

Problem 7

Explain the Mean Value Theorem in terms of instantaneous and average rates of change.

Solution

The MVT states that if f(x) is continuous on [a,b] and differentiable on (a,b), then for some point c in (a,b) f'(c)=((f(b)-f(a))/(b-a)).  The quantity on the right is the average rate of change of f(x) on [a,b] (geometrically the slope of the line joining the 2 points (a,f(a)) and (b, f(b)) ) and the quantity on the left the instantaneous rate of change of f(x) at the point x=c.  So the MVT states that under the right conditions the instantaneous rate of change must equal the average rate of change at least once.  So for example if your average velocity on a trip is 50 miles an hour, at som epoint on the trip you must have been going exactly 50 miles an hour.

Problem 8

Find the linear approximations to y=(Sqrt[x]) at x=1,4,9.  Use these to estimate the square root of 2.  Which is the most accurate?

Solution

The linear approximation formula is L(x)=f(a)+f'(a)(x-a), where f(x) is the function and x=a is the center of the approximation.  For this problem f(x)=(Sqrt[x]) and f'(x)=(1/(2(Sqrt[x])))

For x=1:

f(1)=(Sqrt[1])=1 and f'(1)=(1/(2(Sqrt[1])))=(1/2).  So L(x)=1 +(1/2)(x-1), and (Sqrt[2])≈L(2)=1+(1/2)(2-1)=(3/2)=1.5

For x=4:

f(4)=(Sqrt[4])=2 and f'(4)=(1/(2(Sqrt[2])))=(1/4).  So L(x)=2 +(1/4)(x-4), and (Sqrt[2])≈L(2)=2+(1/4)(2-4)=(3/2)=1.5

For x=9:

f(9)=(Sqrt[9])=3 and f'(3)=(1/(2(Sqrt[9])))=(1/6).  So L(x)=3+(1/6)(x-9), and (Sqrt[2])≈L(2)=3+(1/6)(2-9)=(11/6)≈1.83

Both x=1 and x=4 are tied for accuracy by sheer luck.

Problem 9

Estimate (27.5^(1/3)) by linear approximation.

Solution

The function involved is f(x)=(x^(1/3)), and a nice value for this near 27.5 is a=27 (f(27)=3).  f'(x)=(1/(3(x^(2/3)))), so f'(27)=(1/27). Therefore the linear approximation is given by L(x)=f(a)+f'(a)(x-a)=3+(1/27)(x-27). Therefore (27.5^(1/3))≈L(27.5)=3+(1/27)(27.5-27)=3+(1/54)≈3.01851852.  The true value is (27.5^(1/3))≈3.01840537.

Problem 10

Use linear approximation to estimate the change of volume of a cube as its side length goes from 5 to 5.2 inches.

Solution

The formula for the volume of a cube in terms of its side length is V=(s^3).  The relative change version of linear approximation is Δ f≈f'(x)Δ x.  In this case that would be ΔV≈3(s^2)Δ s.  s=5 and Δ s=.2, so we have the change in volume is about 3 (5^2) ×.2=15 cubic inches.

The true value would be (5.2^3)-(5^3)=15.608 cubic inches.

Problem 11

Use Newton’s method to estimate all of the roots of (x^3)-3x+1=0 to 3 decimal places.

Solution

First note that this equation is in "=0" format, so we can use Newton’s method directly with f(x)=(x^3)-3x+1.  The Newton’s method formula is (Subscript[x, new])=(Subscript[x, old])-((f((Subscript[x, old])))/(f'((Subscript[x, old])))).  As f'(x)=3(x^2)-3, this would become (Subscript[x, new])=(Subscript[x, old])-((((((Subscript[x, old])))^3)-3((Subscript[x, old]))+1)/(3((((Subscript[x, old])))^2)-3)).  We will still need to see how many solutions there should be and roughly what they are to start.  Graphing f(x) we have

Plot[ x^3-3x+1, {x,-3,3}, PlotLabel-> TraditionalForm[y==(x^3)-3x+1]]

Solutions to f(x)=0 would be x-intercepts on this graph.  There are 3 solutions, apparently near x=-2, x=.25, and x=1.5.  Running these through the formula (Subscript[x, new])=(Subscript[x, old])-((((((Subscript[x, old])))^3)-3((Subscript[x, old]))+1)/(3((((Subscript[x, old])))^2)-3)) gives us:

For x=-2:

TableForm[ NestList[ {#1[[1]]+1, N[#1[[2]]-(#1[[2]]^3-3 #1[[2]] + 1)/(3#1[[2]]^2-3),20]}&, {0,N[-2,20]},5], TableHeadings->{None, {"iteration number", "(Subscript[x, new])"}}, TableAlignments-> Center]

For x=.25:

TableForm[ NestList[ {#1[[1]]+1, N[#1[[2]]-(#1[[2]]^3-3 #1[[2]] + 1)/(3#1[[2]]^2-3),20]}&, {0,N[1/4,20]},5], TableHeadings->{None, {"iteration number", "(Subscript[x, new])"}}, TableAlignments-> Center]

For x=1.5:

TableForm[ NestList[ {#1[[1]]+1, N[#1[[2]]-(#1[[2]]^3-3 #1[[2]] + 1)/(3#1[[2]]^2-3),20]}&, {0,N[3/2,20]},5], TableHeadings->{None, {"iteration number", "(Subscript[x, new])"}}, TableAlignments-> Center]

Note that it doesn’t take long to get 3 digits after the decimal point to stay the same.  This is much faster in practice than the Method of Bisection, although there are a few (fairly rare) cases where Newton’s Method can break down.

Problem 12

Use Newton’s method to estimate to estimate the positive roots of the equation (x^2)=cos(x).

Solution

This equation is not in the “=0” format so we will need to move cos(x) to the other side to get (x^2)-cos(x)=0.  So we can treat this as finding the positive roots of f(x)=(x^2)-cos(x).  Graphing f(x) for x≥0 we have

Plot[ x^2-Cos[x], {x,0,3}, PlotLabel-> TraditionalForm[y==x^2-Cos[x]]]

So it looks like there’s only one positive root and it is somewhere near x=0.75. The general Newton’s method formula is (Subscript[x, new])=(Subscript[x, old])-((f((Subscript[x, old])))/(f'((Subscript[x, old])))).  As f'(x)=2x+sin(x), this becomes (Subscript[x, new])=(Subscript[x, old])-((((Subscript[x, old])^2)-cos((Subscript[x, old])))/(2(Subscript[x, old])+sin((Subscript[x, old])))).  Using x=0.75 as the starting value we have

TableForm[ NestList[ {#1[[1]]+1, N[#1[[2]]-(#1[[2]]^2-Cos[#1[[2]]])/(2#1[[2]]+Sin[#1[[2]] ]),30]}&, {0,N[3/4,30]},7], TableHeadings->{None, {"iteration number", "(Subscript[x, new])"}}, TableAlignments-> Center]

So to far too many places it looks like the unique positive root is x≈.824132312302522422960956786.

Note: We can verify that there are no roots past x=1 - as cosine is trapped from -1 to 1, when x>1 we have (x^2)-cos(x)>(1^2)-1=0.

Problem 13

Use Newton’s method to estimate (27.5^(1/3)) to 8 places.

Solution

At first glance this doesn’t look like Newton’s method as there’s no equation involved.  However (27.5^(1/3)) literally means “that which when cubed is 27.5”, or a solution to (x^3)=27.5.  Putting this in “=0” format gives us (x^3)-27.5=0, so we can apply Newton’s Method with f(x)=(x^3)-27.5.  (27.5^(1/3)) is probably close to 3 (as (27^(1/3))=3), we can use x=3 as our initial guess.  f'(x)=3(x^2), so the general Newton’s Method formula (Subscript[x, new])=(Subscript[x, old])-((f((Subscript[x, old])))/(f'((Subscript[x, old])))) becomes (Subscript[x, new])=(Subscript[x, old])-((((((Subscript[x, old])))^3)-27.5)/(3((((Subscript[x, old])))^2))).  This gives us

TableForm[ NestList[ {#1[[1]]+1, N[#1[[2]]-(#1[[2]]^3-275/10)/(3#1[[2]]^2),20]}&, {0,N[3,20]},5], TableHeadings->{None, {"iteration number", "(Subscript[x, new])"}}, TableAlignments-> Center]

So to 8 decimal places (27.5^(1/3))≈3.01840537.  Note that this is much more accurate than the similar result in problem 9 which used linear approximation.

Problem 14

Find the maximum and minimum value of f(x)=(x^2)+(x^(-2)) on the interval [(1/2),3].

Solution

As the function is continuous on the interval, we need only find the critical numbers and then make a table of values that includes the endpoints x=(1/2) and 3.
f'(x)=2x-2(x^(-3))=((2((x^4)-1))/(x^3)).  This is undefined only at x=0, which is not in our interval.  This will be zero only when the numerator is; (x^4)-1=0 implies (x-1)(x+1)((x^2)+1)=0 which has the solutions x=-1,1,i,-i.  As we only count real solutions in [(1/2),3] we only use x=1.  Making a table of values which includes the endpoints we have

TableForm[ {x, x^2+x^-2} /. {{x-> 1/2}, {x-> 1}, {x-> 3}}, TableHeadings-> {None, {TraditionalForm[x], TraditionalForm[HoldForm[x^2+x^-2]]}}, TableAlignments-> Center]

Looking at the table the absolute minimum is 2 (which occurs at x=1) and the maximum us (82/9) (which occurs at x=3).

Problem 15

Find the maximum and minimum values for the function x(Sqrt[1-8(x^2)])

Solution

Let f(x)=x(Sqrt[1-8(x^2)]).  This function is only defined when 1-8(x^2)≥0; that is, when 8(x^2)≤1 ⟹ (x^2)≤ (1/8) ⟹ |x|≤(1/(Sqrt[8])) ⟹  -(1/(2(Sqrt[2])))≤x≤(1/(2(Sqrt[2]))). This is a closed interval so we need only find the critical numbers and then make a table of values (a table which includes the endpoints). Using the product rule and simplifying gives us f'(x)=(Sqrt[1-8(x^2)])-((8(x^2))/(Sqrt[1-8(x^2)]))=((1-16(x^2))/(Sqrt[1-8(x^2)])).  This exists as long as 1-8(x^2)>0 (> and not ≥ to avoid division by 0), which is when -(1/(2(Sqrt[2])))<x<(1/(2(Sqrt[2]))); so f'(x) exists except at the endpoints which must be checked anyway.  f'(x) will be 0 when 1-16(x^2)=0 ⟹ (x^2)=(1/16) ⟹ x=±(1/4) (both of which lie in the interval).  So making our table of values we have

TableForm[ {x, x Sqrt[1-8x^2]} /. {{x-> -1/(2 Sqrt[2])}, {x-> -1/4},{x->1/4}, {x-> 1/(2 Sqrt[2])}}, TableHeadings-> {None, {TraditionalForm[x], TraditionalForm[ x Sqrt[1-8x^2]]}}, TableAlignments-> Center]

From this we can see the maximum value is (1/(4(Sqrt[2]))) (which occurs at x=(1/4)) and the minimum value is -(1/(4(Sqrt[2]))) (which occurs at x=-(1/4)).

In problems 16-25 find where the functions are increasing, decreasing, concave up, and concave down.  Find and classify the critical points using both the first and second derivative tests and find all inflection points.  Graph the functions, including the behavior as x-> ±∞ and any asymptotes.

Problem 16

f(x)=2(x^3)-9(x^2)+12x-5

Solution

For this function we have f'(x)=6(x^2)-18x+12=6((x^2)-3x+2)=6(x-1)(x-2) and f''(x)=12x-18.  From this we can see that f'(x) always exists and is only 0 at x=1,2.  So the only critical points are at (1,f(1))=(1,0) and (2, f(2))=(2,-1).  Likewise f''(x) always exists and is only 0 at x=(3/2), so the only potential inflection point is ((3/2),f((3/2))) = ((3/2), -(1/2)).
Looking at the first derivative sign line with its break points at x=1,2 we have:

NumberLinePlot[ {6x^2-18x+12>0, 6x^2-18x+12<0, 6x^2-18x+12==0}, {x,-1,3}, Spacings-> {2,-4,2}, PlotStyle-> {Directive[DotDashed,Black], Directive[Dotted, Black], Directive[PointSize[Large],Black]}, PlotLegends-> Map[TraditionalForm[Text[#]]&, {y'[x]>0, y'[x]<0, y'[x]==0}], PlotLabel-> Style[ TraditionalForm[ Text[y'[x]==6x^2-18x+12]],Black], PlotRange-> {All,{-3,3}}]

So we have that f(x) is increasing on (-∞,1), decreasing on (1,2), and increasing on (2,∞).  The first derivative test tells us that (1,0) is a local maximum and that (2,-1) is a local minimum.

Looking at the second derivative sign line with its only break at x=(3/2) gives

NumberLinePlot[ {12x-18>0, 12x-18<0,12x-18==0}, {x,-1,3}, Spacings-> {2,-4,2}, PlotStyle-> {Directive[DotDashed,Black], Directive[Dotted, Black], Directive[PointSize[Large],Black]}, PlotLegends-> Map[TraditionalForm[Text[#]]&, {y''[x]>0, y''[x]<0, y''[x]==0}], PlotLabel-> Style[ TraditionalForm[ Text[y''[x]==12x-18]],Black], PlotRange-> {All,{-3,3}}]

So f(x) is concave up on (-∞,(3/2)) and concave down on ((3/2),∞).  Therefore ((3/2),-(1/2)) is the only infection point.  There are no asymptotes, and as x⟶±∞ f(x) is dominated by its lead term 2(x^3); so as x⟶∞, f(x)⟶∞ and as x⟶-∞, f(x)⟶-∞.  Putting this all in a graph,

Plot[ 2x^3-9x^2+12x-5, {x,-0,3}, Epilog->{PointSize[Large],Point[ {{1,0}, {2,-1}, {3/2, -1/2}}]}, PlotLabel-> Style[TraditionalForm[Text[ y==2x^3-9x^2+12x-5]],Black]]

Problem 17

f(x)=(x^4)-2(x^2)+2

Solution

For f(x)=(x^4)-2(x^2)+2 we have f'(x)=4(x^3)-4x=4x((x^2)-1)=4x(x-1)(x+1) and f''(x)=12(x^2)-4=4(3(x^2)-1).  From this we can see both derivatives always exist.  From the factored form we can see f'(x)=0 only at x=-1,0,1 and so the only critical points are (-1, f(-1))=(-1,1), (0, f(0))=(0,2), and (1, f(1))=(1,1). Setting f''(x)=0 gives 4(3(x^2)-1)=0 ⟹ 3(x^2)-1=0 ⟹ (x^2)=(1/3) ⟹ x=±(1/(Sqrt[3])).  So the only potential inflection points are (-(1/(Sqrt[3])), f(-(1/(Sqrt[3]))))=(-(1/(Sqrt[3])), (13/9)) and ((1/(Sqrt[3])), f((1/(Sqrt[3]))))=((1/(Sqrt[3])), (13/9)).

Looking at the first derivative sign line with its break points at x=-1,0,1 gives us

NumberLinePlot[ { 4x^3-4x>0,  4x^3-4x<0,  4x^3-4x==0}, {x,-2,2}, Spacings-> {2,-4,2}, PlotStyle-> {Directive[DotDashed,Black], Directive[Dotted, Black], Directive[PointSize[Large],Black]}, PlotLegends-> Map[TraditionalForm[Text[#]]&, {y'[x]>0, y'[x]<0, y'[x]==0}], PlotLabel-> Style[ TraditionalForm[ Text[y'[x]== 4x^3-4x]],Black], PlotRange-> {All,{-3,3}}]

Therefore f(x) is increasing on (-1,0) and (1,∞) and decreasing on (-∞,-1) and (0,1).  In addition by the first derivative test we have that (0,2) must be a local max and that (-1,1) and (1,1) must be local minima.

Looking at the second derivative test with its break ponts at  x=±(1/(Sqrt[3])):

NumberLinePlot[ {12x^2-4>0,12x^2-4<0,12x^2-4==0}, {x,-2,2}, Spacings-> {2,-6,4}, PlotStyle-> {Directive[DotDashed,Black], Directive[Dotted, Black], Directive[PointSize[Large],Black]}, PlotLegends-> Map[TraditionalForm[Text[#]]&, {y''[x]>0, y''[x]<0, y''[x]==0}], PlotLabel-> Style[ TraditionalForm[ Text[y''[x]==12x^2-4]],Black], PlotRange-> {All,{-5,3}}, Ticks-> {{-2,-1,0,1,2,1/Sqrt[3], -1/Sqrt[3]},None}]

From this line we know the graph is concave up on (-∞, -(1/(Sqrt[3]))) and ((1/(Sqrt[3])), ∞) and concave down on (-(1/(Sqrt[3])),(1/(Sqrt[3]))).  So (-(1/(Sqrt[3])), (13/9)) and ((1/(Sqrt[3])), (13/9)) are inflection points.

In addition, f(x) has no asymptotes as is dominated as x⟶±∞ by its lead term (x^4); therefore as x⟶∞, f(x)⟶∞ and as x⟶-∞, f(x)⟶∞.  Putting this into a graph we have

Plot[ x^4-2x^2+2, {x,-2,2}, Epilog-> {PointSize[Large],Point[ {{0,2}, {-1,1}, {1,1}, {1/Sqrt[3], 13/9}, {-1/Sqrt[3] ,13/9}}]}, PlotLabel-> TraditionalForm[ Text[y==x^4-2x^2+2]], PlotRange-> {-1,6}, LabelStyle->Black]

Problem 18

f(x)=(((x^2)+3x)/(x+4))

Solution

For f(x)=(((x^2)+3x)/(x+4))by using the quotient rule we can get f'(x)=(((x^2)+8x+12)/(((x+4))^2)) and f''(x)=(8/(((x+4))^3)).  Neither f(x) nor its derivative exist at x=-4; it has a vertical asymptote there.

We know f'(x) does not exist as x=-4; if f'(x)=0 then (x^2)+8x+12=0 or (x+6)(x+2)=0, or x=-2, -6.  So the critical points are (-2, f(-2))=(-2,-1) and (-6, f(-6))=(-6, -9). f''(x) does not exist at x=-4 and is never equal to 0 (as the numerator is never 0), so there are no potential inflection points.

Looking at the first derivative sign line with breaks at x=-2, -6, and -4:

NumberLinePlot[ {(((x^2)+8x+12)/(((x+4))^2))>0, (((x^2)+8x+12)/(((x+4))^2))<0, (((x^2)+8x+12)/(((x+4))^2))==0,x+4==0}, {x,-7,0}, Spacings-> {2,-4,2,0}, PlotStyle-> {Directive[DotDashed,Black], Directive[Dotted, Black], Directive[PointSize[Large],Black],Directive[PointSize[Large],GrayLevel[.7]]}, PlotLegends-> Map[TraditionalForm[Text[#]]&, {y'[x]>0, y'[x]<0, y'[x]==0,"y'(x) DNE"}], PlotLabel-> Style[ TraditionalForm[ Text[y'[x]==(((x^2)+8x+12)/(((x+4))^2))]],Black], PlotRange-> {All,{-3,3}}]

From this we know f(x) is increasing on (-∞, -6) and (2,∞).  It is decreasing on (-6,-4) and (-4, -2) (we cannot join these because of the break where f'(x) does not exist at x=-4).  By the first derivative test (-6, -9) must be a local max and (-2, -1) a local min.  By looking at the increasing/decreasing behavior on either side of the vertical asymptote, we can see that as x⟶-(SuperMinus[4]), f(x)⟶-∞ and as x⟶-(SuperPlus[4]), f(x)-> ∞

Looking at the second derivative sign line with its only break at x=-4:

NumberLinePlot[ {8/(x+4)^3>0,8/(x+4)^3<0,8/(x+4)^3==0,x+4==0}, {x,-7,0}, Spacings-> {2,-4,2,0}, PlotStyle-> {Directive[DotDashed,Black], Directive[Dotted, Black], Directive[PointSize[Large],Black],Directive[PointSize[Large],GrayLevel[.7]]}, PlotLegends-> Map[TraditionalForm[Text[#]]&, {y''[x]>0, y''[x]<0, y''[x]==0,"y''(x) DNE"}], PlotLabel-> Style[ TraditionalForm[ Text[y''[x]==8/(x+4)^3]],Black], PlotRange-> {All,{-3,3}}]

From this we can see the graph is concave down on (-∞, -4) and concave up on (-4, ∞).  There is no inflection point however as f(x) itself does not exist at x=-4.

In addition to the vertical asymptote, f(x)=(((x^2)+3x)/(x+4)) has an oblique asymptote as the degree of the top is exactly one higher than the degree of the bottom; after performing long division we see that f(x)=x-1+(4/(x+4)), so the graph gets close to y=x-1 as x⟶±∞.  Putting this all in a graph:

Plot[ {(x^2+3x)/(x+4), x-1}, {x,-9, 1}, Exclusions-> x==-4, ExclusionsStyle-> {{Red ,Dashed}}, PlotStyle-> { Black, {Red, Dashed}}, Epilog-> {PointSize[Large],Point[ {{-6, -9}, {-2,-1}}]}, PlotRange-> {-15, 7}, PlotLabel-> TraditionalForm[Text[ y==(x^2+3x)/(x+4)]], PlotRangePadding-> .5, LabelStyle->Black]

Problem 19

f(x)=((sin(x))/(1+cos(x)))

Solution

First note that f(x) is periodic with perion 2π; so we need only focus on the interval [0,2π] and then repeat the graph.  In that range the demoninator is 0 at x=π (where cos(x)=-1) yet the numerator is 1, so it has a vertical asymtote there.
After using the quotient rule together with the identity (sin^2)(x)+(cos^2)(x)=1,  f'(x)=(1/(1+cos(x))) and f''(x)=((sin(x))/(((1+cos(x)))^2)).  Therefore f'(x) fails to exist at x=π (again, where cos(x)=-1) and is never 0. f''(x) also fails to exist at x=π, but is 0 at x=0, π (where sin(x)=0).  Therefore (0, f(0))=(0,0) and (π, f(π))=(π,0) are potential critical points. Normally we would not consider an endpoint like x=0 a potential inflection point.  However the full graph will extend beyond the interval [0,2π]; we are only restricting it to [0,2π] to start, so it makes sense it could be an inflection point.

Looking at a first derivative sign line:

NumberLinePlot[ {1/(1+Cos[x])>0, 1/(1+Cos[x])<0, 1/(1+Cos[x])==0,1+Cos[x]==0}, {x,0,2Pi}, Spacings-> {2,-4,2,0}, PlotStyle-> {Directive[DotDashed,Black], Directive[Dotted, Black], Directive[PointSize[Large],Black],Directive[PointSize[Large],GrayLevel[.7]]}, PlotLegends-> Map[TraditionalForm[Text[#]]&, {y'[x]>0, y'[x]<0, y'[x]==0,"y'(x) DNE"}], PlotLabel-> Style[ TraditionalForm[ Text[y'[x]==1/(1+Cos[x])]],Black], PlotRange-> {All,{-3,3}}, Ticks-> {{0,Pi/2, Pi, 3Pi/2, 2Pi}, None}]

From this we can tell that the function is increasing on (0, π) and  on (π, 2π).  As x=π is a vertical asymptote the first derivative test does not apply there.  As the graph is increasing on both of x=π, we have that asx⟶(SuperMinus[π]), f(x)⟶∞ and as x⟶(SuperPlus[π]), f(x)⟶-∞. As we relax the restriction to [0,2π] we find the the function is increasing wherever defined - there are no maxima or minima.

Looking at the second derivative sign line:

NumberLinePlot[ {Sin[x]/(1+Cos[x])^2>0, Sin[x]/(1+Cos[x])^2<0,Sin[x]/(1+Cos[x])^2==0,1+Cos[x]==0}, {x,0,2Pi}, Spacings-> {2,-4,2,0}, PlotStyle-> {Directive[DotDashed,Black], Directive[Dotted, Black], Directive[PointSize[Large],Black],Directive[PointSize[Large],GrayLevel[.7]]}, PlotLegends-> Map[TraditionalForm[Text[#]]&, {y'[x]>0, y'[x]<0, y'[x]==0,"y''(x) DNE"}], PlotLabel-> Style[ TraditionalForm[ Text[y''[x]==Sin[x]/(1+Cos[x])^2]],Black], PlotRange-> {All,{-3,3}}, Ticks-> {{0,Pi/2, Pi, 3Pi/2, 2Pi}, None}]

From this we see that the graph is concave up on (0, 2π) and concave down on (π, 2π).  Viewed as a function on [0,2π] this would mean there are no inflection points - but since the real graph repeats, we will switch concavity at x= ...-4π, -2π,0, 2π, 4π,... and therefore the values (2n π, f(2 n π))=(2 n π,0) will be critical points. Putting all this together into a graph on [0,2π], we now have:

Plot[ Sin[x]/(1+Cos[x]), {x,0,2Pi}, PlotRange-> {-8,8}, Exclusions-> x==Pi, ExclusionsStyle-> {{Red, Dashed}}, PlotLabel-> TraditionalForm[Text[ y==Sin[x]/(1+Cos[x])^2]], PlotRangePadding-> .5, Ticks-> {Table[ Pi/2 k, {k,-4,4}], Automatic} , Epilog-> Point[{{0,0}, {2Pi,0}}], LabelStyle-> Black]

Extending this by repetition, we have:

Plot[ Sin[x]/(1+Cos[x]), {x,-4Pi ,4Pi}, PlotRange-> {-8,8}, Exclusions->{x==-3Pi, x==-Pi,x== Pi, x==3Pi}, ExclusionsStyle-> {{Red, Dashed}}, PlotLabel-> TraditionalForm[Text[ y==Sin[x]/(1+Cos[x])]], PlotRangePadding-> .5, Ticks-> {Table[ Pi/2 k, {k,-12,12}], Automatic}, Epilog->{PointSize[Large], Point[ Table[ {2k Pi,0}, {k,-2,2} ]]}, LabelStyle->Black]

The truly astute observer may have noticed we could have avoided all this by remembering from trigonometry that ((sin(x))/(1+cos(x)))=tan((x/2)).

Problem 20

f(x)=(Sqrt[(x^2)-4x+13])

Solution

As a square root is involved we need to check where f(x) is well defined.  Looking at the quantity under the root, (x^2)-4x+13=(x^2)-4x+4+9=(((x-2))^2)+9; as (((x-2))^2)≥ 0 for all x, (((x-2))^2)+9≥ 9≥ 0 for all x, and the square root is well-defined for all x. Using the generalized power rule to find f'(x), we have f'(x)=((x-2)/(Sqrt[(x^2)-4x+13])) and after using a combination of the quotient rule and generalized power rule along with some simplification we have f''(x)=(9/((((x^2)-4x+13))^(3/2))).
Looking at the first derivative, the bottom is never 0 (in fact the argument where we checked that the root is well-defined shows us the root is at least (Sqrt[9])≥ 3) and f'(x) always exists.  So the only critical points are from where f'(x)=0, which becomes x-2=0 or x=2.  So the only critical point is (2, f(2))=(2, 3). Checking the sign line for f'(x) gives us

NumberLinePlot[ {(x-2)/Sqrt[x^2-4x+13]>0,(x-2)/Sqrt[x^2-4x+13]<0,(x-2)/Sqrt[x^2-4x+13]==0}, {x,-1,5}, Spacings-> {2,-4,2}, PlotStyle-> {Directive[DotDashed,Black], Directive[Dotted, Black], Directive[PointSize[Large],Black]}, PlotLegends-> Map[TraditionalForm[Text[#]]&, {y'[x]>0, y'[x]<0, y'[x]==0}], PlotLabel-> Style[ TraditionalForm[ Text[y'[x]==(x-2)/Sqrt[x^2-4x+13]]],Black], PlotRange-> {All,{-3,3}}]

From this we can see that f(x) is increasing on (2,∞) and decreasing on (-∞, 2); the first derivative test would tell us that the critical point (2,3) is an absolute minimum.

Looking at the second derivative f''(x), the bottom is never 0 (for the same reason as for f'(x)) and the top is never 0.  So the second derivative never has a chance to change sign; it must be always positive or always negative (which makes a sign line overkill). f''(2)=(9/27)=(1/3)>0, so the second derivative is always positive and the graph must be concave up on (-∞, ∞).  As the concavity never changes there are no inflection points.

At first glance it is unclear if there will be asymptotes.  However we can write f(x)= (Sqrt[(x^2)-4x+13]) as (Sqrt[(((x-2))^2)+9]); as  x⟶±∞ we know (((x-2))^2)⟶∞.  Therefore the “+9” in the formula may as well not be there when x is large positive or negative, and we have that as x⟶±∞, f(x)=(Sqrt[(x^2)-4x+13])≈(Sqrt[(((x-2))^2)])=|x-2|.  Therefore as x⟶∞ f(x) has the oblique asymptote y=|x-2|=x-2 (the absolute values not being needed as x-2 will be positive as x⟶∞) and as x⟶-∞ it will have the oblique asymptote y=|x-2|=-(x-2)=-x+2 (the absolute values taking effect as x-2 will be negative as x⟶-∞).  So we have a rare example of a graph with 2 different oblique asymptotes.  Here is the graph of f(x) along with the graph of |x-2|:

Plot[ {Sqrt[ x^2-4x+13], Abs[x-2]}, {x,-8,11}, PlotLabel-> TraditionalForm[Sqrt[x^2-4x+13]], PlotStyle-> {Blue, {Red, Dashed}}, Epilog-> {PointSize[Large],Point[ {2,3}]}, LabelStyle->Black]

Problem 21

f(x)=sin(x)cos(x) on [0,2π]

Solution

As we are only graphing this on [0,2π] we will need to identify the endpoints as potential maxima or minima.  After using the product rule f'(x)=(cos^2)(x)-(sin^2)(x) and after the generalized power rule f''(x)=-4sin(x)cos(x).

Looking at the first derivative, f'(x) always exists on (0,2π).  So when finding the critical points we need only worry about where f'(x)=0.  This is the same as (cos^2)(x)-(sin^2)(x)=0, or (sin^2)(x)=(cos^2)(x).  If cos(x)≠ 0, on dividing by (cos^2)(x) this is (tan^2)(x)=1, or tan(x)=±(Sqrt[1])=±1.  On the interval [0,2π] the tangent is 1 at (π/4) and ((5π)/4) and the tangent is -1 at ((3π)/4) and ((7π)/4).  We still need to check what happens if cos(x)=0; however, this only happens at (π/2) and ((3π)/2) and at both those numbers f'(x)=-1≠0.  So there are 4 critical points; ((π/4), f((π/4)))=((π/4),(1/2)), (((3π)/4),f(((3π)/4)))=(((3π)/4), -(1/2)), (((5π)/4),f(((5π)/4)))=(((5π)/4), (1/2)), and (((7π)/4),f(((7π)/4)))=(((7π)/4), -(1/2)).  We should also keep in mind the endpoints (0,f(0))=(0,0) and (2π, f(2π))=(2π,0).  Putting the 4 critical points on a sign line that only goes from 0 to 2π we have

NumberLinePlot[ {Cos[x]^2-Sin[x]^2>0,Cos[x]^2-Sin[x]^2<0, Cos[x]^2-Sin[x]^2==0,x==0||x==2Pi}, {x,0,2Pi}, Spacings-> {2,-4,2,0}, PlotStyle-> {Directive[DotDashed,Black], Directive[Dotted, Black], Directive[PointSize[Large],Black], Directive[PointSize[Large],GrayLevel[.7]]}, PlotLegends-> Map[TraditionalForm[Text[#]]&, {y'[x]>0, y'[x]<0, y'[x]==0, "endpoint"}], PlotLabel-> Style[ TraditionalForm[ Text[y'[x]==Cos[x]^2-Sin[x]^2]],Black], PlotRange-> {All,{-3,3}}, Ticks-> {{0,Pi/3, 3Pi/4, Pi, 5Pi/4, 7Pi/4, 2Pi},None}]

Looking at this line we can see f(x) is increasing on the intervals (0,(π/4)), (((3π)/4),((5π)/4)), and (((7π)/4), 2π) and decreasing on ((π/4),((3π)/4)) and (((5π)/4),((7π)/4)).  The first derivative test then tells is that ((π/4),(1/2)) is a local max, (((3π)/4), -(1/2)) a local min, (((5π)/4), (1/2)) a local max, and (((7π)/4), -(1/2)) a local min.  In addition f(x) is increasing away from x=0 - so (0,0) is a local min.  Likewise f(x) is increasing to x=2π - so (2π,0) is a local max.

Examining the second derivative  f''(x)=-4sin(x)cos(x) always exists, and will be 0 only when either sin(x)=0 or cos(x)=0.  Going from 0 to 2π sine is 0 only at π and the endpoints; cosine is 0 only at (π/2) and ((3π)/2).  So the breakpoints for the second derivative sign line will be x=(π/2),π,((3π)/2) and as points these would be ((π/2),f((π/2)))=((π/2), 0), (π,f(π))=(π,0), and (((3π)/2),f(((3π)/2)))=((π/2),0).  Our sign line for the second derivative now looks like:

NumberLinePlot[ {-4Sin[x]Cos[x]>0,-4Sin[x]Cos[x]<0, -4Sin[x]Cos[x]==0,x==0||x==2Pi}, {x,0,2Pi}, Spacings-> {2,-4,2,0}, PlotStyle-> {Directive[DotDashed,Black], Directive[Dotted, Black], Directive[PointSize[Large],Black], Directive[PointSize[Large],GrayLevel[.7]]}, PlotLegends-> Map[TraditionalForm[Text[#]]&, {y''[x]>0, y''[x]<0, y''[x]==0, "endpoint"}], PlotLabel-> Style[ TraditionalForm[ Text[y'[x]==-4Sin[x]Cos[x]]],Black], PlotRange-> {All,{-3,3}}, Ticks-> {{0,Pi/3, 3Pi/4, Pi, 5Pi/4, 7Pi/4, 2Pi},None}]

So our graph must be concave up on the intervals ((π/2),π) and (((3π)/2),2π) and concave down on (0,(π/2)) and (π,((3π)/2)). This means all 3 of ((π/2), 0), (π,0), and (((3π)/2), 0) are inflection points.  Even though the second derivative is 0 at the endpoints we can’t count them as inflection points as you can’t change concavity at an endpoint (as it turns out if we extended the graph outside the x-range [0,2π] they would be inflection points).

As our graph only goes from 0 to 2π  and is continuous there are no asymptotes.

Putting all this information into the graph gives us:

Plot[ Sin[x] Cos[x], {x,0,2Pi}, Epilog->{PointSize[Large], Point[ Table[ {k, Sin[k] Cos[k]}, {k,0,2Pi, Pi/4}]]}, PlotLabel-> TraditionalForm[ y==Sin[x]Cos[x]], LabelStyle->Black]

All of this could be a lot simpler if you realize at the outset that f(x)=(1/2)sin(2x) (using the double-angle formula for sine).

Problem 22

f(x)=(((x^3)(5-x))^2)  (for inflection points, find the x-values only)

Solution

This function is continuous everywhere, and as you can write the function as f(x)=(x^5)-10(x^4)+25(x^3) you can see it is dominated by (x^5) as x⟶±∞.  So as x⟶∞, f(x)⟶∞ and as x⟶-∞, f(x)⟶-∞. f'(x)=5(x^4)-40(x^3)+75(x^2)=5(x^2)((x^2)-8x+15)=5(x^2)(x-3)(x-5) and f''(x)=20(x^3)-120(x^2)+150x=10x(2(x^2)-12x+15).  Both of these are continuous, so the only critical points and potential inflection points will come from where these are zero.

Focusing on the first derivative, f'(x)=5(x^2)(x-3)(x-5) will be 0 only at x=0, 3,and 5.  So the only critical points will be (0,f(0))=(0,0),  (3,f(3))=(3,108), and (5,f(5))=(5,0).  Putting these on the first derivative sign line gives

NumberLinePlot[ {5(x-5)(x-3)x^2>0, 5(x-5)(x-3)x^2<0, 5(x-5)(x-3)x^2==0}, {x,-1,6}, Spacings-> {2,-4,2}, PlotStyle-> {Directive[DotDashed,Black], Directive[Dotted, Black], Directive[PointSize[Large],Black]}, PlotLegends-> Map[TraditionalForm[Text[#]]&, {y'[x]>0, y'[x]<0, y'[x]==0}], PlotLabel-> Style[ TraditionalForm[ Text[y'[x]==5(x-5)(x-3)x^2]],Black], PlotRange-> {All,{-3,3}}]

So we can see that f(x) is increasing on (-∞,0), (0,3), and (5,∞) and decreasing on (3,5).  By the first derivative test the point (3,108) is a local max, (5,0) a local min, and (0,0) is neither (simply a horizontal tangent).

Looking at the second derivative f''(x)=10x(2(x^2)-12x+15), this will be zero either when x=0 or when 2(x^2)-12x+15=0.  Using the quadratic formula this gives us the set of numbers x=0, ((6-(Sqrt[6]))/2), ((6+(Sqrt[6]))/2).  Plugging these numbers back into f(x) gets ugly (which explains why the directions say to only worry about the x-values!), but the potential inflection points are (0,f(0))=(0,0), (((6-(Sqrt[6]))/2),f(((6-(Sqrt[6]))/2)))=(((6-(Sqrt[6]))/2),(3/8)(138+7(Sqrt[6])))≈(1.78,58.18) and (((6+(Sqrt[6]))/2), f(((6+(Sqrt[6]))/2)))=(((6+(Sqrt[6]))/2),(3/8)(138-7(Sqrt[6])))≈(4.22, 45.32) .  Putting the x-values on a sign line gives us

NumberLinePlot[ {10x(2x^2-12x+15)>0,10x(2x^2-12x+15)<0,10x(2x^2-12x+15)==0}, {x,-1,6}, Spacings-> {2,-8,6}, PlotStyle-> {Directive[DotDashed,Black], Directive[Dotted, Black], Directive[PointSize[Large],Black]}, PlotLegends-> Map[TraditionalForm[Text[#]]&, {y''[x]>0, y''[x]<0, y''[x]==0}], PlotLabel-> Style[ TraditionalForm[ Text[y''[x]==10x(2x^2-12x+15)]],Black],  Ticks-> {{-1,0,1,3,5,6,(6-Sqrt[6])/2, (6+Sqrt[6])/2},None}, PlotRange-> {All,{-7,4}}]

From this we can see the graph is concave up on (0, ((6-(Sqrt[6]))/2)) and (((6+(Sqrt[6]))/2),∞) and concave down on (-∞,0) and (((6-(Sqrt[6]))/2),((6+(Sqrt[6]))/2)).  All 3 of our potential inflection points turn out to be inflection points, including the point (0,0) which also appeared in our examination of the first derivative.

Putting all this in terms of a graph we have:

Plot[ x^3(5-x)^2, {x,-2,7}, PlotLabel-> TraditionalForm[ Text[y== x^3 (5-x)^2]], Epilog-> {PointSize[Large],Point[{{0,0},{3,108},{5,0},{3-(Sqrt[(3/2)]),(3/8) (138+7 (Sqrt[6]))},{3+(Sqrt[(3/2)]),-(3/8) (-138+7 (Sqrt[6]))}}]}, PlotRange-> {-100, 200}, LabelStyle->Black]

Problem 23

f(x)=(x/((x^2)-4))

Solution

As the denominator of the function is 0 at x=±2 and the top is not this has vertical asymptotes at x=2 and x=-2.  Writing the function as f(x)=(x/((x-2)(x+2))) makes it easier to do the limit on each side of those asymptotes; (lim_(x⟶(SuperPlus[2])))f(x)=∞, (lim_(x⟶(SuperMinus[2])))f(x)=-∞, (lim_(x⟶-(SuperPlus[2])))f(x)=∞, and (lim_(x⟶-(SuperMinus[2])))f(x)=-∞.  In addition as the top has degree 1 and the bottom has degree 2, (lim_(x⟶∞))f(x)=0 and (lim_(x⟶-∞))f(x)=0, so the graph has the horizontal asymptote y=0 and goes to it in both directions.

After using the quotient rule and simplifying f'(x)=-(((x^2)+4)/((((x^2)-4))^2)) and f''(x)=((2x((x^2)+12))/((((x^2)-4))^3)).
f'(x) fails to exist at the vertical asymptotes x=±2 and nowhere else.  Its numerator (x^2)+4 is never 0 (as this would imply (x^2)=-4), so the only break points for the first derivative sign line are the vertical asymptotes:

NumberLinePlot[ {-(x^2+4)/(x^2-4)^2>0, -(x^2+4)/(x^2-4)^2<0,-(x^2+4)/(x^2-4)^2==0,(x^2-4)^2==0}, {x,-3,3}, Spacings-> {2,-4,2,0}, PlotStyle-> {Directive[DotDashed,Black], Directive[Dotted, Black], Directive[PointSize[Large],Black],Directive[PointSize[Large],GrayLevel[.7]]}, PlotLegends-> Map[TraditionalForm[Text[#]]&, {y'[x]>0, y'[x]<0, y'[x]==0,"y'(x) DNE"}], PlotLabel-> Style[ TraditionalForm[ Text[y'[x]==-(x^2+4)/(x^2-4)^2]],Black], PlotRange-> {All,{-5,3}}]

The sign line tells us that the function is decreasing on (-∞, -2), (-2,2), and (2,∞) and never increasing; therefore it has no maxima or minima.  You should check that the always decreasing nature of the function reinforces our answers for the limits on either side of the vertical asymptotes.

The second derivative f''(x)=((2x((x^2)+12))/((((x^2)-4))^3)) fails to exist at the vertical asymptotes as well; the only time its numerator is 0 is as x=0.  So the only potential inflection point is (0,f(0))=(0,0).  Creating the sign line for the second derivative we have

NumberLinePlot[ {2x(x^2+12)/(x^2-4)^3>0, 2x(x^2+12)/(x^2-4)^3<0,2x(x^2+12)/(x^2-4)^3==0,(x^2-4)^2==0}, {x,-3,3}, Spacings-> {2,-4,2,0}, PlotStyle-> {Directive[DotDashed,Black], Directive[Dotted, Black], Directive[PointSize[Large],Black],Directive[PointSize[Large],GrayLevel[.7]]}, PlotLegends-> Map[TraditionalForm[Text[#]]&, {y''[x]>0, y''[x]<0, y''[x]==0,"y''(x) DNE"}], PlotLabel-> Style[ TraditionalForm[ Text[y''[x]==2x(x^2+12)/(x^2-4)^3]],Black], PlotRange-> {All,{-5,3}}]

So the graph is concave up on (-2,0) and (2,∞) and concave down on the intervals (-∞,-2) and (0,2), with the point (0,0) being an inflection point.

Putting this all together into a graph gives us

Plot[ x/(x^2-4), {x,-5,5}, PlotRange-> {-6,6}, Exclusions-> x^2==4, ExclusionsStyle-> {{Red, Dashed}}, Epilog->{PointSize[Large], Point[ {0,0}]}, PlotLabel-> TraditionalForm[Text[y==x/(x^2-4)]], LabelStyle-> Black]

Problem 24

f(x)=(x^(1/3))-((x-4)^(1/3)) (warning -  tricky algebra)

Solution

f(x) is continuous on (-∞,∞), so it has no vertical asymptotes.  We can get the behavior as x⟶±∞ through the use of the cubic conjugate trick (a^3)-(b^3)=(a-b)((a^2)+a b+(b^2)):
f(x)=(((x^(1/3))-((x-4)^(1/3)))/1)=(((x^(1/3))-((x-4)^(1/3)))/1)×((((((x^(1/3))))^2)+(x^(1/3))((x-4)^(1/3))+(((((x-4)^(1/3))))^2))/(((((x^(1/3))))^2)+(x^(1/3))((x-4)^(1/3))+(((((x-4)^(1/3))))^2)))=((((((x^(1/3))))^3)-(((((x-4)^(1/3))))^3))/(((((x^(1/3))))^2)+(x^(1/3))((x-4)^(1/3))+(((((x-4)^(1/3))))^2)))=((x-(x-4))/(((((x^(1/3))))^2)+(x^(1/3))((x-4)^(1/3))+(((((x-4)^(1/3))))^2)))=(4/(((((x^(1/3))))^2)+(x^(1/3))((x-4)^(1/3))+(((((x-4)^(1/3))))^2)))
As x⟶±∞, the 3 terms ((((x^(1/3))))^2),(x^(1/3))((x-4)^(1/3)), and (((((x-4)^(1/3))))^2) all go to ∞.  So as x⟶∞, the fraction goes to (4/(∞+∞+∞))=0.  So f(x) has the horiztonal asymptote y=0 and goes to it in both directions.

Looking at the derivatives, f'(x)=(1/3)(x^(-2/3))-(1/3)(((x-4))^(-2/3))=(((((x-4))^(2/3))-(x^(2/3)))/(3(((x^(2/3))(x-4))^(2/3)))) and f''(x)=-(2/9)(x^(-5/3))+(2/9)(((x-4))^(-5/3))=(2/9)((((((x-4))^(5/3))-(x^(5/3)))/(((x^(5/3))(x-4))^(5/3)))).  These both fail to exist where at x=0, 4 where the denomiators are 0 and exist everywhere else.  As neither x=0,4 cause the numerator to be zero I expect these values correspond to vertical tangents in the graph (as f(x) makes sense at both values but the tangent slope is “infinite” at those two spots).  These points are (0,f(0))=(0, (4^(1/3)))≈(0,1.58) and (4,f(4))=(4,(4^(1/3)))≈(4,1.58).

Working on f'(x) first, it can only be 0 where the numerator (((x-4))^(2/3))-(x^(2/3))=0, or (x^(2/3))=(((x-4))^(2/3)).  Raising both sides to the power 3 would give (x^2)=(((x-4))^2), or (x^2)=(x^2)-8x+16.  This simplifies to 8x=16, or x=2.  So the only criticial point is (2,f(2))=(2,(2^(1/3))-((-2)^(1/3)))=(2,2(2^(1/3)))≈(2,2.52).  Using this point and x=0,4 as the breaks in the sign lines we get

NumberLinePlot[ {(CubeRoot[x-4]^2-CubeRoot[x]^2)/(3 CubeRoot[x]^2CubeRoot[x-4]^2)>0, (CubeRoot[x-4]^2-CubeRoot[x]^2)/(3 CubeRoot[x]^2CubeRoot[x-4]^2)<0,(CubeRoot[x-4]^2-CubeRoot[x]^2)/(3 CubeRoot[x]^2CubeRoot[x-4]^2)==0,(3 CubeRoot[x]^2CubeRoot[x-4]^2)==0}, {x,-3,6}, Spacings-> {2,-4,2,0}, PlotStyle-> {Directive[DotDashed,Black], Directive[Dotted, Black], Directive[PointSize[Large],Black],Directive[PointSize[Large],GrayLevel[.7]]}, PlotLegends-> Map[TraditionalForm[Text[#]]&, {y'[x]>0, y'[x]<0, y'[x]==0,"y'(x) DNE"}], PlotLabel-> Style[ TraditionalForm[ Text[y'[x]==(CubeRoot[x-4]^2-CubeRoot[x]^2)/(3 CubeRoot[x]^2CubeRoot[x-4]^2)]],Black, "FontSize"-> 16], PlotRange-> {All,{-5,3}}]

So we have that f(x) is increasing on the intervals (-∞,0) and (0,2) and decreasing on the intervals (2,4) and (4,∞).  The first derivative test tells us that the point (2,2(2^(1/3)))≈(2,2.52) is a local maximum and that the points (0, (4^(1/3)))≈(0,1.58) and (4,(4^(1/3)))≈(4,1.58) are neither maxima nor minima (which agrees with the suspicion that these are vertical tangents).

Examining f''(x)=-(2/9)(x^(-5/3))+(2/9)(((x-4))^(-5/3))=(2/9)((((((x-4))^(5/3))-(x^(5/3)))/(((x^(5/3))(x-4))^(5/3)))), again we easily see that this is undefined at x=0,4 and only those values.  Setting the numerator equal to 0 we have (((x-4))^(5/3))-(x^(5/3))=0 or (((x-4))^(5/3))=(x^(5/3)).  Raising both sides to the power 3 gives (((x-4))^5)=(x^5), and taking the fifth root (which does not involve ± !) yields x-4=x, or 0=4.  This has no solution, so f''(x) is never 0.  So the only break points in the second derivative sign line will be x=0,4 and the only potential inflection points are (0, (4^(1/3)))≈(0,1.58) and (4,(4^(1/3)))≈(4,1.58).

NumberLinePlot[ {2(CubeRoot[x-4]^5-CubeRoot[x]^5)/(9 CubeRoot[x]^5 CubeRoot[x-4]^5)>0, 2(CubeRoot[x-4]^5-CubeRoot[x]^5)/(9 CubeRoot[x]^5 CubeRoot[x-4]^5)<0,2(CubeRoot[x-4]^5-CubeRoot[x]^5)/(9 CubeRoot[x]^5 CubeRoot[x-4]^5)==0,(9 CubeRoot[x]^5 CubeRoot[x-4]^5)==0}, {x,-3,6}, Spacings-> {2,-4,2,0}, PlotStyle-> {Directive[DotDashed,Black], Directive[Dotted, Black], Directive[PointSize[Large],Black],Directive[PointSize[Large],GrayLevel[.7]]}, PlotLegends-> Map[TraditionalForm[Text[#]]&, {y''[x]>0, y''[x]<0, y''[x]==0,"y''(x) DNE"}], PlotLabel-> Style[ TraditionalForm[ Text[y''[x]==2(CubeRoot[x-4]^5-CubeRoot[x]^5)/(9 CubeRoot[x]^5 CubeRoot[x-4]^5)]],Black, "FontSize"-> 16], PlotRange-> {All,{-5,3}}]

From this we can see the function is concave down on (-∞,0) and (4,∞) and concave up on the inverval (0,4).  In addition (0, (4^(1/3)))≈(0,1.58) and (4,(4^(1/3)))≈(4,1.58) are inflection points.  

Putting all this together in the graph gives us

Plot[ CubeRoot[x]-CubeRoot[x-4], {x,-8,10}, Epilog-> {PointSize[Large],Point[{{0, CubeRoot[4]}, {4, CubeRoot[4]}, {2, 2CubeRoot[2]}}]}, PlotLabel-> Style[TraditionalForm[Text[y==CubeRoot[x]-CubeRoot[x-4]]], "FontSize"-> 16], PlotRange-> {0,2.8}, LabelStyle->Black]

Problem 25

f(x)=(Sqrt[9-(x^2)])-(3/4)x

Solution

First note that the square root is only defined when 9-(x^2)≥0.  This is equivalent to (x^2)≤ 9, or |x|≤ 3, or -3≤ x≤ 3.  So this function is defined on a closed interval and has the endpoints (-3,f(-3))=(-3,(9/4)) and (3,f(3))=(3,-(9/4)).  Using the generalized power rule f'(x)=((-x)/(Sqrt[9-(x^2)]))-(3/4)=-((4x+3(Sqrt[9-(x^2)]))/(4(Sqrt[9-(x^2)]))) and after another application of the quotient rule on the first form of f'(x) we get f''(x)=-(9/(((9-(x^2)))^(3/2))).

Using the form f'(x)=-((4x+3(Sqrt[9-(x^2)]))/(4(Sqrt[9-(x^2)]))) we see this is undefined only when (Sqrt[9-(x^2)]) is 0 or undefined.  Since we are only looking at the interval [-3,3] the root is always defined and only 0 at the endpoints x=-3,3.  So the only places f'(x) is undefined is at the endpoints. f'(x) will be zero only where the numerator is zero.  Solving 4x+3(Sqrt[9-(x^2)])=0, we have 4x=-3(Sqrt[9-(x^2)]); squaring both sides gives 16(x^2)=9(9-(x^2))=81-9(x^2), or 25(x^2)=81.  This leads to (x^2)=(81/25) or x=±(9/5).  However since the solution involved squaring both sides we need to make sure these are true solutions to 4x+3(Sqrt[9-(x^2)])=0 and not extraneous ones introduced in the squaring process.  A quick check shows that x=(9/5) is not a true solution but x=-(9/5) is.  So other than the endpoints the only critical point is (-(9/5), f(-(9/5)))=(-(9/5), (15/4)).  Forming the sign line for the first derivative, we have

NumberLinePlot[ {-(4x+3Sqrt[9-x^2])/(4 Sqrt[9-x^2])>0, -(4x+3Sqrt[9-x^2])/(4 Sqrt[9-x^2])<0,-(4x+3Sqrt[9-x^2])/(4 Sqrt[9-x^2])==0,(4 Sqrt[9-x^2])==0}, {x,-3,3}, Spacings-> {2,-4,2,0}, PlotStyle-> {Directive[DotDashed,Black], Directive[Dotted, Black], Directive[PointSize[Large],Black],Directive[PointSize[Large],GrayLevel[.7]]}, PlotLegends-> Map[TraditionalForm[Text[#]]&, {y'[x]>0, y'[x]<0, y'[x]==0,"y'(x) DNE"}], PlotLabel-> Style[ TraditionalForm[ Text[y'[x]==-(4x+3Sqrt[9-x^2])/(4 Sqrt[9-x^2])]],Black], PlotRange-> {All,{-5,3}}]

From this we can see the function is increasing on the interval (-3,-(9/5)) and decreasing on the interval (-(9/5),3).  This tells us that the critical point (-(9/5), (15/4)) is an absolute max, the endpoint (-3,(9/4)) is a local min, and (3,-(9/4)) is a local min.  Notice that the derivative is infinite at the endpoints - this probably means the tangent line at the endpoints will be vertical.

As f''(x)=-(9/(((9-(x^2)))^(3/2))) the only place the second derivative will be undefined will be at the endpoints x=±3; as the numerator is 9 (or -9 depending on if you include the sign out front) the second derivative will never be 0.  So there are no places the second derivative can change sign and therefore no inflection points.

NumberLinePlot[ {-9/Sqrt[9-x^2]^3>0, -9/Sqrt[9-x^2]^3<0,-9/Sqrt[9-x^2]^3==0,Sqrt[9-x^2]^3==0}, {x,-3,3}, Spacings-> {2,-4,2,0}, PlotStyle-> {Directive[DotDashed,Black], Directive[Dotted, Black], Directive[PointSize[Large],Black],Directive[PointSize[Large],GrayLevel[.7]]}, PlotLegends-> Map[TraditionalForm[Text[#]]&, {y''[x]>0, y''[x]<0, y''[x]==0,"y''(x) DNE"}], PlotLabel-> Style[ TraditionalForm[ Text[y''[x]==-9/Sqrt[9-x^2]^3]],Black], PlotRange-> {All,{-5,3}}]

Apparently this graph is concave down on the interval (-3,3) and never concave up.  

Putting all of this into a graph we have

Plot[ Sqrt[9-x^2]-3x/4, {x,-3,3}, Epilog-> {PointSize[Large],Point[{{-3,9/4}, {3,-9/4}, {-9/5, 15/4}}]}, PlotLabel-> TraditionalForm[Text[ y==Sqrt[9-x^2]-3x/4]], LabelStyle->Black]

Problem 26

Below are 3 graphs - one represents y=f(x), another y=f'(x), and the third y=f''(x).  Determine which is which.

Column[ { Plot[ Evaluate[ D[(1+x/2)Sin[Pi x/2],x,x]], {x,0,6}, PlotLabel-> Style[ "Graph A", FontFamily->"TimesNewRoman"], Ticks-> { {1,2,3,4,5,6},None}, ImageSize-> 250],Plot[(1+x/2)Sin[Pi x/2], {x,0,6}, PlotLabel-> Style[ "Graph B", FontFamily->"TimesNewRoman"], Ticks-> { {1,2,3,4,5,6},None}, ImageSize-> 250],Plot[ Evaluate[ D[(1+x/2)Sin[Pi x/2],x]], {x,0,6}, PlotLabel-> Style[ "Graph C", FontFamily->"TimesNewRoman"], Ticks-> {{1,2,3,4,5,6},None}, ImageSize-> 250]}]

The best way to approach this is to use some strategy.  Rather than try to identify all 3 graphs at once, consider each graph in turn and see what its derivative graph would have to look like.  We know that horizontal tangent lines correspond to a derivative value of 0, increasing portions of a graph correspond to a positive derivative, and decreasing portions of a graph correspond to a negative derivative.

For Graph A:  The derivative of Graph A would be 0 at approximately x=1.4, 3.2, and 5.2.  Graph A is also decreasing from 0 to about 1.4, so the derivative would have to be negative there.  Neither Graph B nor Graph C have these properties, so the derivative of Graph A is not present.  Given that the graphs represent f(x), f'(x), and f''(x) the only one of those 3 whose derivative is not in the list is f''(x).  So Graph A must be f''(x).

For Graph B:  Graph B’s derivative would be 0 at roughly x=1.2, 3.1, and 5.1.  By examining where the graph is increasing or decreasing, its derivative would be positive from 0 to 1.2, negative from 1.2 to 3.1, positive from 3.1 to 5.1, and negative therefter.  Graph A does not match this information, but Graph C does.  So the derivative of Graph B must be Graph C.  Given that the 3 graphs are f(x), f'(x), and f''(x) and Graph A is f''(x), this means that Graph B is f(x), Graph C is f'(x), and Graph A is f''(x).

As a check, consider the derivative of Graph C.  It would have to be 0 at roughly x= .4, 2.1, and 4.1.  The derivative would be positive from 0 to .4, negative from .4 to 2.1, positive from 2.1 to 4.1, and negative thereafter. All of this matches Graph A.

Problem 27

Find the maximum area of a rectangle that can be inscribed in (x^2)+((y^2)/9)=1.

Solution

This ellipse itself looks like this:

ContourPlot[ x^2+y^2/9==1, {x,-3,3},{y,-3,3}, Frame-> False, Axes->True, PlotLabel-> TraditionalForm[ x^2+y^2/9==1]]

The upper and lower curves are given by solving for y which yields y=±3(Sqrt[1-(x^2)]).  To inscribe a rectnagle in this ellipse, let (x,0) be a point on the x-axis where 0≤x≤1.  By going straight up and down you get 2 points on the inscribed rectangle: (x,-3(Sqrt[1-(x^2)])) and (x,3(Sqrt[1-(x^2)])).  Going straight left from those points gives you the other 2 corners:  (-x,-3(Sqrt[1-(x^2)])) and (-x,3(Sqrt[1-(x^2)])):

Show[ ContourPlot[ x^2+y^2/9==1, {x,-1.1,1.1},{y,-3,3}, Frame-> False, Axes->True, PlotLabel-> TraditionalForm[ x^2+y^2/9==1]], Graphics[ {PointSize[Medium], Point[ { {1/2,0}, {1/2, 3Sqrt[3/4]}, {-1/2, 3 Sqrt[3/4]}, {-1/2, -3Sqrt[3/4]}, {1/2, -3 Sqrt[3/4]}}], Line[{ {1/2, 3Sqrt[3/4]}, {-1/2, 3 Sqrt[3/4]}, {-1/2, -3Sqrt[3/4]}, {1/2, -3 Sqrt[3/4]},{1/2, 3Sqrt[3/4]}}], Text["(x,0)", {.7,.2}],Text["(x,3(Sqrt[1 - (x^2)]))", {1,3}],Text["(-x,3(Sqrt[1 - (x^2)]))", {-1,3}],Text["(-x,-3(Sqrt[1 - (x^2)]))", {-1,-3}],Text["(x,-3(Sqrt[1 - (x^2)]))", {1,-3}]}], PlotRangePadding-> .5]

The height of this rectange is the top y-coordinate minus the bottom one; so h=3(Sqrt[1-(x^2)])-(-3(Sqrt[1-(x^2)]))=6(Sqrt[1-(x^2)]).  The width is the right x-coordinate minus the left one; that is w=x-(-x)=2x.  So we are trying to find the maximum area A=h w=12x(Sqrt[1-(x^2)]) where x goes from 0 to 1 (0 and 1 will have an area of 0 which is obviously not the maximum but by including them we get a closed interval which has an easier max-min technique). Using the product rule and simplifying, we get A'=12(Sqrt[1-(x^2)])-((12(x^2))/(Sqrt[1-(x^2)]))=-((12(2(x^2)-1))/(Sqrt[1-(x^2)])).  This exists as long as 1-(x^2)>0, or 0≤x<1 (and we will need to check x=1 anyway), so we need only find there this is zero.  This will happen only when 2(x^2)-1=0 or x=(1/(Sqrt[2])) (the negative value being outside the range 0≤x≤1).  Making a table of values,

TableForm[ { x, 12x Sqrt[1-x^2]}/. {{x->0}, {x-> 1/Sqrt[2]}, {x-> 1}}, TableHeadings-> {None, {TraditionalForm[x], TraditionalForm[ area==12x Sqrt[1-x^2]]}},TableAlignments-> Center]

So the maximum area is 6, and this happens when x=(1/(Sqrt[2])).  This correspeonds to a width of w=2(1/(Sqrt[2]))=(Sqrt[2]) and a height of h=6(Sqrt[1-((((1/(Sqrt[2]))))^2)])=3(Sqrt[2]).

Problem 28

A conical tent with no floor is to have a volume of 500 cubic feet.  What dimensions (i.e. height and radius) minimize the amount of material needed to make the tent?

Solution

The area of a cone with radius r and height h is A=π r(Sqrt[(r^2)+(h^2)]).  This is too many variables for us to handle at once.  But the volume of a cone with radius r and height h is V=(1/3) π (r^2)h.  As the volume is 500, this gives us 500=(1/3) π (r^2)h or h=(1500/(π (r^2))).  Substitiuting this into the area formula gives A=π r(Sqrt[(r^2)+(h^2)])=π r(Sqrt[(r^2)+((((1500/(π (r^2)))))^2)])=π r(Sqrt[(r^2)+((1500^2)/((π^2) (r^4)))])=(Sqrt[(π^2) (r^4)+((1500^2)/( (r^2)))]).  Minimizing this is the same as minimizing the quantity f(r)=(π^2) (r^4)+((1500^2)/( (r^2))) under the square root.  There is no restriction on the value r except for r>0.  As this is not a closed interval the best attack on this is to treat it like a graphing problem and hope the first derivative sign line will help identify the minimum.  f'(r)=4(π^2)(r^3)-((2 (1500^2))/(r^3)).  This exists provided r>0, so the only critical number will come from where this is zero.  4(π^2)(r^3)-((2 (1500^2))/(r^3))=0 ⟹ 2(π^2)(r^6)-(1500^2)=0 ⟹ (r^6)=((1500^2)/(2(π^2))) ⟹ r=(((1500^2)/(2(π^2)))^(1/6))=5(Sqrt[2])((3/π)^(1/3)) (technically r=±5(Sqrt[2])((3/π)^(1/3)) but we know r>0 so the negative is not needed).  Putting this on a sign line gives

NumberLinePlot[ {4Pi^2 r^3-4500000/r^3>0 && r>0, 4Pi^2 r^3-4500000/r^3<0,4Pi^2 r^3-4500000/r^3==0,r^3==0}, {r,0,9}, Spacings-> {2,-6,4,0}, PlotStyle-> {Directive[DotDashed,Black], Directive[Dotted, Black], Directive[PointSize[Large],Black],Directive[PointSize[Large],GrayLevel[.7]]}, PlotLegends-> Map[TraditionalForm[Text[#]]&, {f'[r]>0, f'[r]<0, f'[r]==0,"DNE"}], PlotLabel-> Style[ TraditionalForm[ Text[f'[r]==4Pi^2 r^3-4500000/r^3]],Black], PlotRange-> {{0,10},{-7,3}}, PlotRangePadding->None, Ticks-> {{0,2,4,5,9,10,5 Sqrt[2] CubeRoot[3/Pi]},None}]

From the sign line f(r) is decreasing until the lone critical point r=5(Sqrt[2])((3/π)^(1/3)) and increasing thereafter; therefore the critical point corresponds to an absolute minimum.  So the minimum radius is  r=5(Sqrt[2])((3/π)^(1/3)) and the minimum height is h=(1500/(π (r^2)))=(1500/(π (((5(Sqrt[2])((3/π)^(1/3))))^2)))=10((3/π)^(1/3)).  Interestingly the optimal tent has a height:radius ratio of (h/r)=(Sqrt[2]); that is h=r(Sqrt[2]).  This is true for any open-bottomed tent (to see this just replace the volume 1500 with a letter V that represents a number and not a variable).

Problem 29

Equal squares are cut off of each corner of a rectangular piece of cardboard 16 inches wide and 30 inches long.  The flaps formed by the cuts are folded upwards to form an open-topped box.  What is the largest volume of such a box?

Solution

Let x be the length of the sides of the removed squares.  Then 0≤ x≤ 8 as the original cardboard is only 16 wide (technically 0 and 8 would not leave you with a box but allowing these numbers makes the closed).  We then have the following picture:

Show[ Graphics[  {Line[{{16,5},{16,-5},{13,-5}, {13,-8},{-13,-8},{-13,-5},{-16,-5},{-16,5},{-13,5},{-13,8},{13,8},{13,5},{16,5}}], Red ,Dashed, Line[{{13,8},{16,8},{16,5}}] ,Line[{{13,-8},{16,-8},{16,-5}}], Line[{{-13,-8},{-16,-8},{-16,-5}}], Line[{{-13,8},{-16,8},{-16,5}}], Black, Dashing[0], Text["x",{14.5,9}],Text["x",{-14.5,9}],Text["x",{17,6.5}],Text["x",{17,-6.5}],Text["x",{-17,6.5}],Text["x",{-17,-6.5}],Text["x",{14.5,-9}],Text["x",{-14.5,-9}], Dashed, Line[{{13,5},{-13,5},{-13,-5},{13,-5},{13,5}}], Text[30-2x,{0,4}],Text[16-2x,{11,0}]}  ]]

The length of this box is l=30-2x, the width is w=16-2x, and the depth is d=x.  So the volume to be maximized is V=l w d=x(16-2x)(30-2x)=4x(8-x)(15-x)=480x-92(x^2)+4(x^3) on the interval [0,8]. V'=480-184x+12(x^2).  This exists everywhere, so the only critical numbers must come from where V'=0. Using the quadratic formula, 12(x^2)-184x+480=0 ⟹ x=(10/3),12.  As 12 is outside the interval, we need only look at the volume at x=0,(10/3),8.

TableForm[ {x, x(16-2x)(30-2x)}/. {{x-> 0}, {x-> 10/3}, {x-> 8}}, TableHeadings-> {None, {x, TraditionalForm[ V==x(16-2x)(30-2x)]}}, TableAlignments-> Center]

So the maximum volume you can get is (19600/27)(in^3)≈725.93 (in^3), and this is obtained by removing squares of side length (10/3)in.

Problem 30

A cylinder with an open top and closed bottom is to have a fixed volume of 27 cubic inches.  What is the smallest surface area of such a cylinder?

Solution

Let r be the radius of the cylinder and h its height.  Then the surace area is given by A=π (r^2)+2π r h (the area of the bottom circle plus the area of a rectange with height h and width 2π r, the circumference of the bottom circle).  This is too many variables for us to handle.  The volume of any cylinder of radius r and height h is V=π (r^2)h.  Knowing the volume is 27 gives 27=π (r^2)h ⟹ h=(27/(π (r^2))).  Substituting this into the area formula gives A=π (r^2)+2π r h=π (r^2)+2π r ((27/(π (r^2))))=π (r^2)+(54/( r)).  The only restriction on the radius is r>0. So we will need to treat this like a graphing problem and hope this sheds light on the minimum.  A'=2π r-(54/(r^2)); this exists as long as r>0, so the only critical points will come from when this is zero. 2π r-(54/(r^2))=0 ⟹ π (r^3)-27=0 ⟹(r^3)=(27/π) ⟹r=(3/(π^(1/3))).  Using this as the break point on a sign line gives

NumberLinePlot[ {2Pi r-54/r^2>0 && r>0, 2Pi r-54/r^2<0 && r>0,2Pi r-54/r^2==0,r^3==0}, {r,0,6}, Spacings-> {2,-6,4,0}, PlotStyle-> {Directive[DotDashed,Black], Directive[Dotted, Black], Directive[PointSize[Large],Black],Directive[PointSize[Large],GrayLevel[.7]]}, PlotLegends-> Map[TraditionalForm[Text[#]]&, {f'[r]>0, f'[r]<0, f'[r]==0,"DNE"}], PlotLabel-> Style[ TraditionalForm[ Text[f'[r]==2Pi r-54/r^2]],Black], PlotRange-> {{0,10},{-7,3}}, PlotRangePadding->None, Ticks-> {{0,1,3,4,5,6, 3/CubeRoot[Pi]},None}, LabelStyle-> Black]

Looking at where this implies A is increasing and decreasing tells us that an absolute minimum occurs at r=(3/(π^(1/3))).  Putting this into the formula  A=π (r^2)+(54/( r)) gives the minimum area is A=27(π^(1/3)).  If you check the height you will find the height is also h=(3/(π^(1/3))).

Problem 31

A company prints Calculus texts, each of which costs $20 to print.  When the company charges $70 for the book, it will sell 100,000 copies.  For each dollar over $70 is charges it will sell 1000 fewer copies.  Given that the company must charge at least $70, what price should the company charge to maximize its profit?

Solution

The company cannot charge more than $170, as at that price it would be selling 100,000-100×1000=0 books.  So if x is the price the company charges over $70, x can range from 0 to 100.  When they charge x dollars over $70, the number of copies sold will be 100,000-x.  The revenue obtained when the prince is x dollars over 70 will be  (price)×(# sold)=(70+x)(100000-1000x)=7000000+30000x-1000(x^2).  The costs will be (cost)×(# sold)=(20)(100000-1000x)=2000000-20000x.  So the profit is P(x)=revenue-cost=(7000000+30000x-1000(x^2))-(2000000-20000x)=5000000+50000x-1000(x^2)

P'(x)=50000-2000x; this exists everywhere.  So outside of the endpoints x=0,100 the critical numbers must come from P'(x)=0.  This happens when 50000-2000x=0 or x=25. Making a table of values we have

TableForm[ {x, 5000000+50000x-1000x^2}/. {{x-> 0}, {x-> 25} , {x-> 100}}, TableHeadings-> {None, {x, TraditionalForm[ P[x]==5000000+50000x-1000x^2]}}, TableAlignments-> Center]

So the maximum profit is $5,625,000 and this will occur when the price of the book is $70+$25=$95.

Problem 32

A laddrer 3 meters long leans against  house.  The upper end slips down the wall at a rate of 1.5 meters/second.  How fast is the angle the ladder makes with the wall changing when it is (π/3)?

Solution

First draw a picture of the ladder at any time t. As the angle the ladder makes with the wall changes over time, we can label it as θ(t).  The height of the top end of ladder changes over time, so call it y(t).  How far out the bottom of the ladder is from the wall also changes over time, so call it x(t):

Show[Graphics[ {Line[ {{6,0},{0,0},{0,6}} ], PointSize[Medium], Thick, Line[{{0,5},{3,0}}], Point[{{0,5},{3,0}}], Style[Text[θ(t), {.44,3.6}], FontSize-> 14],Style[Text[y(t), {-.5,5}], FontSize-> 14],Style[Text[x(t), {3,-.5}], FontSize-> 14],Style[Text[3, {2,2.5}], FontSize-> 14], Arrow[{{0,5},{0,3}}], Arrow[ {{3,0}, {5,0}}]}], ImageSize-> 250]

If (Subscript[t, 1])is time in question, we have that θ((Subscript[t, 1]))=(π/3) and y'((Subscript[t, 1]))=-1.5.  As the unknown rate is θ'(t) and the known rate is y'(t), we need to find an equation that relates θ(t) and y(t) and then differentiate it.  From right-triangle trigonometry we have y(t)=3 cos(θ(t)).  So at the time in question y((Subscript[t, 1]))=3cos(θ((Subscript[t, 1])))=3cos((π/3))=(3/2).  Taking the derivative of y(t) by the chain rule gives us y'(t)=-3 sin(θ(t))(d/(dt))θ(t).  Plugging in t=(Subscript[t, 1]) yields y'((Subscript[t, 1]))=-3 sin(θ((Subscript[t, 1])))(d/(dt))θ((Subscript[t, 1])) or -(3/2)=-3sin((π/3))(d/(dt))θ((Subscript[t, 1]))=-((3(Sqrt[3]))/2)(d/(dt))θ((Subscript[t, 1])).  Solving this for (d/(dt))θ((Subscript[t, 1])) gives (d/(dt))θ((Subscript[t, 1]))=((3/2)/((3(Sqrt[3]))/2))=(1/(Sqrt[3])).  As the angle is measured in radians and time in seconds, this means that the angle is increasing at the rate of (1/(Sqrt[3])) radians per second, or approximately 33.08 degrees per second

Problem 33

Sand is falling on to a conical pile at the rate of 2 cubic meters per second, and the pile maintains the shape of a cone whose height is the same as the radius of the base.  How fast is the height of the pile increasing when it is 6 meters tall?

Solution

The volume of a cone is given by V=(1/3)π (r^2)h; as the radius is the same as the height we can use r=h to reduce this to V = (1/3) π (h^3).  Both the volume and the height are changing over time; to show another (easier) notation for this kind of problem I will continue to write them as V and h rather than V(t) and h(t).  We know at the time in question V'=2, h=6, and we would like to find h'.  So we take the formula that relates the parent quantities V and h and differentiate it.   This gives us (d/(dt))V=(d/(dt))((1/3)π (h^3)), or V'=π (h^2)h'.  Substituting in V'=2 and h=6 gives us 2=π (6^2) h', or h'=(2/(36π))=(1/(18π)).  Keeping in mind the units (meters and seconds), this means that when the height of the pile is 6 meters it is increasing at a rate of (1/(18π))≈.0177 meters per second.

Problem 34

A radar station is tracking a vertically launched rocket.  The radar dish is located 5 miles from the launch site.  When the angle from the dish to the rocket is 45 degrees, it is changing at the rate of 10 degrees per minute. How fast is the rocket moving?

Solution

Since we know derivatives will be involved we need to convert the degrees to radians by multiplying my (π/180); so when the angle is (π/4) radians it is increasing at (π/18) radians per minute.  If h is the height of the rocket at any time and θ is the angle the dish is inclined at, we have the following picture:

Show[Graphics[{Line[{{0,0},{5,0}, {5,6},{0,0}}], PointSize[Medium], Point[{{0,0}, {5,0}, {5,6}}], Style[Text["5 miles", {2.5, -.5}], FontSize-> 14],Style[Text["θ", {.65, .25}], FontSize-> 14],Style[Text["h", {5.5, 3}], FontSize-> 14]}]]

At the specific time in question we know θ=(π/4), θ'=(π/18), and we would like to find h'.  So we need to find an equation that relates the parent quantities θ and h.  Right triangle trigonometry tells us that tan(θ)=(h/5), or h=5 tan(θ).  Taking the derivative of both sides gives h'=5(sec^2)(θ) ((dθ)/(dt)); evaluating this at the given instant allows us to substitute in θ=(π/4) and ((dθ)/(dt))=(π/18) to give us h'=5 (sec^2)((π/4))(π/18)=5 ((((Sqrt[2])))^2)(π/18)=((5π)/9) miles per minute, or approximately 1.75 miles per minute.

Problem 35

A hydraulics system is leaking fluid ino a circular pool on the floor.  If the area of the spill is increasing at 10 square inches per minute, how fast is the diameter of the pool changing when it is 1 foot?

Solution

The area of a circle is given by A=π (r^2), and since the radius is half the diameter we can rewrite this as A=π (r^2)=((π((D/2)))^2)=(π/4)(D^2).  At the time in question ((dA)/(dt))=10 and D=12 (we have to be careful to convert D to inches as those are the units for ((dA)/(dt))); we want to find ((dD)/(dt)).  The formula A=(π/4)(D^2) relates the parent quantities, and taking the derivative of this equation gives ((dA)/(dt))=(π/2)D((dD)/(dt)). Substituting in  ((dA)/(dt))=10 and D=12 yields 10=(π/2)×12×((dD)/(dt)), or ((dD)/(dt))=(5/(3π))≈.531 inches per minute.

Problem 36

A snowball is handed to someone in hell. Due to the extreme heat the snowball melts at a rate of 100 cubic inches a minute.  How fast is the radius of the snowball changing when it is 4 inches?

Solution

The formula for the volume of a sphere is V=(4/3)π (r^3).  At the instant in question we know r=4 and ((dV)/(dt))=-100; we are looking for ((dr)/(dt)).  V=(4/3)π (r^3) relates the parent quantities, so differentiating with respect to time gives ((dV)/(dt))=4π (r^2)((dr)/(dt)).  Substituting in r=4 and ((dV)/(dt))=-100 leads us to -100=4 π (4^2)((dr)/(dt)), or ((dr)/(dt))=-(100/(64π))=-(25/(16π)).  So the radius is decreasing at -(25/(16π))≈-.497 inches per minute.