1.3 Functions Given by Graphs

The stuff before Example 1.7 will not make sense unless you watch the video first!!!!

Plot[{1.3x,0.75x^2,2Sqrt[x],1},{x,0,2}]

Plot[{1.5*(x-1)+0.75,0.75x^2},{x,0,2}, PlotRange->{{0,2},{-1,3}}]

Plot[{(x-1)+2,2Sqrt[x]},{x,0,2}, PlotRange->{{0,2},{0,3}}]

Plot[x*Sin[x],{x,0,10}]

Intervals of increase: [0, 2], [5, 8] approximately
Intervals of decrease: [2, 5], [8, 10] approximately
Intervals of concave up: [0, 1], [3.6, 6.5], [9.5, 10]
Intervals of concave down: [1, 3.6],[6.5, 9.5]
Inflection point is a change in concavity.
Inflection points at: x=1, 3.6, 6.5, 9.5

Plot[{-1.3x,-0.75x^2,-2Sqrt[x]},{x,0,2}]

Example 1.7

We are gonna be cheap here and simply paste in the image of the graph... sorry....

Part 1

Clearly the distance stays fixed from about 1.5 hours to 5 hours or so, thus this is the time you were at your friend’s house. (This is time after noon, so 1:30 to 5:00 PM)

Part 2

The graph is decreasing from about 5 hours until 8 hours, and corresponds to traveling back to home base. (So 5:00 to 8:00 PM)

Part 3

The location of steepest increase occurs at inflection points, and in this case, it appears to be near t=1 hour, which is 1:00 PM. This is where you attained your maximum speed on your journey to your friend’s house.

Part 4

You get back home 8 hours after noon, thus 8:00 PM.

Example 1.8

Another pic and paste job...  Here E(d) is the percentage of the Americans 25 or older which who had not completed at least 12 years of school.

Part 1

In function notation E(1940) = 73%, E(1980) = 30%, and E(2010) = 15%.  These are approximations obviously.  We plot them together on one graph.  Pay attention to the extra options in the ListPlot command that follows.

ListPlot[{{1940,73},{1980,30},{2010,15}},AxesOrigin->{1940,0},PlotRange->{{1937,2013},{0,80}},PlotStyle->Blue, PlotMarkers->{"▽",16}]

Part 2

The idea is clearly to connect the points in a rather “smooth” fashion.  Our first attempt has a sharp corner at the point (1980, 30).

ListPlot[{{1940,73},{1980,30},{2010,15}},AxesOrigin->{1940,0},PlotRange->{{1937,2013},{0,80}},PlotStyle->Blue, PlotMarkers->{"▽",16}, Joined-> True]

If we add the InterpolationOrder-> 2 option, it smooths out!  The more points we have, the higher the interpolation order we can use, but for three points, 2 is all we can use.

ListPlot[{{1940,73},{1980,30},{2010,15}},AxesOrigin->{1940,0},PlotRange->{{1937,2013},{0,80}},PlotStyle->Blue, PlotMarkers->{"▽",16}, Joined-> True, InterpolationOrder->2]

Exercise 2

Here the function is v(d), where d is the date and v is the value.

Part a.

In function notation, v(2010) = $5000

Part b.

The graph is concave up, and since it is increasing, this means that the rate of growth in the account is increasing over time.

Part c.

The value of the investment reaches $55000 in 2040 (simply draw a horizontal line at the $55000 mark of the graph, and determine where it intersects the function, then drop that line down to the x-axis).

Part d.

Average yearly increase is given by (v(2060)-v(2050))/(60-50).  Here, we estimate v(2060) = $285000 and v(2050) = $105000.

(285000-105000)/(60-50.0)

So the average yearly increase over the 2050 to 2060 decade is $18000.

Part e.

We do not even have to perform calculations here.  Since the function is concave up and increasing, the average yearly increase from 2050 to 2060 must be greater than that of the average yearly increase over the 2010 to 2020 time frame.

Exercise 8

Yet another cheap picture of a graph from the textbook.  Here the function is W = W(y) where y is the year and W is the number of weddings (in millions).

Part a.

It appears that in 1990, W reached its maximum value of 2.44 million weddings.

Part b.

In 1960, W reaches its minimum value of 1.52 million weddings.

Part c.

So we look for the steepest line segment between points, which appears to be between 1965 and 1970 where it goes from 1.80 to 2.16 million weddings.  The average rate of change in this case would be (2.16-1.80)/5 which is 0.072 million weddings/year is the greatest rate of increase.

(2.16-1.80)/5.0