Calculus: The Language of Change

Exercise Set 9.1
Graphs from Formulas

by John Robeson

Exercise 1) Three Points Are Not Enough

This exercise shows you a simple reason why points alone are not enough.

(a)  The cartesian pairs (-2,-2), (0,0), (2,2) are recorded in the table on page 147.  Plot these points.

(b)  Show that the graph of y = x contains all three points from part (a) and sketch the graph.

(c)  Show that the graph of y = (x^3) - 3 x contains all three points from part (a).  Can you sketch it without more points?

(d)  Find a point on the graph of y = x that is not on the graph of y =  (x^3) - 3 x and plot it.

(e)  Use the graphing program in aComputerIntro to plot both functions on the same graph.

(a)

(a)  The cartesian pairs (-2,-2), (0,0), (2,2) are recorded in the table on page 147.  Plot these points.

Solution to (a)

Evaluate the cell below.

\(plot\  = \ 
      ListPlot[{{\(-2\), \(-2\)}, \ {0, 0}, \ {2, 2}}, 
\ \ \ \ \ \ \ 
        Prolog\  -> \ AbsolutePointSize[5]]\)

(b)

(b)  Show that the graph of y = x contains all three points from part (a) and sketch the graph.

Solution to (b)

y = x means that every y coordinate is equal to its x coordinate.  Since this is the case for all three of our points in part (a), we know that they will all be on the graph of y = x.

Now evaluate the cell below to see the graphs.

\(Clear[x, y]; 
y\  = x; 

    line\  = \ 
      Plot[y, {x, \(-3\), 3}, AxesLabel -> {"x", "y"}, 
        PlotRange -> {\(-3\), 3}, PlotStyle -> {RGBColor[1, 0, 0]}]; 

    Show[line, plot\ , Prolog\  -> \ AbsolutePointSize[5]]; \)

(c)

(c)  Show that the graph of y = (x^3) - 3 x contains all three points from part (a).  Can you sketch it without more points?

Solution to (c)

For x = -2, y = (((-2))^3) - 3 (-2) = -8 + 6 = -2 

For x = 0, y = (0^3) - 3 (0)  = 0

For x = 2, y = (2^3) - 3 (2) = 8 - 6 = 2

So all three of our points are also on the graph of  y = (x^3) - 3 x.  However, since this function is not linear, these three points do not give us enough information to know how they are connected.

(d)

(d)  Find a point on the graph of y = x that is not on the graph of y =  (x^3) - 3 x and plot it.

Solution to (d)

The point (1, 1) is on the graph of y = x, but when x = 1, the value of y for  y =  (x^3) - 3 x is y =  (1^3) - 3 (1) = 1 - 3 = -2.  So the point (1, 1) is not on the graph of y =  (x^3) - 3 x.

\(Clear[x, y]; 
y\  = \ x\^3 - 3\ x; 

    point\  = \ 
      ListPlot[{{1, 1}}, 
\ \ \ \ \ \ \ Prolog\  -> \ AbsolutePointSize[5]]; 
    
curve\  = \ 
      Plot[y, {x, \(-3\), 3}, AxesLabel -> {"x", "y"}, 
        PlotStyle -> {RGBColor[1, 0, 0]}]; 

    Show[curve, point, \ Prolog\  -> \ AbsolutePointSize[5]]\)

(e)

(e)  Use the graphing program in aComputerIntro to plot both functions on the same graph.

Solution to (e)

Show[line,  curve,  plot, point, Prolog  ->  AbsolutePointSize[5]]

Exercise 2) Find Equations

Find the equations of the lines shown in Figure 9.3

Solution

All three lines in the first graph have the same slope, which is(1/2).  This is determined by examining the middle line and noticing that it passes through the points (0,0) and (1, (1/2)).  The y-intercepts are (top to bottom) 1, 0 and -1.5 so the equations of the three lines are

y = (1/2)x + 1


		y = (1/2)x 
		
		
and		y = (1/2)x - 1.5

The lines in the second graph have slopes of -(1/4) and +(2/3).  Their y-intercepts are 1 and -2 respectively.  So their equations are

y = -(1/4)x + 1


and		y = (2/3)x - 2

Exercise 3) Scale of the Plot

Use the computer to make several plots on different scales.  First, replot
		y = (x^5) + 4 (x^4) + x
at the four scales described in Example 9.1 above, but leave the default "Ticks" on so that the computer puts the scales on the plots.  This will show you more clearly what we described in the previous paragraph.

Second, plot the functions below on the different suggested scales:

y =(x^x)	0 ≺ x ≺ 2	and	0 ≺ x ≺ 5

y = 3 (x^4) - 4 (x^3) - 36 (x^2)	-10 ≺ x ≺ 10  and	-3 ≺ x ≺ 5

Explain why the pairs of graphs appear to be different even though they are of the same function.

Solution to the first part

Evaluate the cell below.

Clear[x, f]; f[x_] := (x^5) + 4 (x^4) + x\(scale = 100; 

    \(large = Plot[f[x], {x, \(-scale\), scale}, PlotStyle -> {Red}]; 

    scale = 10; 

    medium = Plot[f[x], {x, \(-scale\), scale}, PlotStyle -> {Blue}]; 

    scale = 1; 

    small = Plot[f[x], {x, \(-scale\), scale}, PlotStyle -> {DarkGreen}]; 

    scale = 0.1; 

    tiny = Plot[f[x], {x, \(-scale\), scale}, PlotStyle -> {Brown}]; 

    Show[GraphicsArray[{{large, medium}, {small, tiny}}]]; \)\)

Solution to the second part

Evaluate the cells below.

Clear[x, f]; f[x_] := (x^x)\(plot1 = Plot[f[x], {x, 0, 2}, PlotStyle -> {Red}]; 

    plot2 = Plot[f[x], {x, 0, 5}, PlotStyle -> {Blue}]; 
\)

Both graphs begin at (0,0) and appear to get very large as x gets large.  But the first graph is negative for x less than 1and the second graph appears to be positive for all values of x.  The vertical scale is deceiving us.  Notice that the range of the second graph is 0 to 250 whereas the range of the first is only -1 to 2.  So the second graph is also negative from 0 to 1, but -1 on the much larger scale cannot be perceived.

For the second function evaluate the cell below.

\(Clear[x, f]; 
g[x_] := 3\ x\^4 - 4\ x\^3 - 36\ x\^2\)\(plot3 = Plot[g[x], {x, \(-10\), 10}, PlotStyle -> {Red}]; 

    plot4 = Plot[g[x], {x, \(-3\), 5}, PlotStyle -> {Blue}]; 
\)

These two graphs actually appear to be similar.

Exercise 4) Which is the Graph?

Which of the graphs shown on pages 148 and 149 is the graph of

		y = 3 (x^5) - 5 (x^3)  ?
		
Which shape features of the incorrect ones makes each one wrong?

Solution

Evaluate the cell below to see four graphs of our same function plotted on different scales.

Clear[x, f]; f[x_] := 3 (x^5) - 5 (x^3)\(scale = 100; 

    large = Plot[f[x], {x, \(-scale\), scale}, PlotStyle -> {Red}]; 

    scale = 10; 

    medium = Plot[f[x], {x, \(-scale\), scale}, PlotStyle -> {Blue}]; 

    scale = 2; 

    small = Plot[f[x], {x, \(-scale\), scale}, PlotStyle -> {DarkGreen}]; 

    scale = 0.1; 

    tiny = Plot[f[x], {x, \(-scale\), scale}, PlotStyle -> {Brown}]; 

    Show[GraphicsArray[{{large, medium}, {small, tiny}}]]; \)

Graph (a) appears incorrect since it is symmetrical with respect to the y axis.

Graph (b) appears correct since it is similar to the above dark green graph.

Graphs (c) and (d) appear incorrect since they do not have a "flat" spot at the origin.  Since y'[0] = 0, the correct graph must have a slope of zero at the origin.