Calculus: The Language of Change

Exercise Set 9.3
Ups and Downs of the Derivative

by John Robeson

Exercise 1) Fill In a Curve

This exercise has a row of microscopic views filled out.  Just looking across the y' row we see that the graph goes up - over - down - over - down - over - up.  Plot the points given in the x and y row.  add little tangent segments given in the y' row at the (x,y) points.  Then fill in a curve that goes through the points and is tangent to the segments.

The table below contains x and y coordinates of the point as well as microscopic views of y = f[x] at those dotted points.  Sketch the graph.

x	-3	-2	-1	0	1	2	3

y	(1/2)	1	0	-1	-(3/2)	-2	0

y'	↗	→	↘	→	↘	→	↗

Solution

Your graph should "connect the dots" and have the following general shape.

Exercise 2) Fill Out The Table

Fill out the table in Figure 9.11 to correspond to the given graph.

Solution

If we assume each tick mark is 1 unit, then our table is

x	-6	-4	-2	0	2	4	6

y	2	0	2	4	2	-2	-4

y'	↘	→	↗	→	↘	→	↘

Exercise 3) Graphing With Slopes Drill

Use the above first derivative procedure to sketch the graphs of

a)  y = f[x] = 2 (x^2) - (x^4)

b)  y = f[x] = (x^3) - 3 x

c)  y = f[x] = 6 (x^5) - 10 (x^3)

d)  y = f[x] = (((x + 2))^(2/3))

e)  y = f[x] = Sin[x] + Cos[x]

f)  y = f[x] = Sin[x] * Cos[x]

g)  y = f[x] = x Exp[x]

h)  y = f[x] = Exp[-(x^2)]

i)  y = f[x] = x Log[x]

j)  y = f[x] = x - Log[x]

You may check your graphs with the computer after you sketch by hand.  Later, you will need the skills you develop in this exercise together with the computer to understand complicated graphs like Planck's Formula in the projects.

Solution to (a)

a)  y = f[x] = 2 (x^2) - (x^4)

First we find f'[x] and all values of x where f'[x] = 0.

Clear[x, f];
f[x_] :=  2 (x^2) - (x^4);
f'[x]Factor[f'[x]]NSolve[f'[x]==0,x]

So f'[x] = 0 when x = -1, 0, 1

Next we check the sign of f'[x] at -2, -(1/2), (1/2), and 2

f'[-2]f'[-1/2]f'[1/2]f'[2]

So	f'[-2]  is positive
	f'[-1/2] is negative
	f'[1/2] is positive
	f'[2] is negative

Now we determine some key (x,y) pairs.

f[-2]f[-1]f[-1/2]f[0]f[1/2]f[1]\(f[2]
\)

Finally we use the following table to draw our graph.

x	-2	-1	-1/2	0	1/2	1	2

y	-8	1	7/16	0	7/16	1	-8

y'	↗	→	↘	→	↗	→	↘

Plot[f[x], {x, -3, 3}];

Solution to (b)

b)  y = f[x] = (x^3) - 3 x

First we find f'[x] and all values of x where f'[x] = 0.

Clear[x, f];
f[x_] := (x^3) - 3 x;
f'[x]Factor[f'[x]]NSolve[f'[x]==0,x]

So f'[x] = 0 when x = -1,  1

Next we check the sign of f'[x] at -2, 0, and 2

f'[-2]f'[0]f'[2]

So	f'[-2]  is positive
	f'[0] is negative
	f'[2] is positive

Now we determine some key (x,y) pairs.

f[-2]f[-1]f[0]f[1]\(f[2]
\)

Finally we use the following table to draw our graph.

x	-2	-1	0	1	2

y	-2	2	0	-2	2

y'	↗	→	↘	→	↗

Plot[f[x], {x, -3, 3}];

Solution to (c)

c)  y = f[x] = 6 (x^5) - 10 (x^3)

First we find f'[x] and all values of x where f'[x] = 0.

Clear[x, f];
f[x_] :=  6 (x^5) - 10 (x^3);
f'[x]Factor[f'[x]]NSolve[f'[x]==0,x]

So f'[x] = 0 when x = -1, 0, 1

Next we check the sign of f'[x] at -2, -(1/2), (1/2), and 2

f'[-2]f'[-1/2]f'[1/2]f'[2]

So	f'[-2]  is positive
	f'[-1/2] is negative
	f'[1/2] is negative
	f'[2] is positive

Now we determine some key (x,y) pairs.

f[-2]f[-1]f[-1/2]f[0]f[1/2]f[1]\(f[2]
\)

Finally we use the following table to draw our graph.

x	-2	-1	-1/2	0	1/2	1	2

y	-112	4	17/16	0	-17/16	4	112

y'	↗	→	↘	→	↘	→	↗

Plot[f[x], {x, -2, 2}];

Solution to (d)

d)  y = f[x] = (((x + 2))^(2/3))

First we find f'[x] and all values of x where f'[x] = 0.

Clear[x, f];
f[x_] :=  (((x + 2))^(2/3));
f'[x]NSolve[f'[x]==0,x]

So f'[x] = 0 for no x.

Next we check the sign of f'[x] at -3, -2, 0, and 2

N[f'[-3]]f'[-2]f'[0]f'[2]

So the domain of f is all x greater than -2 and
	f'[0] is positive
	f'[2] is positive

Now we determine some key (x,y) pairs.

f[-1.99]f[-1]N[f[0]]N[f[2]]\(N[f[5]]

\)

Finally we use the following table to draw our graph.

x	-1.99	-1	0	2	5

y	.046	1	1.59	2.52	3.66

y'	↗	↗	↗	↗	↗

Plot[f[x], {x, -2, 6}];

Solution to (e)

e)  y = f[x] = Sin[x] + Cos[x]

First we find f'[x] and all values of x where f'[x] = 0.

Clear[x, f];
f[x_] :=   Sin[x]  +  Cos[x];
f'[x]NSolve[f'[x]==0,x]

So f'[x] = 0 when x = -2.36, 0.785 and we know this is cyclical.
Next we check the sign of f'[x] at -2.5, 0, and 1

f'[-2.5]f'[0]N[f'[1]]

So	f'[-2.5]  is negative
	f'[0] is positive
	f'[1] is negtive

Now we determine some key (x,y) pairs.

f[-2.5]f[-2.35619449019234483]f[0]f[0.785398163397448278`]\(N[f[1]]

\)

Finally we use the following table to draw our graph.

x	-2.5	-2.36	0	0.785	1

y	-1.4	-1.41	1	1.41	1.38

y'	↘	→	↗	→	↘

Plot[f[x], {x, -3, 1.5}];

Solution to (f)

f)  y = f[x] = Sin[x] * Cos[x]

First we find f'[x] and all values of x where f'[x] = 0.

Clear[x, f];
f[x_] :=  Sin[x] Cos[x];
f'[x]NSolve[f'[x]==0,x]

So f'[x] = 0 when x = -2.36, -0.785, 0.785, 2.36 and f is cyclical.
Next we check the sign of f'[x] at -2.4, -1, 0, 1, and 2.4

f'[-2.4]N[f'[-1]]f'[0]N[f'[1]]N[f'[2.4]]

So	f'[-2.4] is positive
	f'[-1] is negative
	f'[0] is positive
	f'[1] is negative
	f'[2.4] is positive

Now we determine some key (x,y) pairs.

f[-2.5]f[-2.35619449019234483`]N[f[-1]]f[-0.785398163397448278`]f[0]f[0.785398163397448278`]N[f[1]]f[2.35619449019234483`]\(f[2.5]
\)

Finally we use the following table to draw our graph.

x	-2.5	-2.36	-1	-0.785	0	0.785	1	2.36

y	0.48	0.5	-0.45	-0.5	0	0.5	0.45	-0.5

y'	↗	→	↘	→	↗	→	↘	→

Plot[f[x], {x, -3, 3}];

Solution to (g)

g)  y = f[x] = x Exp[x]

First we find f'[x] and all values of x where f'[x] = 0.

Clear[x, f];
f[x_] := x Exp[x];
f'[x]NSolve[f'[x]==0,x]

So f'[x] = 0 when x = -1

Next we check the sign of f'[x] at -2, and 0

f'[-2]f'[0]

So	f'[-2]  is negative
	f'[0] is positive

Now we determine some key (x,y) pairs.

N[f[-5]]N[f[-2]]N[f[-1]]N[f[0]]\(N[f[1]]
\)

Finally we use the following table to draw our graph.

x	-5	-2	-1	0	1

y	-0.03	-0.27	-0.37	0	2.78

y'	↘	↘	→	↗	↗

Plot[f[x], {x, -5, 1}];

Solution to (h)

h)  y = f[x] = Exp[-(x^2)]

First we find f'[x] and all values of x where f'[x] = 0.

Clear[x, f];
f[x_] := Exp[-(x^2)];
f'[x]NSolve[f'[x]==0,x]

So f'[x] = 0 when x = 0

Next we check the sign of f'[x] at -1 and 1

f'[-1]\(\(f'\)[1]
\)

So	f'[-1]  is positive
	f'[1] is negative

Now we determine some key (x,y) pairs.

N[f[-2]]N[f[-1]]f[0]N[f[1]]\(N[f[2]]
\)

Finally we use the following table to draw our graph.

x	-2	-1	0	1	2

y	0.018	0.37	1	0.37	0.018

y'	↗	↗	→	↘	↘

Plot[f[x], {x, -3, 3}];

Solution to (i)

i)  y = f[x] = x Log[x]

First we find f'[x] and all values of x where f'[x] = 0.

Clear[x, f];
f[x_] := x Log[x];
f'[x]NSolve[f'[x]==0,x]

So f'[x] = 0 when x = 0.368

Next we check the sign of f'[x] at 0.1 and 1

f'[0.1]f'[1]

So	f'[0.1]  is negative
	f'[1] is positive

Now we determine some key (x,y) pairs.

f[0.1]f[0.367879441171442334`]f[1]\(N[f[2]]

\)

Finally we use the following table to draw our graph.

x	0.1	0.37	1	2

y	-0.23	-0.37	0	1.39

y'	↘	→	↗	↗

Plot[f[x], {x, 0, 3}];

Solution to (j)

j)  y = f[x] = x - Log[x]

First we find f'[x] and all values of x where f'[x] = 0.

Clear[x, f];
f[x_] :=  x  -  Log[x] ;
f'[x]NSolve[f'[x]==0,x]

So f'[x] = 0 when x = 1

Next we check the sign of f'[x] at 0.1 and 2

f'[0.1]f'[2]

So	f'[0.1] is negative
	f'[2] is positive

Now we determine some key (x,y) pairs.

f[0.1]f[1]N[f[2]]\(N[f[3]]

\)

Finally we use the following table to draw our graph.

x	0.1	1	2	3

y	2.4	1	1.3	1.9

y'	↘	→	↗	↗

Plot[f[x], {x, 0, 3}];