Calculus: The Language of Change

Exercise Set 9.4
Bending and the Second Derivative

by John Robeson

Exercise 1) Which Graph?

Identify bends in the figures on page 162 and then compute y'' from the given formula to see which graph is right.

Which graph is nearest y = 4 (x^2) - ((x^5)/5)  ?

Solution

We observe that the bends of graph

	(a) are smile - frown.
	
	(b) are smile - frown
	
	(c) are frown - smile - frown
	
	(d) are frown - smile - frown - smile

We now determine y' and y'' and all values of x where f'[x] = 0 and f''[x] = 0.

\(Clear[x, f]; 
f[x_] := 4\ x\^2 - x\^5\/5\ ; 
\(f'\)[x]\)f''[x]NSolve[f'[x] == 0, x]NSolve[f''[x] == 0, x]

We see from the first derivative that the slope is zero when x = 0 and 2.  This rules out graph (b).

Now let's determine the value of the second derivative at x = 0 and 2.

f''[0]\(\(\(f'\)'\)[2]
\)

Since f''[0] is positive and f''[2] is negative, this tells us our graph pattern is smile - frown.  This rules out graphs (c) and (d).  Therefore, graph (a) is nearest to y = 4 (x^2) - ((x^5)/5) .

\(\(\(Plot[f[x], {x, \(-4\), 4}]; \)
\)\)

Exercise 2) Graphing Using 2nd Derivative

Use first and second 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] =Log[x]/x

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

k)  y = f[x] = (1/(1 + (x^2)))

l)  y = f[x] = (1/(1 - (x^2)))

Check your graphs with the computer, but remember that these are simple practice problems for you to work by hand.  Later, messy real-world problems (like Planck's Law in the Projects) will require this calculus effort before you start the computer.

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

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

So f''[x] = 0 when x = -0.57735, 0.57735

Next we check the sign of f''[x] when x = -1, 0, 1

f''[-1]f''[0]f''[1]

So	f''[-1] is negative (a frown)
	f''[0] is positive (a smile)
	f''[1] is negative (a frown)

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	-0.57	0	0.57	1	2

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

y'	↗	→	↘	→	↗	→	↘

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

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

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

Next we check the sign of f''[x] when x = -1, and 1

f''[-1]f''[1]

So	f''[-1] is negative (a frown)
	f''[1] is positive (a smile)

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'	↗	→	↘	→	↗

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 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]\)NSolve[f''[x] == 0, x]

So f''[x] = 0 when x =-0.707107, 0, 0.707107

Next we check the sign of f''[x] when x = -1, -1/2, 1/2, and 1

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

So	f''[-1] is negative (a frown)
	f''[-1/2] is positive (a smile)
	f''[1/2] is negative (a frown)
	f''[1] is positive (a smile)

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'	↗	→	↘	→	↘	→	↗	

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 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, but is undefined when x = -2

Next we check the sign of f''[x] when x = 0

\(\(\(\(f'\)'\)[0]
\)\)

So	f''[-1] is negative (a frown)
and the function is a frown for all x.

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'	↗	↗	↗	↗	↗	

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 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 = -0.785398, 2.35619

Next we check the sign of f''[x] when x = -1, 0, 3

N[f''[-1]]f''[0]N[f''[3]]

So	f''[-1] is positive (a smile)
	f''[0] is negative (a frown)
	f''[3] is positive (a smile)

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'	↘	→	↗	→	↘	

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 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 = -1.5708, 0, 1.5708

Next we check the sign of f''[x] when x = -2, -1, 1, 2

N[f''[-2]]N[f''[-1]]N[f''[1]]N[f''[2]]

So	f''[-2] is negative (a frown)
	f''[-1] is positive (a smile)
	f''[1] is negative (a frown)
	f''[2] is positive (a smile)

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'	↗	→	↘	→	↗	→	↘	→

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 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 = -2

Next we check the sign of f''[x] when x = -5, 0

f''[-5]f''[0]

So	f''[-5] is negative (a frown)
	f''[0] is positive (a smile)

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'	↘	↘	→	↗	↗

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 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 = -0.707107, 0.707107

Next we check the sign of f''[x] when x = -1, 0, 1

f''[-1]f''[0]f''[1]

So	f''[-1] is positive (a smile)
	f''[0] is negative (a frown)
	f''[1] is positive (a smile)

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'	↗	↗	→	↘	↘	

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 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 for no x but is undefined when x = 0

Next we check the sign of f''[x] when x =  1

f''[1]

So	f''[1] is positive (a smile)

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'	↘	→	↗	↗	

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 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 for no x, but is undefined when x = 0
Next we check the sign of f''[x] when x = -1, 0, 1

f''[-1]f''[0]f''[1]

So	f''[-1] is negative (a frown)
	f''[0] is positive (a smile)
	f''[1] is negative (a frown)

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'	↘	→	↗	↗

y''	∧	∧	→	∨	→	∧	∧

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

Solution to (k)

k)  y = f[x] = (1/(1 + (x^2)))

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

Clear[x, f];
f[x_] := (1/(1 + (x^2)));
f'[x]Factor[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 find f''[x] and all values of x where f''[x] = 0.

\(Clear[x, f]; 
f[x_] := \ 1\/\(1 + x\^2\); 
\(\(f'\)'\)[x]\)NSolve[f''[x] == 0, x]

So f''[x] = 0 when x = -0.57735, 0.57735

Next we check the sign of f''[x] when x = -1, 0, 1

f''[-1]f''[0]f''[1]

So	f''[-1] is positive (a smile)
	f''[0] is negative (a frown)
	f''[1] is positive (a smile)

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	1/5	1/2	4/5	1	4/5	1/2	1/5

y'	↗	↗	↗	→	↘	↘	↘

y''	∨	∨	∧	∧	∧	∨	∨

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

Solution to (l)

l)  y = f[x] = (1/(1 - (x^2)))

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

Clear[x, f];
f[x_] :=  (1/(1 - (x^2)));
f'[x]Factor[f'[x]]NSolve[f'[x]==0,x]

So f'[x] = 0 when x = 0 and is undefined when x = -1. 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 negative
	f'[-1/2] is negative
	f'[1/2] is positive
	f'[2] is positive

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

\(Clear[x, f]; 
f[x_] := \ \ 1\/\(1 - x\^2\); 
\(\(f'\)'\)[x]\)NSolve[f''[x] == 0, x]

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

Next we check the sign of f''[x] when x = -2, 0, 2

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

So	f''[-2] is negative (a frown)
	f''[0] is positive (a smile)
	f''[2] is negative (a frown)

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

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

Finally we use the following table to draw our graph.

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

y	-1/3	∞	4/3	1	4/3	∞	-1/3

y'	↗	*	↘	→	↗	*	↘

y''	∧	*	∨	∨	∨	*	∧

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