# The value h ′ ( 2 ) .

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 3.4, Problem 56E

(a)

To determine

Expert Solution

## Answer to Problem 56E

The value h(2) is approximately 1.

### Explanation of Solution

Given:

The function is h(x)=f(f(x)).

Result used: Chain Rule

If h is differentiable at x and g is differentiable at h(x), then the composite function F=gh defined by F(x)=g(h(x)) is differentiable at x and F is given by the product,

F(x)=g(h(x))h(x) (1)

The slope of the line passing through the points (x1,y1) and (x2,y2) is y2y1x2x1.

Calculation:

Obtain the derivative of h(x)=f(f(x)).

h(x)=ddx(h(x))=ddx(f(f(x)))

Apply the chain rule as shown in equation (1),

h(x)=f(f(x))f(x)

Substitute x=2 in the above equation,

h(2)=f(f(2))f(2)

From the given graph it is observed that, f(2)=1.

h(2)=f(1)f(2) (2)

Obtain the values f(1) and f(2).

From the given graph, draw the tangent line at x=1 is approximately joining the points (1,(f(1))) and (0.9,(f(0.9))).

Use the slope formula stated above and compute the slope of the line passing through the points (1,f(1)) and (0.9,(f(0.9))).

f(1)=f(0.9)f(1)0.91

From the given graph it is observed that,f(0.9)=2.3 and f(1)=2.2.

f(1)=2.32.20.91=0.10.1=1

Thus, the value is f(1)=1.

Similarly, the value f(2) is f(2)=1.

Substitute 1 for f(1) and 1 for f(2) in equation (2),

h(1)(1)(1)=1

Therefore, the value h(2) is approximately 1.

(b)

To determine

Expert Solution

## Answer to Problem 56E

The value g(2) is approximately 8.

### Explanation of Solution

Given:

The function is g(x)=f(x2).

Calculation:

Obtain the derivative of g(x)=f(x2).

g(x)=ddx(g(x))=ddx(f(x2))

Let h(x)=x2 and g(u)=f(u)  where u=h(x).

Apply the chain rule as shown in equation (1),

g(x)=g(h(x))h(x) (3)

The derivative of g(h(x)) is computed as follows,

g(h(x))=g(u)=ddu(g(u))=ddu(f(u))=f(u)

Substitute u=x2 in the above equation,

g(h(x))=f(x2)

Thus, the derivative is g(h(x))=f(x2).

The derivative of h(x) is computed as follows,

h(x)=ddx(x2)=(2x21)=2x

Thus, the derivative is h(x)=2x.

Substitute f(x2) for g(h(x)) and 2x  for h(x) in equation (3),

g(x)=f(x2)2x=2xf(x2)

Substitute x=2 in the above equation,

g(2)=2(2)f(22)

=4f(4) (4)

Obtain the value of f(4).

From the given graph, draw the tangent line at x=4 is approximately joining the points (4,(f(4))) and (4.1,(f(4.1))).

Use the slope formula stated above and compute the slope of the line passing through the points (4,(f(4))) and (4.1,(f(4.1))).

f(4)=f(4.1)f(4)4.14

From the given graph observe that,f(4)=2 and f(4.1)=2.2.

f(4)=2.224.14=0.20.1=2

Thus, the value f(4) is f(4)=2.

Substitute 2 for f(4) in equation (4),

g(2)42=8

Therefore, the value g(2) is approximately 8.

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