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Math
CalculusSingle Variable Calculus: Early Transcendentals, Volume I8th Edition

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Chapter 2.8, Problem 57E
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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

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Problem 1E
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Problem 52E
Problem 53E
Problem 54E
Problem 55E
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Problem 58E
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Problem 63E
Problem 64E
Problem 65E
Problem 66E
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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

Let f ( x ) = x 3 .

(a) If a ≠ 0, use Equation 2 .7 .5 to find f'(a).

(b) Show that f'(0) does not exist.

(c) Show that f ( x ) = x 3 . has a vertical tangent line at (0, 0).

(Recall the shape of the graph of f. See Figure 1.2.13.)

(a)

To determine

To find: The derivative of the function f(x) at x=a if a≠0.

Explanation

Given:

The given function is, f(x)=x3.

Result Used:

The derivative of a function f at x=a, denoted by f′(x), is

f′(a)=limx→af(x)−f(a)x−a (1)

Difference of cube formula: x3−y3=(x−a)(x2+y2+xy).

Calculation:

Obtain the derivative of the function f(x) at x=a.

Compute f′(a) by using the equation (1),

f′(a)=limx→af(x)−f(a)x−a=limx→ax3−a3x−a=limx→a(x13−a13)x−a=limx→a(x13−a13)x−a×(x23+a13x13+a23)(x23+a13x13+a23)

Apply the difference of cube formula in the numerator as follows,

f′(a)=limx→a(x13)3−(a13

(b)

To determine

To Show: The function f′(0) does not exist.

(c)

To determine

To show: The y=x3 has a vertical tangent line at (0,0)

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