Chapter 2: Q. 42 (page 184)

Use the definition of the derivative to find $f'$ for each function f in Exercises 39-54
$f (x) = - x^{3} - x + 1$

Short Answer

Expert verified

The value of $f' (x) = - 3 x^{2} + 1$

Step by step solution

Step 1. Given information

The given function $f (x) = - x^{3} - x + 1$

Step 2. Finding the value of f'(x)

We know that $f' (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} . . . . . . . . . (1)$

Given $f (x) = - x^{3} - x + 1$ then

$f (x + h) = - {(x + h)}^{3} - (x + h) + 1$

Putting the values in (1)

$f' (x) = \lim_{h \to 0} \frac{(- {(x + h)}^{3} - (x + h) + 1) - (- x^{3} - x + 1)}{h} = \lim_{h \to 0} \frac{(- (x^{3} + h^{3} + 3 x^{2} h + 3 x h^{2}) - (x + h) + 1)) + x^{3} + x - 1}{h} [∵ {(a + b)}^{3} = a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2}] = \lim_{h \to 0} \frac{- x^{3} - h^{3} - 3 x^{2} h - 3 x h^{2} - x - h + x^{3} + x - 1}{h} = \lim_{h \to 0} \frac{- h^{3} - 3 x^{2} h - 3 x h^{2} + h}{h} = \lim_{h \to 0} \frac{h (- h^{2} - 3 x^{2} - 3 x h + 1)}{h} = \lim_{h \to 0} - h^{2} - 3 x^{2} - 3 x h + 1$

Putting $h = 0$

$= - {(0)}^{2} - 3 x^{2} - 3 x (0) + 1 = - 3 x^{2} + 1$

Hence, $f' (x) = - 3 x^{2} + 1$

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Use the definition of the derivative to find $f'$ for each function f in Exercises 39-54
$f (x) = - x^{3} - x + 1$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Finding the value of f'(x)

One App. One Place for Learning.

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Use the definition of the derivative to find f' for each function f in Exercises 39-54f(x)=-x3-x+1

Short Answer

Step by step solution

Step 1. Given information

Step 2. Finding the value of f'(x)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Theoretical and Mathematical Physics

Probability and Statistics

Pure Maths

Calculus

Mechanics Maths

Study anywhere. Anytime. Across all devices.

Use the definition of the derivative to find $f'$ for each function f in Exercises 39-54
$f (x) = - x^{3} - x + 1$