f'(x) = f(x) = Find a formula for f'(x). Note: f(x) is differentiable at x = -1, but is NOT differentiable at x = 1. Can you see why that is so? if x < -1 if if x > 1 - 1 1
f'(x) = f(x) = Find a formula for f'(x). Note: f(x) is differentiable at x = -1, but is NOT differentiable at x = 1. Can you see why that is so? if x < -1 if if x > 1 - 1 1
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Differentiation of Piecewise Functions
We are given a piecewise function \( f(x) \) defined as follows:
\[ f(x) =
\begin{cases}
-(12x + 6) & \text{if } x < -1 \\
6x^2 & \text{if } -1 \le x \le 1 \\
6x & \text{if } x > 1
\end{cases}
\]
#### Objective
Find a formula for the derivative of the function \( f'(x) \).
#### Note
\( f(x) \) is differentiable at \( x = -1 \), but it is NOT differentiable at \( x = 1 \). Investigate why this differentiation is not possible at \( x = 1 \).
#### Solution
To determine the derivative for each piece:
1. For \( x < -1 \):
\[
f(x) = -(12x + 6) \\
f'(x) = \frac{d}{dx} [-(12x + 6)] = -12
\]
2. For \( -1 \le x \le 1 \):
\[
f(x) = 6x^2 \\
f'(x) = \frac{d}{dx} [6x^2] = 12x
\]
3. For \( x > 1 \):
\[
f(x) = 6x \\
f'(x) = \frac{d}{dx} [6x] = 6
\]
Thus, the derivative \( f'(x) \) is given by:
\[ f'(x) =
\begin{cases}
-12 & \text{if } x < -1 \\
12x & \text{if } -1 \le x < 1 \\
6 & \text{if } x > 1
\end{cases}
\]
### Explanation
\( f(x) \) is differentiable at \( x = -1 \) because both the function itself and its derivative are continuous at this point. However, at \( x = 1 \), although \( f(x) \) is continuous, the derivatives from the left (\(12x\)) and the right (\(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F114d112a-89da-40ea-8ec8-a25f26317aff%2F3846cb01-a437-49e6-82cd-eb0e6f39b410%2F353c4sh_processed.png&w=3840&q=75)
Transcribed Image Text:### Differentiation of Piecewise Functions
We are given a piecewise function \( f(x) \) defined as follows:
\[ f(x) =
\begin{cases}
-(12x + 6) & \text{if } x < -1 \\
6x^2 & \text{if } -1 \le x \le 1 \\
6x & \text{if } x > 1
\end{cases}
\]
#### Objective
Find a formula for the derivative of the function \( f'(x) \).
#### Note
\( f(x) \) is differentiable at \( x = -1 \), but it is NOT differentiable at \( x = 1 \). Investigate why this differentiation is not possible at \( x = 1 \).
#### Solution
To determine the derivative for each piece:
1. For \( x < -1 \):
\[
f(x) = -(12x + 6) \\
f'(x) = \frac{d}{dx} [-(12x + 6)] = -12
\]
2. For \( -1 \le x \le 1 \):
\[
f(x) = 6x^2 \\
f'(x) = \frac{d}{dx} [6x^2] = 12x
\]
3. For \( x > 1 \):
\[
f(x) = 6x \\
f'(x) = \frac{d}{dx} [6x] = 6
\]
Thus, the derivative \( f'(x) \) is given by:
\[ f'(x) =
\begin{cases}
-12 & \text{if } x < -1 \\
12x & \text{if } -1 \le x < 1 \\
6 & \text{if } x > 1
\end{cases}
\]
### Explanation
\( f(x) \) is differentiable at \( x = -1 \) because both the function itself and its derivative are continuous at this point. However, at \( x = 1 \), although \( f(x) \) is continuous, the derivatives from the left (\(12x\)) and the right (\(
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