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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Question
![**Problem Statement:**
Let \( h(x) = f(x)g(x) \). If \( f(x) = -5x^2 + 5x - 4 \), \( g(2) = 3 \), and \( g'(2) = -4 \), what is \( h'(2) \)?
**Provide your answer below:**
\( h'(2) = \) \[\_\_\_\_\_\]
**Explanation:**
This problem involves finding the derivative of a product of functions using the product rule. The product rule states:
\[
h'(x) = f'(x)g(x) + f(x)g'(x)
\]
Given:
- \( f(x) = -5x^2 + 5x - 4 \)
- \( g(2) = 3 \)
- \( g'(2) = -4 \)
First, find \( f'(x) \):
\[
f'(x) = \frac{d}{dx}(-5x^2 + 5x - 4) = -10x + 5
\]
Evaluate \( f'(2) \):
\[
f'(2) = -10(2) + 5 = -20 + 5 = -15
\]
Use the product rule to find \( h'(2) \):
\[
h'(2) = f'(2)g(2) + f(2)g'(2)
\]
Given \( g(2) = 3 \) and \( g'(2) = -4 \). Calculate \( f(2) \):
\[
f(2) = -5(2)^2 + 5(2) - 4 = -20 + 10 - 4 = -14
\]
Substitute the values into the product rule formula:
\[
h'(2) = (-15)(3) + (-14)(-4) = -45 + 56 = 11
\]
**Answer:**
\( h'(2) = 11 \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcea36964-4b42-42e5-a5c3-5622db384909%2F7a87a064-2a67-400e-9a04-17f92b80b9ee%2Fge0b7ru_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Let \( h(x) = f(x)g(x) \). If \( f(x) = -5x^2 + 5x - 4 \), \( g(2) = 3 \), and \( g'(2) = -4 \), what is \( h'(2) \)?
**Provide your answer below:**
\( h'(2) = \) \[\_\_\_\_\_\]
**Explanation:**
This problem involves finding the derivative of a product of functions using the product rule. The product rule states:
\[
h'(x) = f'(x)g(x) + f(x)g'(x)
\]
Given:
- \( f(x) = -5x^2 + 5x - 4 \)
- \( g(2) = 3 \)
- \( g'(2) = -4 \)
First, find \( f'(x) \):
\[
f'(x) = \frac{d}{dx}(-5x^2 + 5x - 4) = -10x + 5
\]
Evaluate \( f'(2) \):
\[
f'(2) = -10(2) + 5 = -20 + 5 = -15
\]
Use the product rule to find \( h'(2) \):
\[
h'(2) = f'(2)g(2) + f(2)g'(2)
\]
Given \( g(2) = 3 \) and \( g'(2) = -4 \). Calculate \( f(2) \):
\[
f(2) = -5(2)^2 + 5(2) - 4 = -20 + 10 - 4 = -14
\]
Substitute the values into the product rule formula:
\[
h'(2) = (-15)(3) + (-14)(-4) = -45 + 56 = 11
\]
**Answer:**
\( h'(2) = 11 \)
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