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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![### Problem Statement
Given :
\[ f(-6) = -2 \]
\[ f'(-6) = -18 \]
\[ g(-6) = -12 \]
\[ g'(-6) = 7 \]
find the value of \( h'(-6) \) based on the function below:
\[ h(x) = \frac{f(x)}{g(x)} \]
### Solution
To find \( h'(-6) \), we will use the quotient rule for derivatives. The quotient rule states that if you have a function \( h(x) = \frac{f(x)}{g(x)} \), then its derivative \( h'(x) \) is given by:
\[ h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \]
Now, we substitute the given values into this formula:
1. \( f'(-6) = -18 \)
2. \( f(-6) = -2 \)
3. \( g'(-6) = 7 \)
4. \( g(-6) = -12 \)
Substitute these values into the quotient rule formula:
\[ h'(-6) = \frac{(-18)(-12) - (-2)(7)}{(-12)^2} \]
Calculate the numerator and denominator separately:
- Numerator: \((-18)(-12) - (-2)(7) = 216 + 14 = 230\)
- Denominator: \((-12)^2 = 144\)
Thus:
\[ h'(-6) = \frac{230}{144} = \frac{115}{72} \]
Therefore, the value of \( h'(-6) \) is \( \frac{115}{72} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F323c4818-e3ec-479b-a7ba-1a514193e117%2F39032218-0b98-4cb9-88c5-c450e88b863f%2F7rs2kj3_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Given :
\[ f(-6) = -2 \]
\[ f'(-6) = -18 \]
\[ g(-6) = -12 \]
\[ g'(-6) = 7 \]
find the value of \( h'(-6) \) based on the function below:
\[ h(x) = \frac{f(x)}{g(x)} \]
### Solution
To find \( h'(-6) \), we will use the quotient rule for derivatives. The quotient rule states that if you have a function \( h(x) = \frac{f(x)}{g(x)} \), then its derivative \( h'(x) \) is given by:
\[ h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \]
Now, we substitute the given values into this formula:
1. \( f'(-6) = -18 \)
2. \( f(-6) = -2 \)
3. \( g'(-6) = 7 \)
4. \( g(-6) = -12 \)
Substitute these values into the quotient rule formula:
\[ h'(-6) = \frac{(-18)(-12) - (-2)(7)}{(-12)^2} \]
Calculate the numerator and denominator separately:
- Numerator: \((-18)(-12) - (-2)(7) = 216 + 14 = 230\)
- Denominator: \((-12)^2 = 144\)
Thus:
\[ h'(-6) = \frac{230}{144} = \frac{115}{72} \]
Therefore, the value of \( h'(-6) \) is \( \frac{115}{72} \).
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