Suppose f(x) and g9(z) have the following values: • f(- 7) = - 6 and f'( – 7) = 1, • 9( – 7) = 1 and g'( – 7) = – 5. Assume all functions are continuous. Find the following. [f(z))² – 36 lim z-7 In g(z) (f(z) + 6)² lim 1-1 (g(z) – 1)²
Suppose f(x) and g9(z) have the following values: • f(- 7) = - 6 and f'( – 7) = 1, • 9( – 7) = 1 and g'( – 7) = – 5. Assume all functions are continuous. Find the following. [f(z))² – 36 lim z-7 In g(z) (f(z) + 6)² lim 1-1 (g(z) – 1)²
College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter4: Exponential And Logarithmic Functions
Section: Chapter Questions
Problem 3CC: If xis large, which function grows faster, f(x)=2x or g(x)=x2?
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![### Limits and Function Values
#### Given Information:
Suppose \( f(x) \) and \( g(x) \) have the following values:
- \( f(-7) = -6 \) and \( f'(-7) = 1 \),
- \( g(-7) = 1 \) and \( g'(-7) = -5 \).
Assume all functions are continuous. Find the following limits:
#### Limit Problems:
1. \[
\lim_{x \to -7} \frac{[f(x)]^2 - 36}{\ln g(x)} = \_\_\_\_\_\_
\]
2. \[
\lim_{x \to -7} \frac{(f(x) + 6)^2}{(g(x) - 1)^2} = \_\_\_\_\_\_
\]
#### Explanation of Each Limit Problem:
1. **First Limit:**
The limit to be solved is:
\[
\lim_{x \to -7} \frac{[f(x)]^2 - 36}{\ln g(x)}
\]
- \( f(x) \) is given as \( -6 \) when \( x \) is \( -7 \).
- \( [f(x)]^2 - 36 \) simplifies as follows at \( x = -7 \):
\[
[-6]^2 - 36 = 36 - 36 = 0
\]
- \( g(x) \) is given as \( 1 \) when \( x \) is \( -7 \).
- \( \ln g(x) \) simplifies since \( \ln 1 = 0 \).
Using the provided derivative values and L'Hopital's Rule might be necessary to resolve the \( \frac{0}{0} \) indeterminate form.
2. **Second Limit:**
The limit to be solved is:
\[
\lim_{x \to -7} \frac{(f(x) + 6)^2}{(g(x) - 1)^2}
\]
- \( f(x) \) is given as \( -6 \) when \( x \) is \( -7 \).
- \( (f(x) + 6) \) simplifies as follows at \( x = -7](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc38c2b2e-c8f5-4a38-8774-d53e7e70016b%2F5dc92a85-ccd1-4df7-acf7-b38cafa38b32%2Ff27cxt_processed.png&w=3840&q=75)
Transcribed Image Text:### Limits and Function Values
#### Given Information:
Suppose \( f(x) \) and \( g(x) \) have the following values:
- \( f(-7) = -6 \) and \( f'(-7) = 1 \),
- \( g(-7) = 1 \) and \( g'(-7) = -5 \).
Assume all functions are continuous. Find the following limits:
#### Limit Problems:
1. \[
\lim_{x \to -7} \frac{[f(x)]^2 - 36}{\ln g(x)} = \_\_\_\_\_\_
\]
2. \[
\lim_{x \to -7} \frac{(f(x) + 6)^2}{(g(x) - 1)^2} = \_\_\_\_\_\_
\]
#### Explanation of Each Limit Problem:
1. **First Limit:**
The limit to be solved is:
\[
\lim_{x \to -7} \frac{[f(x)]^2 - 36}{\ln g(x)}
\]
- \( f(x) \) is given as \( -6 \) when \( x \) is \( -7 \).
- \( [f(x)]^2 - 36 \) simplifies as follows at \( x = -7 \):
\[
[-6]^2 - 36 = 36 - 36 = 0
\]
- \( g(x) \) is given as \( 1 \) when \( x \) is \( -7 \).
- \( \ln g(x) \) simplifies since \( \ln 1 = 0 \).
Using the provided derivative values and L'Hopital's Rule might be necessary to resolve the \( \frac{0}{0} \) indeterminate form.
2. **Second Limit:**
The limit to be solved is:
\[
\lim_{x \to -7} \frac{(f(x) + 6)^2}{(g(x) - 1)^2}
\]
- \( f(x) \) is given as \( -6 \) when \( x \) is \( -7 \).
- \( (f(x) + 6) \) simplifies as follows at \( x = -7
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