Verify that f and g are inverse functions algebraically and graphically. f(x) = X+8 X-7' g(x) = 7x + 8 X-1
Verify that f and g are inverse functions algebraically and graphically. f(x) = X+8 X-7' g(x) = 7x + 8 X-1
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![## Verification of Inverse Functions
### Verify algebraically and graphically that \( f \) and \( g \) are inverse functions.
Given:
\[ f(x) = \frac{x + 8}{x - 7} \]
\[ g(x) = \frac{7x + 8}{x - 1} \]
### (a) Algebraically
#### Step 1: Check if \( f(g(x)) = x \)
\[ f(g(x)) = f \left( \frac{7x + 8}{x - 1} \right) \]
Substitute \( g(x) \) into \( f(x) \):
\[ = \frac{\frac{7x + 8}{x - 1} + 8}{\frac{7x + 8}{x - 1} - 7} \]
Simplify the expression step-by-step:
\[ = \frac{\frac{7x + 8 + 8(x - 1)}{x - 1}}{\frac{7x + 8 - 7(x - 1)}{x - 1}} \]
\[ = \frac{7x + 8 + 8x - 8}{7x + 8 - 7x + 7} \]
\[ = \frac{15x}{15} = x \]
Thus, \( f(g(x)) = x \).
#### Step 2: Check if \( g(f(x)) = x \)
\[ g(f(x)) = g \left( \frac{x + 8}{x - 7} \right) \]
Substitute \( f(x) \) into \( g(x) \):
\[ = \frac{7 \left( \frac{x + 8}{x - 7} \right) + 8}{\frac{x + 8}{x - 7} - 1} \]
Simplify the expression step-by-step:
\[ = \frac{\frac{7(x + 8)}{x - 7} + 8}{\frac{x + 8}{x - 7} - \frac{x - 7}{x - 7}} \]
\[ = \frac{\frac{7x + 56 + 8(x - 7)}{x - 7}}{\frac{x + 8 - (x - 7)}{x -](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff2367e66-1ae4-4371-aa7b-db391deab840%2F0c6756ca-4bb8-4420-af23-a584f1a2e7f0%2Fzwbay5m_processed.png&w=3840&q=75)
Transcribed Image Text:## Verification of Inverse Functions
### Verify algebraically and graphically that \( f \) and \( g \) are inverse functions.
Given:
\[ f(x) = \frac{x + 8}{x - 7} \]
\[ g(x) = \frac{7x + 8}{x - 1} \]
### (a) Algebraically
#### Step 1: Check if \( f(g(x)) = x \)
\[ f(g(x)) = f \left( \frac{7x + 8}{x - 1} \right) \]
Substitute \( g(x) \) into \( f(x) \):
\[ = \frac{\frac{7x + 8}{x - 1} + 8}{\frac{7x + 8}{x - 1} - 7} \]
Simplify the expression step-by-step:
\[ = \frac{\frac{7x + 8 + 8(x - 1)}{x - 1}}{\frac{7x + 8 - 7(x - 1)}{x - 1}} \]
\[ = \frac{7x + 8 + 8x - 8}{7x + 8 - 7x + 7} \]
\[ = \frac{15x}{15} = x \]
Thus, \( f(g(x)) = x \).
#### Step 2: Check if \( g(f(x)) = x \)
\[ g(f(x)) = g \left( \frac{x + 8}{x - 7} \right) \]
Substitute \( f(x) \) into \( g(x) \):
\[ = \frac{7 \left( \frac{x + 8}{x - 7} \right) + 8}{\frac{x + 8}{x - 7} - 1} \]
Simplify the expression step-by-step:
\[ = \frac{\frac{7(x + 8)}{x - 7} + 8}{\frac{x + 8}{x - 7} - \frac{x - 7}{x - 7}} \]
\[ = \frac{\frac{7x + 56 + 8(x - 7)}{x - 7}}{\frac{x + 8 - (x - 7)}{x -
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