Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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**Problem Statement:** Find a formula for \( f^{-1}(x) \) and \((f^{-1})'(x)\) if \( f(x) = \sqrt{\frac{1}{x - 6}} \). **Solution:** 1. **Finding the inverse function, \( f^{-1}(x) \):** Given \( f(x) = \sqrt{\frac{1}{x - 6}} \). To find the inverse, solve for \( x \) in terms of \( y \) where \( y = f(x) \). \[ y = \sqrt{\frac{1}{x - 6}} \] Square both sides to eliminate the square root: \[ y^2 = \frac{1}{x - 6} \] Rearrange to solve for \( x \): \[ x - 6 = \frac{1}{y^2} \] \[ x = \frac{1}{y^2} + 6 \] Thus, the inverse function is: \[ f^{-1}(x) = \frac{1}{x^2} + 6 \] 2. **Finding the derivative of the inverse, \( (f^{-1})'(x) \):** We use the formula \((f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}\). First, differentiate \( f(x) \): \[ f(x) = (x - 6)^{-\frac{1}{2}} \] Using the chain rule: \[ f'(x) = -\frac{1}{2}(x - 6)^{-\frac{3}{2}} \] Substitute \( f^{-1}(x) \) into \( f'(x) \): \( f'(f^{-1}(x)) = -\frac{1}{2}\left(\frac{1}{x^2}\right)^{-\frac{3}{2}} = -\frac{1}{2}x^3 \) Thus, \((f^{-1})'(x)\) is: \[ (f^{-1})'(x) = \frac{1}{-\frac{1}{2}x^
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Transcribed Image Text:**Problem Statement:** Find a formula for \( f^{-1}(x) \) and \((f^{-1})'(x)\) if \( f(x) = \sqrt{\frac{1}{x - 6}} \). **Solution:** 1. **Finding the inverse function, \( f^{-1}(x) \):** Given \( f(x) = \sqrt{\frac{1}{x - 6}} \). To find the inverse, solve for \( x \) in terms of \( y \) where \( y = f(x) \). \[ y = \sqrt{\frac{1}{x - 6}} \] Square both sides to eliminate the square root: \[ y^2 = \frac{1}{x - 6} \] Rearrange to solve for \( x \): \[ x - 6 = \frac{1}{y^2} \] \[ x = \frac{1}{y^2} + 6 \] Thus, the inverse function is: \[ f^{-1}(x) = \frac{1}{x^2} + 6 \] 2. **Finding the derivative of the inverse, \( (f^{-1})'(x) \):** We use the formula \((f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}\). First, differentiate \( f(x) \): \[ f(x) = (x - 6)^{-\frac{1}{2}} \] Using the chain rule: \[ f'(x) = -\frac{1}{2}(x - 6)^{-\frac{3}{2}} \] Substitute \( f^{-1}(x) \) into \( f'(x) \): \( f'(f^{-1}(x)) = -\frac{1}{2}\left(\frac{1}{x^2}\right)^{-\frac{3}{2}} = -\frac{1}{2}x^3 \) Thus, \((f^{-1})'(x)\) is: \[ (f^{-1})'(x) = \frac{1}{-\frac{1}{2}x^
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Step 1: Given function

f open parentheses x close parentheses equals square root of fraction numerator 1 over denominator x minus 6 end fraction end root

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