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Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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![**Step 4**
To find \((f^{-1})'(x)\), use the theorem that states that if \(f\) is differentiable on an interval \(I\) and \(f\) has an inverse function \(g\), that is \(g = f^{-1}\), then \(g\) is differentiable at any \(x\) for which \(f'(g(x)) \neq 0\) and \(g'(x) = \frac{1}{f'(g(x))}\).
Find the values of \(f^{-1}(121)\) and then \(f'(f^{-1}(121))\),
\[ f^{-1}(x) = (x + 4)^{1/3} \]
\[ f^{-1}(121) = 5 \]
\[ f'(f^{-1}(121)) = f'(5) \]
\[ = 3(5)^2 \]
\[ = 75 \]
**Step 5**
Substitute the above values in the expression for \((f^{-1})'(121) = \frac{1}{f'(f^{-1}(121))}\).
\[ (f^{-1})'(121) = \frac{1}{75} \]
Thus, \((f^{-1})'(121) = \frac{1}{75}\).](https://content.bartleby.com/qna-images/question/679cf9fe-7ea4-4759-a28d-7fe07c30cdca/36d90f50-e676-43a0-94f0-f87f7a03b138/8ikavd_thumbnail.png)
Transcribed Image Text:**Step 4**
To find \((f^{-1})'(x)\), use the theorem that states that if \(f\) is differentiable on an interval \(I\) and \(f\) has an inverse function \(g\), that is \(g = f^{-1}\), then \(g\) is differentiable at any \(x\) for which \(f'(g(x)) \neq 0\) and \(g'(x) = \frac{1}{f'(g(x))}\).
Find the values of \(f^{-1}(121)\) and then \(f'(f^{-1}(121))\),
\[ f^{-1}(x) = (x + 4)^{1/3} \]
\[ f^{-1}(121) = 5 \]
\[ f'(f^{-1}(121)) = f'(5) \]
\[ = 3(5)^2 \]
\[ = 75 \]
**Step 5**
Substitute the above values in the expression for \((f^{-1})'(121) = \frac{1}{f'(f^{-1}(121))}\).
\[ (f^{-1})'(121) = \frac{1}{75} \]
Thus, \((f^{-1})'(121) = \frac{1}{75}\).
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