Explanation:
To find \((f^{-1})'(a)\) where \(f(x) = 5 - 2x^3\) and \(a = 7\), you'll need to follow these steps:
1. Find the inverse function of \(f(x)\).
2. Evaluate \((f^{-1})'(x)\) at the point \(a\).
Step 1: Finding the inverse function \(f^{-1}(x)\):
Let \(y = 5 - 2x^3\), then solve for \(x\):
\[y - 5 = -2x^3\]
\[\frac{5 - y}{2} = x^3\]
\[\sqrt[3]{\frac{5 - y}{2}} = x\]
So, the inverse function is:
\[f^{-1}(x) = \sqrt[3]{\frac{5 - x}{2}}\]
Step 2: Evaluate \((f^{-1})'(a)\) at \(a = 7\):
\[(f^{-1})'(a) = \frac{d}{dx}\left(\sqrt[3]{\frac{5 - x}{2}}\right)\]
Plug in \(a = 7\) to get:
\[(f^{-1})'(7) = \frac{d}{dx}\left(\sqrt[3]{\frac{5 - 7}{2}}\right) = \frac{d}{dx}(-1) = 0\]
Therefore, \((f^{-1})'(a)\) at \(a = 7\) is 0.following me