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Verify f has an inverse function. Then use the function f & the given real number a to find (f^-1)'(a). F(x)=sqrt(x-4); a =2
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to get inverse, flip the variables and solve for "y" \[x = \sqrt{y-4}\] \[x^2 = y-4\] \[y = x^2 + 4\] now for derivative of inverse \[y' = 2x\] plug in x=2 \[y'(2) = 4\]
So from there would you go \[inverse f \prime (2)= \frac{ 1 }{ f \prime(4) }\] or would 4 be the final answer?
no 4 would be the answer since they want the derivative of inverse and we already found the inverse :) to do it the other way, you would just take derivative of f(x) and use relation above \[(f^{-1})'(2) = \frac{1}{f'(8)}\] 2 = sqrt(x-4) ----> x = 8
\[f'(x) = \frac{1}{2 \sqrt{x-4}}\]
Okay. Thank you.
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