Question

webwork 2 - topics 4 - 5: problem 10 (1 point) suppose ( f ) is a one - to - one, differentiable function and its inverse function ( f^{-1} ) is also differentiable. one can show using implicit differentiation (do it), that ( (f^{-1})(x)=\frac{1}{f(f^{-1}(x))} ). find ( (f^{-1})(3) ) if ( f(5)=3 ) and ( f(5)=3/11 ). ( (f^{-1})(3)=square )

Explanation:

Step1: Identify the formula

We are given the formula for the derivative of the inverse - function: $(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}$.

Step2: Substitute the given values

We want to find $(f^{-1})'(3)$. We know that $f(5) = 3$ (which means $f^{-1}(3)=5$) and $f'(5)=\frac{3}{11}$.
Substitute $x = 3$ into the formula $(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}$. When $x = 3$, $f^{-1}(3)=5$, so $(f^{-1})'(3)=\frac{1}{f'(f^{-1}(3))}=\frac{1}{f'(5)}$.

Step3: Calculate the result

Since $f'(5)=\frac{3}{11}$, then $(f^{-1})'(3)=\frac{1}{\frac{3}{11}}=\frac{11}{3}$.

Answer:

$\frac{11}{3}$