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))(-4) if f(1)=-4 and f(1)=2/11. (f^(-1))(-4)=

Explanation:

Step1: Identify the values of $x$, $f^{-1}(x)$ and $f'(f^{-1}(x))$

Given $x = - 4$, since $f(1)=-4$, then $f^{-1}(-4)=1$. Also given $f'(1)=\frac{2}{11}$, so $f'(f^{-1}(-4)) = f'(1)=\frac{2}{11}$.

Step2: Apply the formula for the derivative of the inverse - function

The formula for the derivative of the inverse function is $(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}$. Substitute $x = - 4$ into the formula. We get $(f^{-1})'(-4)=\frac{1}{f'(f^{-1}(-4))}$.

Step3: Calculate the value of $(f^{-1})'(-4)$

Substitute $f'(f^{-1}(-4))=\frac{2}{11}$ into the formula. Then $(f^{-1})'(-4)=\frac{1}{\frac{2}{11}}$. Using the rule of dividing by a fraction ($a\div\frac{b}{c}=a\times\frac{c}{b}$), we have $(f^{-1})'(-4)=\frac{11}{2}$.

Answer:

$\frac{11}{2}$