Let $f(x)$ be a one-to-one continuous function that satisfy $f(3)=5$. The tangent line at $(3,5)$ is given by $y=2x-1$. I have to find the tangent line of $f^{-1}(x)$ at $(5,3)$.
I know that $\exists f^{-1}(x) $. Furthermore, I know that $f^{-1}(x)$ and $f(x)$ are symmetric with respect to $y=x$, hence $(5,3)$ is on the graph of $f^{-1}(x)$, i.e., $f^{-1}(5)=3$. But I don't know how to use all that in order to find the tangent line at $(5,3)$. Will appreciate any help.
Here's two ways to do it,
(1) Calculate inverse directly. $f^{-1}(x) = \frac{1}{2} \left( x+1 \right)$. Then the slope of the tangent line at any point is clearly $1/2$ after taking a derivative
(2) Alternatively, the inverse function theorem says that if $f$ is a continuously differentiable function with nonzero derivative at the point $a$, then $$ (f^{-1})'(f(a)) = \frac{1}{f'(a)} $$
You need the slope of the tangent line of $f^{-1}(x)$ at $(5,3)$. Note that $f(3)=5$, so by Inverse Function Theorem $$ (f^{-1})'(5) =(f^{-1})'(f(3)) = \frac{1}{f'(3)} = \dfrac{1}{2}. $$
in either case, the tangent line has equation $$ y-3 = \frac{1}{2} \left(x-5 \right) $$