Let $f(x)= e^{2x} + x^5 + 1$
- Find $(f^{-1})'(2)$
- Find $(f^{-1})''(2)$
There is a missing link in my brain with regards to dealing with a function containing exponential and algebra. :/
I'm guessing I probably do not need to find the equation for $f^{-1}$ and I tried using $(f^{-1})'(b)= 1/f'(a)$
So for $b=2, f(a)=2$ then $e^{2a} + a^5 + 1 =2$
and I'm not able to proceed from here. Please help! :(
Let $g = f^{-1}$, and $h(x) = f(g(x)) = x$. Then $h'(x) = f'(g(x))g'(x) = 1$. Therefore $g'(x) = \frac{1}{f'(g(x))}$.
Since $f(x) = e^{2x} + x^5 + 1$, $f'(x) = 2e^{2x} + 5x^4$.
Now we need to figure out $g(2)$. After some guess-work we see that $g(2)= 0$, since $f(0) = 2$.
Therefore $g'(2) = \frac{1}{2e^{2\cdot 0} + 5\cdot 0^4} = \frac{1}{2}$.
To find $g''(2)$, I would use the quotient rule on $\frac{1}{f'(g(x))}$.