$h(x)=e^{2x}+x^3$. Find $(h^{-1})'(1)$.

Question

Let $h:\mathbb{R}\to\mathbb{R}$ be a function, $h(x)=e^{2x}+x^3$.

Find $(h^{-1})'(1)$.

First thing I did was find $h'(x)$:

$h'(x)=2e^{2x}+3x^{2}$

And now I'm so lost, because I know that $(h^{-1})'(1)=\frac{1}{(h'(h^{-1})'(1))}$ but I don't know how to find $(h^{-1})'(1)$. I'm having real trouble with these kind of problems, and I always get stuck at that part.

Answer 1

That should be $$(h^{-1})'(1) = \frac{1}{h'(h^{-1}(1))}$$ So you need to find $h^{-1}(1)$, i.e. $x$ such that $h(x) = 1$.

Hint: It's easy to guess. Try some small integers.

Answer 2

hint

$$h(0)=e^0+0=1$$

$$\implies h^{-1}(1)=\color{red}{0}$$

$$(h^{-1})'(1)=\frac{1}{h'(\color{red}{0})}$$

$$h'(x)=2e^{2x}+3x^2$$

I am sure you can finish it.

Answer 3

$$ h(x) = e^{2x}+x^3 $$

has inverse because $h'(x) = 2e^{2x}+3x^2 > 0 \ \ \forall x\in \mathbb{R}$ then

$$ h^{-1}(h(x)) = x\Rightarrow \left(h^{-1}(h(x))\right)' = \frac{1}{h'(x)} = \frac{1}{2e^{2x}+3x^2} $$

and

$$ h(0) = 1\Rightarrow \left(h^{-1}(1)\right)'=\frac 12 $$