For complete context to this question, I am attempting to prove the following:
Suppose $f:\mathbb{R}\to\mathbb{R}$ is continuously differentiable such that $f'(x)>0$ for all $x\in\mathbb{R}$. Show that:
- $f$ is invertible on the interval $J=f(\mathbb{R})$
- The inverse is continuously differentiable
- $(f^{-1})'(y)>0$ for all $y\in\mathbb{R}$
I've pieced together so far that since $f'(x)>0$ it means $f$ is strictly increasing. I also understand that any strictly increasing function $f$ is injective. What I am struggling with is part (1) of this proposition.
When we say invertible on $J=f(\mathbb{R})$ (i.e. the image of $f$) is that the same as stating that $f$ has a left-inverse? (by definition injective)
If so, what I have above should be enough for part (1), correct?
Any injective function is bijective onto its image, that's all there is to it. Since $f$ is continuously differentiable with non-vanishing derivative, the Inverse Function Theorem says that $f^{-1}$ is also continuously differentiable (unless the point of the exercise is to show the IFT for $n=1$? Then there's some work to do here). The last assertion follows from the chain rule: $$(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}>0,$$since $f'>0$.