Left hand derivative and Right hand derivative of the inverse of a non-differentiable function

Question

Let $f(x)$ be an injective function with domain $[a,b]$ and range $[c,d]$.

If $\alpha$ is a point in $(a,b)$ such that $f$ has left hand derivative $l$ and right hand derivative $r$ at $x=\alpha$

with both $l$ and $r$ being non-zero different and negative ,

then prove that the left hand derivative and the right hand derivative of $f^{-1}(x)$ at $x=f(\alpha)$ are $\frac{1}{r}$ and $\frac{1}{l}$ respectively.

My Attempt:

If $g(x)$ be the inverse of $f((x)$ then $f(g(x))=x$ which gives $g'(x)=\frac{1}{f'(g(x))}$.

So if $x=f(\alpha)$ then $g'(f(\alpha))=\frac{1}{f'(g(f(\alpha)))}=\frac{1}{f'(\alpha)}$.

But now how do I get the left hand derivative and the right hand derivative.

I could frame an example $f(x)=\left\{ \begin{array}{ll} \frac{1}{x} & 0<x<2 \\ \frac{2}{x^2} & x\geq 2 \\ \end{array} \right.$

Left hand derivative at $x=2$ equals $\frac{-1}{4}=l$

Right hand derivative at $x=2$ equals $\frac{-1}{2}=r$

$f^{-1}(x)=\left\{ \begin{array}{ll} \\\sqrt {\frac{2}{x}} & 0<x\leq \frac{1}{2} \\ \frac{1}{x} & x> \frac{1}{2} \\ \end{array} \right.$

Left hand derivative of $f^{-1}(x)$ at $x=\frac{1}{2}$ equals $-2=\frac{1}{r}$

Right hand derivative of $f^{-1}(x)$ at $x=\frac{1}{2}$ equals $-4=\frac{1}{l}$

But I am not able to get a proper method.

Answer 1

We have $f'_-(\alpha)=l < 0$ and $f'_+(\alpha)=r < 0$, with $l \neq r$. That implies, for $h > 0$ small enough, that $f(\alpha - h) > f(\alpha) > f(\alpha + h)$.

Since $f$ has a left and a right derivative at $\alpha$, it is continuous at $\alpha$.

Let $\beta = f(\alpha)$ and given $h > 0$, let $k_1 = f(\alpha + h) - f(\alpha)$ and $k_2 = f(\alpha - h) - f(\alpha)$. Note that $k_1 < 0$ and $k_2 > 0$.

Then $\frac{g(\beta + k_1) - g(\beta)}{k_1}=\frac{h}{f(\alpha + h) - f(\alpha)} \to_{h \to 0^+} \frac{1}{r}$,

and $\frac{g(\beta + k_2) - g(\beta)}{k_2}=\frac{-h}{f(\alpha - h) - f(\alpha)} \to_{h \to 0^+} \frac{1}{l}$.

Answer 2

Let $\beta=f(\alpha)$.. First, since $f$ is differentiable at $x=\alpha$ then $f$ is continuous at $x=\alpha$ and consequently

$$x\rightarrow \alpha^{+}\Longrightarrow f(x)\rightarrow f(\alpha)=\beta.$$

Since $f^{\prime+}(\alpha)=\lim_{x\rightarrow \alpha^{+}}{\frac{f(x)-f(\alpha)}{x-\alpha}}=r$ then

$$\frac{1}{{(f^{-1})}^{\prime}(\beta)}=\lim_{y\rightarrow \beta}{\frac{y-\beta}{f^{-1}(y)-f^{-1}(\beta)}}=r$$.

Observe that we denoted $y=f(x)$, $x\in [a,b]$ or equivalently $x=f^{-1}(y)$, $y\in [c,d]$. More importantly, we have that $f^{-1}(y)-f^{-1}(\beta)\neq 0$ since $f$ is injective and $x\neq \alpha$.

The left derivative of the inverse can be obtained analogously.

Answer 3

Picking up from both answers given I was able to understand and consequently plotted the graph depicting $y=f(x)$ and $y=f^{-1}(x)$