Chapter 2.2, Problem 1E

To determine

To show: The function is not differentiable.

Explanation of Solution

Given information:

The point is $P\left(0,0\right)$ and right hand derivative function is $f\left(x\right)=x$ and left hand derivative is $f\left(x\right)=x^2$

Concept used:

The left hand derivative of a function is $\underset{h\to 0^{-}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}$ and right hand derivative is $\underset{h\to 0^{+}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}$

Calculation:

Value of limit $\underset{h\to 0^{-}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}$ is as shown below.

  $\begin{array}{c}\underset{h\to 0^{-}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}=\underset{h\to 0^{-}}{\lim }\frac{\left(0+h\right)-\left(0\right)}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{h}{h}\\ =1\end{array}$

Value of limit $\underset{h\to 0^{+}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}$ is as shown below.

  $\begin{array}{c}\underset{h\to 0^{+}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{{\left(0+h\right)}^2-\left(0\right)}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{h^2}{h}\\ =\underset{h\to 0^{+}}{\lim }h\\ =0\end{array}$

Since the left hand derivatives and right hand derivatives are not equal.

Conclusion: So, the function is not differentiable at $P\left(0,0\right)$ .