Q. 70 - Calculus

Question

In Exercises 69-80, determine whether or not $f$ is continuous and/or differentiable at the given value of $x$. If not, determine any left or right continuity or differentiability. For the last four functions, use graphs instead of the definition of the derivative.

$f\left(x\right)=x^{2/3},x=0$

Step-by-Step Solution

Answer

The function is continuous and differentiable at $x=0$ and the graph is 

Step 1. Given information

Given function $f\left(x\right)=x^{2/3},x=0$

Step 2: Calculate the LHL and RHL and calculate

Calculating, we get

$$\begin{gathered} \underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{-}}{\lim }{x}^{\frac{2}{3}} \\ \underset{x\to 0^{-}}{\lim }f\left(x\right)=0^{\frac{2}{3}} \\ =0 \\ \underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }{x}^{\frac{2}{3}} \\ \underset{x\to 0^{+}}{\lim }f\left(x\right)=0^{\frac{2}{3}} \\ =0 \\ f\left(0\right)=0^{\frac{2}{3}} \\ =0 \end{gathered}$$

So the function is continuous at $x=0$

Step 3: Checking the differentiability

Checking, we get

$$\begin{gathered} \underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{x\to 0^{-}}{\lim }\frac{x^{\frac{3}{2}}-0}{x} \\ \underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{x\to 0^{-}}{\lim }{x}^{\frac{3}{2}-1} \\ =\underset{x\to 0^{-}}{\lim }{x}^{\frac{1}{2}} \\ =0 \\ \underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{x\to 0^{+}}{\lim }\frac{x^{\frac{3}{2}}-0}{x} \\ =\underset{x\to 0^{+}}{\lim }{x}^{\frac{3}{2}-1} \\ =\underset{x\to 0^{+}}{\lim }{x}^{\frac{1}{2}} \\ =0 \end{gathered}$$

So the function is differentiable at $x=0$

Step 4: Sketching the graph

The graph is