Q. 71 - 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.

$f\left(x\right)=\left|x^2-4\right|,x=2$

Step-by-Step Solution

Answer

The function is continuous at $x=2$ but not differentiable and the graph is 

Step 1. Given information

Given function $f\left(x\right)=\left|x^2-4\right|,x=2$

Step 2: Calculate the LHL and RHL and calculate

Calculating, we get

$$\begin{gathered} \underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{-}}{\lim }\left(4-x^2\right) \\ =4-2^2 \\ =0 \\ \underset{x\to 2^{+}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }{x}^2-4 \\ =2^2-4 \\ =0 \\ f\left(2\right)=\left|2^2-4\right| \\ =0 \end{gathered}$$

So the function is continuous at $x=2$

Step 3: Checking for differentiability

Checking, we get

$$\begin{gathered} \underset{x\to 2^{-}}{\lim }\frac{f\left(x\right)-f\left(2\right)}{x-2}=\underset{x\to 2^{-}}{\lim }\frac{4-x^2-0}{x-2} \\ \underset{x\to 2^{-}}{\lim }\frac{f\left(x\right)-f\left(2\right)}{x-2}=\underset{x\to 2^{-}}{\lim }\frac{-\left(x^2-4\right)}{x-2} \\ =\underset{x\to 2^{-}}{\lim }\frac{-(x+2)(x-2)}{x-2} \\ =\underset{x\to 2^{-}}{\lim }(-(x+2) \\ =-(2+2) \\ =-4 \end{gathered}$$

The function is not differentiable at $x=2$

Step 4: Checking the graph

The graph is