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.

$f\left(x\right)=\left\{\begin{array}{rl}x+4,& \text{if}x<2\\ 3x,& \text{if}x\ge 2,\end{array}x=2\right.$

Short answer

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

Step-by-step solution

Step 1. Given information

Given function$f\left(x\right)=\left\{\begin{array}{rl}x+4,& \text{if}x<2\\ 3x,& \text{if}x\ge 2,\end{array}x=2\right.$

Calculate the LHL and RHL and calculate

Calculating, we get

$$\begin{gathered} \underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{-}}{\lim }(x+4) \\ \underset{x\to 2^{-}}{\lim }f\left(x\right)=2+4 \\ =6 \\ \underset{x\to 2^{+}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }3x \\ =3\times 2 \\ =6 \\ f\left(2\right)={\left.3x\right|}_{x=2} \\ =3\times 2 \\ =6 \end{gathered}$$

So the function is continuous at$x=2$

Checking for differentiability

Checking, we get

$$\begin{gathered} \underset{x\to 2^{-}}{\lim }\frac{f\left(x\right)-f\left(2\right)}{x-2}=\underset{x\to 2^{-}}{\lim }\frac{x+4-(2+4)}{x-2} \\ \underset{x\to 2^{-}}{\lim }\frac{f\left(x\right)-f\left(2\right)}{x-2}=\underset{x\to 2^{-}}{\lim }\frac{x+4-6}{x-2} \\ =\underset{x\to 2^{-}}{\lim }\frac{x-2}{x-2} \\ =\underset{x\to 2^{-}}{\lim }1 \\ =1 \\ \underset{x\to 2^{+}}{\lim }\frac{f\left(x\right)-f\left(2\right)}{x-2}=\underset{x\to 2^{+}}{\lim }\frac{3x-6}{x-2} \\ \underset{x\to 2^{+}}{\lim }\frac{f\left(x\right)-f\left(2\right)}{x-2}=\underset{x\to 2^{+}}{\lim }\frac{3(x-2)}{x-2} \\ =\underset{x\to 2^{+}}{\lim }3 \\ =3 \end{gathered}$$

So the function is not differentiable at$x=2$