Question

Show that the function f defined as follows, is continuous at x = 2, but not differentiable thereat: $f\left(x\right)=\left\{\begin{array}{ll}3x-2,& 0<x\le 1\\ 2x^2-x,& 1<x\le 2\\ 5x-4,& x>2\end{array}\right.$

Answer 1

Given:
$f\left(x\right)$ = $\left\{\begin{array}{l}3x-2,0<x\le 1\\ \begin{array}{l}2x^2-x,1<x\le 2\\ 5x-4,x>2\end{array}\end{array}\right.$

First , we will show that f(x) is continuos at $x=2$.

We have,

(LHL at x=2)

$$\begin{gathered} \underset{x\to 2^{-}}{=\lim }f\left(x\right) \\ =\underset{h\to 0}{\lim }f(2-h) \\ =\underset{h\to 0}{\lim }2(2-h{)}^2-(2-h) \\ =\underset{h\to 0}{\lim }(8+2h^2-8h-2+h) \\ =6 \end{gathered}$$

(RHL at x = 2)

$$\begin{gathered} =\underset{x\to 2^{+}}{\lim }f\left(x\right) \\ =\underset{h\to 0}{\lim }f(2+h) \\ =\underset{h\to 0}{\lim }5(2+h)-4 \\ =\underset{h\to 0}{\lim }(10+5h-4) \\ =6 \end{gathered}$$

and $f\left(2\right)=2\times 4-2=6.$

Thus, $\underset{x\to 2^{-}}{\lim }f\left(x\right)$ = $\underset{x\to 2^{+}}{\lim }f\left(x\right)$ = $f\left(2\right)$.

Hence the function is continuous at x=2.


Now, we will check whether the given function is differentiable at x = 2.

We have,

(LHD at x = 2)

$$\begin{gathered} \underset{x\to 2^{-}}{\lim }\frac{f\left(x\right)-f\left(2\right)}{x-2} \\ =\underset{h\to 0}{\lim }\frac{f(2-h)-f\left(2\right)}{-h} \\ =\underset{h\to 0}{\lim }\frac{2h^2-7h+6-6}{-h} \\ =\underset{h\to 0}{\lim }-2h+7 \\ =7 \end{gathered}$$

(RHD at x = 2)

$$\begin{gathered} \underset{x\to 2^{+}}{\lim }\frac{f\left(x\right)-f\left(2\right)}{x-2} \\ =\underset{h\to 0}{\lim }\frac{f(2+h)-f\left(2\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{10+5h-4-6}{h} \\ =5 \end{gathered}$$

Thus, LHD at x=2 ≠ RHD at x = 2.

Hence, function is not differentiable at x = 2.