Question

f(x) = $\{\begin{array}{c}2x,ifx<0\\ 0,if0\le x\le 1\\ 4x,ifx>1\end{array}$

Answer 1

Here, f(x) = $\{\begin{array}{c}2x,ifx<0\\ 0,if0\le x\le 1\\ 4x,ifx>1\end{array}$

for x <0,f(x)=2x;0<x<, f(x)=0 and x>1,f(x)=4x are polynomial and constant funtion, so it is a continuous in the given interval. So, we have to check the continuity at x =0, 1.

At x = 0 LHL = $\underset{x\to 0^{-}}{lim}$ f(x) = $\underset{x\to 0^{-}}{lim}$ (2x)

Putting x=0-h as $x\to 0^{-}$ when $h\to 0$

$\therefore \underset{x\to 0}{lim}$ [2(0-h)] $\underset{x\to 0}{lim}$ (-2h)=$-2ltimes0=0,$

RHL = $\underset{x\to 0^{+}}{lim}$ f(x) = $\underset{x\to 0^{+}}{lim}$ (0)=0

Also, f(0)=0 $\therefore$LHL=RHL=f(0)

Thus, f(x) is continuous for all values of x.

At x = 1, LHL = $\underset{x\to 1^{+}}{lim}$ f(x) = $\underset{x\to 1^{+}}{lim}$ (0)=0, RHL = $\underset{x\to 1^{+}}{lim}$ f(x) = $\underset{x\to 1^{+}}{lim}$ (4x)

Putting x=1+h as $h\to 1^{+}$ when $h\to 0$

$\underset{h\to 0}{lim}$[4(1+h)] = (4+4h)=$4+4\times$0=4

$\therefore LHL\ne RHL.$ Thus, f(x) is continuous everywhere except at x=1.