Question

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

Answer 1

Here, f(x) $\{\begin{array}{c}-2,ifx\le -1\\ 2x,if-1<x\le 1\\ 2,ifx>1\end{array}$

We have to check the continuity only at x=-1 and 1.

At x=-1, LHL = $\underset{x\to 1^{-}}{lim}$ f(x) = $\underset{x\to 1^{-}}{lim}$ (-2)=-2

RHL = $\underset{x\to -1^{-}}{lim}$ f(x) = $\underset{x\to -1^{-}}{lim}$ (2x)

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

$\therefore$ $\underset{h\to 0}{lim}$ 2(-1+h) = $\underset{h\to 0}{lim}$ (-2+2h)=-2+0=-2

Also, f(-1)=-2

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

At x=1, LHL = $\underset{x\to -1^{-}}{lim}$ f(x) = $\underset{x\to -1^{-}}{lim}$ (2x)

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

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

RHL $\underset{x\to 1^{+}}{lim}$ f(x)= $\underset{h\to 1^{+}}{lim}$(2)=2

Also, f(1)=$2\times$1=2 [$\therefore$f(x)=2x]

LHL=RHL=f(1)

Thus, f(x) is continuous at x=1.

Hence, f(x) is continuous for every value of x.