Question

f(x) = $\{\begin{array}{c}x+1,ifx\ge 1\\ x^2+1,ifx<1\end{array}$

Answer 1

Here, f(x) = $\{\begin{array}{c}x+1,ifx\ge 1\\ x^2+1,ifx<1\end{array}$

for x > 1, f(x) = x + 1 and x < 1, f(x) = $x^2$ + 1 is a polynomial funtion, so f(x) is a continuous in the given interval. Therefore, we have to check the continuity at x = 1.

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

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

$\therefore$ $\underset{h\to 0}{lim}\left[\right(1-h{)}^2+1]$ = $\underset{h\to 0}{lim}[1+h^2-2h+1]$ = $\underset{h\to 0}{lim}[2+h^2-2h]$= 2+0-0=2

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

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

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

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

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

Hence, there is no point of discontinuity for this function f(x).