Question

f(x) $\{\begin{array}{c}{x}^3-3,ifx\le 2\\ x^2+1,ifx>2\end{array}$

Answer 1

Here, f(x) $\{\begin{array}{c}{x}^3-3,ifx\le 2\\ x^2+1,ifx>2\end{array}$

for x < 2, f(x) = $(x^2+1)$ and for x > 2, 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 = 2.

LHL = $\underset{x\to 2^{-}}{lim}$ f(x) = $\underset{x\to 2^{-}}{lim}$ $(x^3-3)$

Putting x=2-h as $x\to 2^{-}$ when $h\to 0$

$\therefore$ $\underset{h\to 0}{lim}\left)\right[(2-h{)}^3-3]$ = $\underset{h\to 0}{lim}[8+h^3-12h+6h^2-3]$ = $\underset{h\to 0}{lim}[5+h^3-12h+6h^2]=5$

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

Putting x=2+h as $x\to 2^{+}$ when $x\to 0$

$\underset{h\to 0}{lim}\left[\right(2+h{)}^2+1]$ \)= $\underset{h\to 0}{lim}$ (4+h^2+4h+1) = $\underset{h\to 0}{lim}(5+h^2+4h)$ = 5

Also f(2)=$(2{)}^3$-3=8-3=5 $[\therefore f(x)=x^3-3]$

$\therefore$ LHL = RHL = f(2). Thus, f(x) is continuous at x=2.

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