Answer to Question #258927 in Calculus for Pankaj

Question

Let f(x) =

1 + 2x, x <= 0

3x - 2, 0 < x <= 1 2x ^ 2 - 1, x > 1

Check whether f is discontinuous. If yes, find where? ii) Give a rough sketch of the graph of f

Answer 1

$f(x)=\begin{cases} 1+2x,\qquad x\leq 0 \\ 3x-2,\qquad 0<x\leq 1 \\2x^2-1,\qquad x>1 \end{cases}$

1) $x=0$

$\lim\limits_{x\rightarrow 0-0}f( x)= \lim\limits_{x\rightarrow 0-0}(1+2x)=1$

$\lim\limits_{x\rightarrow 0+0}f( x)= \lim\limits_{x\rightarrow 0+0}(3x-2)=-2$

$\lim\limits_{x\rightarrow 0-0}f(x)\neq \lim\limits_{x\rightarrow 0+0}f(x)$

$f(x)$ is discontinuous at $x=0$

2) $x=1$

$\lim\limits_{x\rightarrow 1-0}f( x)= \lim\limits_{x\rightarrow 1-0}(3x-2)=1$

$\lim\limits_{x\rightarrow 1+0}f( x)= \lim\limits_{x\rightarrow 1+0}(2x^2-1)=1$

$\lim\limits_{x\rightarrow 1-0}f(x)=\lim\limits_{x\rightarrow 1+0}f(x)=f(1)$

$f(x)$ is continuous at $x=1$