Question

The continuity on an interval has a geometric interpretation. namely, a function f defined on an interval I is continuous on I if its graph has no 'holes' or 'jumps' .f is said to have a removable discontinuity at c if f(x) has a limit at c but lim ${lim}_{x\to c}f\left(x\right)\ne f\left(c\right)$. If ${lim}_{x\to c+}f\left(x\right)and{lim}_{x\to c-}f\left(x\right)$ exist but are not equal then c is called jump discontinuity. If ${lim}_{x\to c+}f\left(x\right)and{lim}_{x\to c-}f\left(x\right)$ fail to exist then c is called infinite discontinuity.

$g\left(x\right)=\{\begin{array}{cc}x+7& x<-3\\ |x-2|& -3\le x<-1;\\ x^2-2x& -1\le x<3\\ 2x-3& x\ge 3\end{array}$ then

• A. $x=-1$ is a jump discontinuity
• B. $x=3$ is an infinite discontinuity
• C. $x=-3$ is a jump discontinuity
• D. $x=-1$ is a removable discontinuity

Answer 1

The correct option is C $x=-3$ is a jump discontinuity

$\underset{x\to 1+}{lim}f\left(x\right)=\underset{x\to 1+}{lim}{x}^2-2x=3$

$\underset{x\to -1-}{lim}f\left(x\right)=\underset{x\to -1+}{lim}|x-2|=3$

So $\underset{x\to -1}{lim}f\left(x\right)$ exists and is equal to $f(-1)=3$

Thus $f$ is continuous at $x=-1$
$\underset{x\to -3+}{lim}f\left(x\right)=\underset{x\to -3+}{lim}|x-2|=5$

$\underset{x\to -3-}{lim}f\left(x\right)=\underset{x\to -3-}{lim}x+7=4$
So $x=-3$ is a jump discontinuity