Question

The function $f(x)=\{\begin{array}{ccc}|x-3|,& if& x\ge 1\\ \frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4},& if& x<1\end{array}$ is

• A. continuous and differentiable at $x=3$
• B. continuous at $x=3,$ but not differentiable at $x=3$
• C. continuous and differentiable everywhere
• D. continuous at $x=1,$ but not differentiable at $x=1,$

Answer 1

Answer: (B)

Clearly, $f(x)$ is not differentiable at $x=3.$
Now, $\underset{x\to 3^{-}}{\lim }f(x)=\underset{h\to 0}{\lim }f(3-h)$
$=\underset{h\to 0}{\lim }|3-h-3|$
$\underset{x\to 3^{+}}{\lim }f(x)=\underset{h\to 0}{\lim }f(3+h)$
$=\underset{h\to 0}{\lim }|3+h-3|=0$
and $f(3)=|3-3|=0$
$\therefore$ $f(x)$ is continuous at $x=3.$