Question

The function $f\left(x\right)=\{\begin{array}{cc}|x-3|& \text{for}x\ge 1\\ \frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4}& \text{for}x<1\end{array}$ is

• A. continuous at x=1
• B. continuous at x=3
• C. differentiable at x=1
• D. differentiable at x=3

Answer 1

Answer: (A), (B), (C)

continuous at x=3
continuous at x=1
differentiable at x=1

The given function can be written as

$f\left(x\right)=\{\begin{array}{cc}x-3& \text{for}x\ge 3\\ 3-x& \text{for}1\le x\le 3\\ \frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4}& \text{for}x<1\end{array}$

$\Rightarrow$ $f^{\prime }(1+)=\underset{h\to 0+}{lim}\frac{f(1+h)-f\left(1\right)}{h}$

$=\underset{h\to 0+}{lim}\frac{3-(1+h)-2}{h}=-1$
and $f^{\prime }(1-)=\underset{h\to 0-}{lim}\frac{f(1+h)-f\left(1\right)}{h}$

$=\underset{h\to 0-}{lim}\frac{\frac{{(1+h)}^2}{4}-\frac{3(1+h)}{2}+\frac{13}{4}-2}{h}$
$=\underset{h\to 0+}{lim}\frac{(\frac{1}{4}-\frac{3}{2}+\frac{13}{4}-2)+(\frac{h}{2}-\frac{3h}{2})+\frac{h^2}{4}}{h}$
$=\underset{h\to 0+}{lim}\frac{-h+\frac{h^2}{4}}{h}=-1$

Hence f is differentiable at $x=1$ and so also continuous,
Since $g\left(x\right)=|x|$ is continuous everywhere but not differentiable at $x=0$, f is continuous at $x=3$ but not differentiable.