Question

Given $f\left(x\right)=\{\begin{array}{ccc}\sqrt{10-x^2}& if& -3<x<3\\ 2-e^{x-3}& if& x\ge 3\end{array}$ The graph of f(x) is

• A. continuous and differentiable at $x=3$
• B. continuous but not differentiable at $x=3$
• C. differentiable but not continuous at $x=3$
• D. neither differentiable nor continuous at $x=3$

Answer 1

The correct option is B continuous but not differentiable at $x=3$

$LHL={lim}_{x\to 3^{-}}f\left(x\right)={lim}_{x\to 3^{-}}\sqrt{10-x^2}=1$
$RHL={lim}_{x\to 3^{+}}f\left(x\right)={lim}_{x\to 3^{+}}2-e^{x-3}=1$
Also, $f\left(3\right)=1$
$LHL=RHL=f\left(3\right)$
Hence, $f\left(x\right)$ is continuous at $x=3$

$R{f}^{\prime }\left(3^{+}\right)=li{m}_{h\to 0}\frac{f(3+h)-f\left(3\right)}{h}$
$=li{m}_{h\to 0}\frac{(2-e^h)-1}{h}$
$=li{m}_{h\to 0}-\left(\frac{e^h-1}{h}\right)=-1$
$L{f}^{\prime }\left(3^{-}\right)=li{m}_{h\to 0}\frac{f(3-h)-f\left(3\right)}{-h}$
$=li{m}_{h\to 0}\frac{\sqrt{10-(3-h{)}^2}-1}{-h}$
$li{m}_{h\to 0}\frac{\sqrt{1+6h-h^2}-1}{-h}$
$li{m}_{h\to 0}\frac{(6h-h^2)}{-h\sqrt{1+6h-h^2+1}}$
$=li{m}_{h\to 0}\frac{h(h-6)}{h(\sqrt{1+6h-h^2}+1)}$
$=\frac{-6}{2}=-3$

$\Rightarrow f^{\prime }\left(3^{+}\right)\ne f^{\prime }\left(3^{-}\right)$

Hence, $f\left(x\right)$ is not differentiable at $x=3$