Question

(a) Find $a$ and $b$ if the function
$f\left(x\right)=\{\begin{array}{cc}\frac{\sin x}{x}& -2\le x<0\\ a\cdot 2^x& 0\le x\le 1\\ b+x& 1<x\le 2\end{array}$
is a continuous function on $[-2,2]$.
(b) How many of the function $f\left(x\right)=|x|,g\left(x\right)=|x{|}^2$ and $h\left(x\right)=|x{|}^3$ are not differentiable at $x=0$?
(i) 0
(ii) 1
(iii) 2
(iv) 3

Answer 1

(a)

$f\left(x\right)$ is continuous on $[-2,2]$

$⟹f\left(0^{-}\right)=f\left(0^{+}\right)$

$⟹\underset{x\to 0}{lim}\frac{\sin x}{x}=a\cdot 2^0$

$⟹1=a$

Since, $f$ is continuous on $[-2,2]$

$⟹f\left(1^{-}\right)=f\left(1^{+}\right)$

$⟹a\cdot 2^1=b+1$

$⟹2=b+1$

$⟹b=1$

(b)

$f\left(x\right)=|x|$

$f\left(x\right)=\{\begin{array}{cc}x& ifx>0\\ -x& ifx<0\\ 0& ifx=0\end{array}$

To check differentiability,

$\underset{h\to 0^{+}}{lim}\frac{|0+h|-|0|}{h}=\underset{h\to 0^{+}}{lim}\frac{|h|}{h}=1$

$\underset{h\to 0^{-}}{lim}\frac{|0-h|-|0|}{-h}=\underset{h\to 0^{-}}{lim}\frac{|h|}{-h}=-1$

$\therefore \underset{h\to 0^{+}}{lim}\ne \underset{h\to 0^{-}}{lim}$

Thus, $f$ is not differentiable at $x=0$

$g\left(x\right)=|x{|}^2=x^2$

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

$=\underset{h\to 0}{lim}\frac{2xh+h^2}{h}$

$=\underset{h\to 0}{lim}2x+h=2x$

Thus, the limit exist.

Hence, $g$ is differentiable at $x=0$ and $f^{\prime }\left(x\right)=0$ at $x=0$

$h\left(x\right)=|x{|}^3$

$h\left(x\right)=\{\begin{array}{cc}{x}^3& ifx>0\\ -x^3& ifx<0\\ 0& ifx=0\end{array}$

Now, consider

$\underset{t\to 0^{+}}{lim}\frac{h(0+t)-h\left(0\right)}{t}$

$=\underset{t\to 0^{+}}{lim}\frac{|0+t{|}^3-|0{|}^3}{t}=\underset{h\to 0}{lim}\frac{t^3}{t}=0$

and $\underset{t\to 0^{-}}{lim}\frac{|0+t{|}^3-|0|}{t}=\underset{t\to 0}{lim}\frac{-t^3}{t}=0$

$\therefore \underset{t\to 0^{+}}{lim}=\underset{t\to 0^{-}}{lim}$

Thus, $h\left(x\right)$ is differentiable at $x=0$

Hence only one function $f\left(x\right)=|x|$ is not differentiable at $x=0$