Show that the function f(x)= |2x-3||x|, & x 1 ( 2 ),& x>1 is continuous but not differentiable at x = 1.

Given: f(x)= |2x-3||x|, & x 1 ( 2 ),& x>1 Continuity at x = 1: (LHL at x = 1) = _ x 1^ - f(x)= _ h 0 f(1-h)= _ h 0 ( (1-h) 2 )= 2 =1 (RHL at x = 1) = _ x 1^ + f(x)= _ h 0 f(1+h) _ h 0 |2(1+h)-3|[1+h]= _ h 0 |2(1+h)-3|=1 Hence, (LHL at x = 1) = (RHL at x = 1) Differentiable at x = 1: (LHL at x = 1) = _ x 1^ - f(x)-f(1) x-1 (LHL at x = 1) = _ h 0 f(1-h)-f(1) 1-h-1 (LHL at x = 1) = _ h 0 f(1-h)-f(1) -h (LHL at x = 1) = _ h 0 ( (1-h) 2 )-1 -h (LHL at x = 1) = _ h 0 2 -1 -h (LHL at x = 1) =- 2 _ h 0 2 -1 2 h =0 (RHL at x = 1) = _ x 1^ + f(x)-f(1) x-1 (RHL at x = 1) = _ h 0 f(1+h)-f(1) 1+h-1 (RHL at x = 1) = _ h 0 f(1+h)-f(1) h (RHL at x = 1) = _ h 0 -(2(1+h)-3)-1 h (RHL at x = 1) = _ h 0 -2h h =-2 LHL Hence, the function is continuous but not differentiable at x = 1.

Show that the function f(x)= |2x-3||x|, & x 1 ( 2 ),& x>1 is cont…

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Question

Show that the function $\text{f(x)}=\begin{cases}|2\text{x}-3||\text{x}|, & \text{x}\geq1\\\sin\Big(\frac{\pi\text{x}}{2}\Big),& \text{x}>1\end{cases}$ is continuous but not differentiable at x = 1.

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Answer

Given: $\text{f(x)}=\begin{cases}|2\text{x}-3||\text{x}|, & \text{x}\geq1\\\sin\Big(\frac{\pi\text{x}}{2}\Big),& \text{x}>1\end{cases}$
Continuity at x = 1:
(LHL at x = 1) = $\lim_\limits{\text{x}\rightarrow1^{-}}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1-\text{h})=\lim_\limits{\text{h}\rightarrow0}\sin\Big(\frac{\pi(1-\text{h})}{2}\Big)=\sin\frac{\pi}{2}=1$
(RHL at x = 1) $=\lim_\limits{\text{x}\rightarrow1^{+}}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1+\text{h})$
$\Rightarrow\ \lim_\limits{\text{h}\rightarrow0}|2(1+\text{h})-3|[1+\text{h}]=\lim_\limits{\text{h}\rightarrow0}|2(1+\text{h})-3|=1$
Hence, (LHL at x = 1) = (RHL at x = 1)
Differentiable at x = 1:
(LHL at x = 1) $=\lim_\limits{\text{x}\rightarrow1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
(LHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{1-\text{h}-1}$
(LHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{-\text{h}}$
(LHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Big(\frac{\pi(1-\text{h})}{2}\Big)-1}{-\text{h}}$
(LHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{\cos\frac{\pi\text{h}}{2}-1}{-\text{h}}$
(LHL at x = 1) $=-\frac{\pi}{2}\lim_\limits{\text{h}\rightarrow0}\frac{\cos\frac{\pi\text{h}}{2}-1}{\frac{\pi}{2}\text{h}}=0$
(RHL at x = 1) $=\lim_\limits{\text{x}\rightarrow1^{+}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
(RHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-\text{f}(1)}{1+\text{h}-1}$
(RHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-\text{f}(1)}{\text{h}}$
(RHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{-(2(1+\text{h})-3)-1}{\text{h}}$
(RHL at x = 1) $=\lim_\limits{\text{h}\rightarrow0}\frac{-2\text{h}}{\text{h}}=-2$
$\text{LHL}\neq\text{RHL}$
Hence, the function is continuous but not differentiable at x = 1.