Show that the function f defined as follows, f(x)= 3x-2, & 0 2 is countinuous at x = 2, but not differentiable there at x = 2.

Show that the function f defined as follows, f(x)= 3x-2, & 0<x 1 …

Question

Show that the function f defined as follows,
$\text{f(x)}=\begin{cases}3\text{x}-2, & 0<\text{x}\leq1\\2\text{x}^2-\text{x,} & 1<\text{x}\leq2\\5\text{x}-4,&\text{x}>2\end{cases}$
is countinuous at x = 2, but not differentiable there at x = 2.

✓

Answer

Given:
$\text{f(x)}=\begin{cases}3\text{x}-2, & 0<\text{x}\leq1\\2\text{x}^2-\text{x,} & 1<\text{x}\leq2\\5\text{x}-4,&\text{x}>2\end{cases}$
First, we will show that f(x) is continuos at x = 2.
We have,
(LHL at x = 2)
$=\lim_\limits{\text{x}\rightarrow2^{-}}\text{f(x)}$
$=\lim_\limits{\text{x}\rightarrow0}\text{f}(2-\text{h)}$
$=\lim_\limits{\text{x}\rightarrow0}2(2-\text{h)}^2-(2-\text{h})$
$=\lim_\limits{\text{x}\rightarrow0}(8+2\text{h}^2-8\text{h}-2+\text{h})$
$=6$
(RHL at x = 2)
$=\lim_\limits{\text{x}\rightarrow2^{+}}\text{f(x)}$
$=\lim_\limits{\text{x}\rightarrow0}\text{f}(2+\text{h)}$
$=\lim_\limits{\text{x}\rightarrow0}5(2+\text{h)}-4$
$=\lim_\limits{\text{x}\rightarrow0}(10+5\text{h}-4)$
$=6$
and $\text{f}(2)=2\times4-2=6$
Thus, $=\lim_\limits{\text{x}\rightarrow2^{-}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow2^{+}}\text{f(x)}=\text{f}(2)$
Hence the function is continuous at x = 2.
Now, we will check whether the given function is differerentiable at x = 2.
We have,
(LHL at x = 2)
$\lim_\limits{\text{x}\rightarrow2^{-}}\frac{\text{f(x)}-\text{f}(2)}{\text{x}-2}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\text{f}(2-\text{h})-\text{f}(2)}{-\text{h}}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{2\text{h}^2-7\text{h}+6-6}{-\text{h}}$
$=\lim_\limits{\text{x}\rightarrow0}-2\text{h}+7$
$=7$
(RHL at x = 2)
$\lim_\limits{\text{x}\rightarrow2^{+}}\frac{\text{f(x)}-\text{f}(2)}{\text{x}-2}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\text{f}(2+\text{h})-\text{f}(2)}{\text{h}}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{10+5\text{h}-4-6}{\text{h}}$
$=5$
Thus, LHL at x = 2 $\neq$ RHL at x = 2.
Hence, function is no differentiable at x = 2.