Question
Show that f(x) = |x − 2| is continuous but not differentiable at x = 2.
✓
Answer
Given: $\text{f}(\text{x})=|\text{x}-3|=\begin{vmatrix}\text{x}-3,&\text{x}\geq3 \\-\text{x}+3, & \text{x}<3 \end{vmatrix}$
Continuity at X = 2: We have,
(LHL at x = 3)
$=\lim_\limits{\text{x} \rightarrow 2}\text{f}(\text{x})$
$=\lim_\limits{\text{x} \rightarrow 0}\text{f}(3-\text{h})$
$=\lim_\limits{\text{x} \rightarrow 0}0(-3+\text{h})+3$
$=0.$
(RHL at x = 3)
$=\lim_\limits{\text{x} \rightarrow 2^{-}}\text{f}(\text{x})$
$=\lim_\limits{\text{x} \rightarrow 0}\text{f}(3+\text{h})$
$=\lim\limits_{\text{x} \rightarrow 0}3+\text{h}-3$
$=0.$
and f(2) = 0
Thus, $\lim_\limits{\text{x} \rightarrow 2^{-}}\text{f}(\text{x})=\text{f}(2).$
Hence,
f(x) is continuous at x = 3
We have,
(LHL at x = 3)
$=\lim_\limits{\text{x} \rightarrow 2^{-}}-\frac{\text{f}(\text{x})-\text{f}(2)}{\text{x}-3}$
$=\lim_\limits{\text{x} \rightarrow 2}\frac{(-\text{x}+3)-0}{\text{x}-3}$
$=\lim_\limits{\text{x} \rightarrow 2}\text{f}(\text{x}-3)\text{x}-2$
$\lim_\limits{\text{x} \rightarrow 2}(-1)$
$=-1$
(RHL at x = 3)
$=\lim_\limits{\text{x} \rightarrow 2^{-}}\frac{\text{f}(\text{x})-\text{f}(2)}{\text{x}-3}$
$=\lim_\limits{\text{x} \rightarrow 2}\frac{(\text{x}-3)-0}{\text{x}-3}$
$=\lim_\limits{\text{x}\rightarrow2}1=1$
Thus, $=\lim_\limits{\text{x}\rightarrow2^{-}}\text{f}(\text{x})\neq\lim_\limits{\text{x}\rightarrow2^{-}}\text{f}(\text{x}).$
Hence, f(x) is not differentiable at x = 3.
Continuity at X = 2: We have,
(LHL at x = 3)
$=\lim_\limits{\text{x} \rightarrow 2}\text{f}(\text{x})$
$=\lim_\limits{\text{x} \rightarrow 0}\text{f}(3-\text{h})$
$=\lim_\limits{\text{x} \rightarrow 0}0(-3+\text{h})+3$
$=0.$
(RHL at x = 3)
$=\lim_\limits{\text{x} \rightarrow 2^{-}}\text{f}(\text{x})$
$=\lim_\limits{\text{x} \rightarrow 0}\text{f}(3+\text{h})$
$=\lim\limits_{\text{x} \rightarrow 0}3+\text{h}-3$
$=0.$
and f(2) = 0
Thus, $\lim_\limits{\text{x} \rightarrow 2^{-}}\text{f}(\text{x})=\text{f}(2).$
Hence,
f(x) is continuous at x = 3
We have,
(LHL at x = 3)
$=\lim_\limits{\text{x} \rightarrow 2^{-}}-\frac{\text{f}(\text{x})-\text{f}(2)}{\text{x}-3}$
$=\lim_\limits{\text{x} \rightarrow 2}\frac{(-\text{x}+3)-0}{\text{x}-3}$
$=\lim_\limits{\text{x} \rightarrow 2}\text{f}(\text{x}-3)\text{x}-2$
$\lim_\limits{\text{x} \rightarrow 2}(-1)$
$=-1$
(RHL at x = 3)
$=\lim_\limits{\text{x} \rightarrow 2^{-}}\frac{\text{f}(\text{x})-\text{f}(2)}{\text{x}-3}$
$=\lim_\limits{\text{x} \rightarrow 2}\frac{(\text{x}-3)-0}{\text{x}-3}$
$=\lim_\limits{\text{x}\rightarrow2}1=1$
Thus, $=\lim_\limits{\text{x}\rightarrow2^{-}}\text{f}(\text{x})\neq\lim_\limits{\text{x}\rightarrow2^{-}}\text{f}(\text{x}).$
Hence, f(x) is not differentiable at x = 3.
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