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Show that F(X) = |X − 2| is Continuous but Not Differentiable at X = 2. - Mathematics

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Question

Show that f(x) = |x − 2| is continuous but not differentiable at x = 2. 

Answer in Brief
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Solution

Given:  

\[f(x) = |x - 2| = \] `{(x-2,x≥,2),(-x+2,x<,2) :}`
 

Continuity at x=2: We have,
(LHL at x = 2) 

\[{= \lim}_{x \to 2^-} f(x) \]
\[ = \lim_{h \to 0} f(2 - h) \]
\[ = \lim_{h \to 0} ( - 2 + h) + 2\]
\[ = 0\]

(RHL at x = 2) 

\[{= \lim}_{x \to 2^+} f(x) \]
\[ = \lim_{h \to 0} f(2 + h) \]
\[ = \lim_{h \to 0} 2 + h - 2 \]
\[ = 0\]
and 

\[f(2) = 0\]
\[\lim_{x \to 2^-} f(x)\]
\[\lim_{x \to 2^+} f(x)\]
\[f(2)\]

Hence, 

\[f(x)\]  is continuous at 
\[x = 2\]
Differentiability at x = 2: We have,
(LHD at x = 2)
\[{=lim}_{x \to 2^-} \frac{f(x) - f(2)}{x - 2} \]
\[ = \lim_{x \to 2} \frac{( - x + 2) - 0}{x - 2} \]
\[ = \lim_{x \to 2} \frac{- (x - 2)}{x - 2} \]
\[ = \lim_{x \to 2} ( - 1) \]
\[ = - 1\]
(RHD at x=2)
\[= \lim_{x \to 2^+} \frac{f(x) - f(2)}{x - 2} \]
\[ = \lim_{x \to 2} \frac{(x - 2) - 0}{x - 2} \]
\[ = \lim_{x \to 2} 1 \]
\[ = 1\]
Thus, 
\[\lim_{x \to 2^-} f(x)\]
\[\lim_{x \to 2^+} f(x)\]
Hence,  
\[f(x)\] is not differentiable at x=2 .
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Chapter 10: Differentiability - Exercise 10.1 [Page 10]

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RD Sharma Mathematics [English] Class 12
Chapter 10 Differentiability
Exercise 10.1 | Q 1 | Page 10

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