Question

Discuss the continuity and differentiability of f (x) = e^|x| .

Answer 1

Discuss the continuity and differentiability of f (x) = e^|x| .

Given: 

\[f(x) = e^\left| x \right|\]

⇒f`(x) = {(e^x, ,xge0),(e^(-x), ,x<0):}`

Continuity: 

(LHL at x = 0) 

\[\lim_{x \to 0^-} f(x) \]
\[ = \lim_{h \to 0} f(0 - h) \]
\[ = \lim_{h \to 0} e^{- (0 - h)} \]
\[ = \lim_{h \to 0} e^h \]
\[ = 1\]

(RHL at x = 0) 

\[\lim_{x \to 0^+} f(x) \]
\[ = \lim_{h \to 0} f(0 + h) \]
\[ = \lim_{h \to 0} e^{(0 + h)} \]
\[ = 1\]

\[f(0) = e^0 = 1\]

Thus,  

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

Hence,function is continuous at x = 0 .
Differentiability at x = 0.
(LHD at x = 0)

\[\lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0}\]
\[ = \lim_{h \to 0} \frac{f(0 - h) - f(0)}{0 - h - 0}\]
\[ = \lim_{h \to 0} \frac{e^{- (0 - h)} - 1}{- h}\]
\[ = \lim_{h \to 0} \frac{e^h - 1}{- h} \]
\[ = - 1 \left[ \because \lim_{h \to 0} \frac{e^h - 1}{h} = 1 \right]\]

(RHD at x = 0)

\[\lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0}\]
\[ = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{0 + h - 0}\]
\[ = \lim_{h \to 0} \frac{e^{(0 + h)} - 1}{h}\]
\[ = \lim_{h \to 0} \frac{e^h - 1}{h} \]
\[ = 1 \left[ \because \lim_{h \to 0} \frac{e^h - 1}{h} = 1 \right]\]

LHD at (x = 0)

\[\neq\]RHD at (x = 0)

Hence the function is not differentiable at x = 0.