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  • The function f(x) = e|x| is A. continuous everywhere but not differentiable at x = 0 B. continuous and differentiable everywhere C. not continuous at x = 0 D. none of these

The function f(x) = e^{|x|} is
A. continuous everywhere but not differentiable at x = 0
B. continuous and differentiable everywhere
C. not continuous at x = 0
D. none of these

Answers (1)

A)

Given that, f(x) = e^{|x|}

Let g(x) = |x| and h(x) = e^x

Then, f(x) = hog(x)

We know that, modulus and exponential functions are continuous everywhere.

Since, composition of two continuous functions is a continuous function.

Hence, f(x) = hog(x) is continuous everywhere.

Now, v(x)=|x| is not differentiable at x=0.

\begin{aligned}\\ &\lim _{=h \rightarrow 0} \frac{|0-h|-|0|}{-h} \quad(\because \vee(x)=|x|)\\ &\lim _{h \rightarrow 0} \frac{|-h|}{-h}\\ &\lim _{h \rightarrow 0} \frac{h}{-h}\\ &\lim _{=h \rightarrow 0}-1=-1\\ &\lim _{\mathrm{R} \mathrm{W}^{\prime}(0)}=\lim _{x \rightarrow 0} \frac{\mathrm{v}(\mathrm{x})-\mathrm{v}(0)}{\mathrm{x}-0}\\ &\lim _{h \rightarrow 0} \frac{v(0+h)-v(0)}{(0+h)-0}\\ &\lim _{n \rightarrow 0} \frac{|0+h|-|0|}{h} \quad(\because v(x)=|x|)\\ &=\lim _{h \rightarrow 0} \frac{|h|}{h} \end{aligned}

\lim _{h \rightarrow 0} \frac{h}{h}
\lim _{h \rightarrow 0} 1=1
\\\Rightarrow \mathrm{LX}^{\prime}(0) \neq \mathrm{Bx}^{\prime}(0)
So,\mathrm{e}^{|x|}is not differentiable at x=0.
Hence,  f(x)  continuous everywhere but not differentiable at x=0.

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infoexpert22

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