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Question:
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Consider the following statements:
1- The function $f(x)=|x|$ is not differentiable at $x=1$.
2- The function $f(x)=e^{x}$ is not differentiable at $x=0$. Which of the above statements is/are correct?
Options:
1- The function $f(x)=|x|$ is not differentiable at $x=1$.
2- The function $f(x)=e^{x}$ is not differentiable at $x=0$. Which of the above statements is/are correct?
Solution:
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Verified Answer
The correct answer is:
2 only
Statement $1: \mathrm{f}(\mathrm{x})=\mid \mathrm{x}$
From the graph, the curve has sharp turn at $\mathrm{x}=0$. Therefore, the function $\mathrm{f}(\mathrm{x})=|\mathrm{x}|$ is not differentiable only $\mathrm{x}=0$, it is differentiable at $\mathrm{x}=1$ Statement $2: \mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}$
$\mathrm{Rf}^{\prime}(0)=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0+\mathrm{h})-\mathrm{f}(0)}{\mathrm{h}}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{e}^{(0+\mathrm{h})}-\mathrm{e}^{\circ}}{\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{e}^{\mathrm{h}}-1}{\mathrm{~h}}$
Use L'Hospital rule
$=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{e}^{\mathrm{h}}-0}{1}=\mathrm{e}^{\mathrm{o}}=1$
$f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{1-\mathrm{e}^{-\mathrm{h}}}{\mathrm{h}}$
Use L' Hospital rule
$=\lim _{h \rightarrow 0} \frac{\mathrm{e}^{-\mathrm{h}}}{1}=\mathrm{e}^{-0}=1$
Therefore $\mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}$ is differentiable at $\mathrm{x}=1$.
From the graph, the curve has sharp turn at $\mathrm{x}=0$. Therefore, the function $\mathrm{f}(\mathrm{x})=|\mathrm{x}|$ is not differentiable only $\mathrm{x}=0$, it is differentiable at $\mathrm{x}=1$ Statement $2: \mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}$
$\mathrm{Rf}^{\prime}(0)=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0+\mathrm{h})-\mathrm{f}(0)}{\mathrm{h}}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{e}^{(0+\mathrm{h})}-\mathrm{e}^{\circ}}{\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{e}^{\mathrm{h}}-1}{\mathrm{~h}}$
Use L'Hospital rule
$=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{e}^{\mathrm{h}}-0}{1}=\mathrm{e}^{\mathrm{o}}=1$
$f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{1-\mathrm{e}^{-\mathrm{h}}}{\mathrm{h}}$
Use L' Hospital rule
$=\lim _{h \rightarrow 0} \frac{\mathrm{e}^{-\mathrm{h}}}{1}=\mathrm{e}^{-0}=1$
Therefore $\mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}$ is differentiable at $\mathrm{x}=1$.
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