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A function $f: R \rightarrow R$ is defined as $f(x)=x^{2}$ for $x \geq 0$ and $f(x)=-x$ for $x < 0 .$
Consider the following statements in respect of the above function:
1- The function is continuous at $x=0$.
2- The function is differentiable at $x=0$.
Which of the above statements is/are correct?
Options:
Consider the following statements in respect of the above function:
1- The function is continuous at $x=0$.
2- The function is differentiable at $x=0$.
Which of the above statements is/are correct?
Solution:
1773 Upvotes
Verified Answer
The correct answer is:
1 only
$\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}\mathrm{x}^{2} & , \quad \mathrm{x} \geq 0 \\ -\mathrm{x} & , \quad \mathrm{x} < 0\end{array}\right.$
For continuity at $\mathrm{x}=0$ $\mathrm{f}(0-0)=\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(0-\mathrm{h})$
$=\lim _{\mathrm{h} \rightarrow 0}[(0-\mathrm{h})]=\lim _{\mathrm{h} \rightarrow 0} \mathrm{~h}=0$
$\mathrm{f}(0+0)=\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(0+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}(0+\mathrm{h})^{2}=0$ and $\mathrm{f}(0)=0$
$\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$
For differentiability at $\mathrm{x}=0$ $\lim _{h \rightarrow 0} \frac{-(-h)-0}{-h}=\lim _{h \rightarrow 0}=\frac{h}{-h}=-1$
and $\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} h=0$
$\mathrm{f}(\mathrm{x})$ is not differentiable at $\mathrm{x}-0$
For continuity at $\mathrm{x}=0$ $\mathrm{f}(0-0)=\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(0-\mathrm{h})$
$=\lim _{\mathrm{h} \rightarrow 0}[(0-\mathrm{h})]=\lim _{\mathrm{h} \rightarrow 0} \mathrm{~h}=0$
$\mathrm{f}(0+0)=\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(0+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}(0+\mathrm{h})^{2}=0$ and $\mathrm{f}(0)=0$
$\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$
For differentiability at $\mathrm{x}=0$ $\lim _{h \rightarrow 0} \frac{-(-h)-0}{-h}=\lim _{h \rightarrow 0}=\frac{h}{-h}=-1$
and $\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} h=0$
$\mathrm{f}(\mathrm{x})$ is not differentiable at $\mathrm{x}-0$
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