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  • Consider the function in the interval <img alt="\mathrm{-1< x<1}" src="

Consider the function \mathrm{f(x)=|x|} in the interval \mathrm{-1< x<1}. At the point \mathrm{x=0, f(x)} is

Option: 1

continuous and differentiable.
 


Option: 2

non-continuous and differentiable.
 


Option: 3

continuous and non-differentiable.
 


Option: 4

neither continuous nor differentiable.


Answers (1)

best_answer

\mathrm{f(x)=|x|}

\mathrm{\Rightarrow}                      \mathrm{f(x)= \begin{cases}x & x \geq 0 \\ -x & x<0\end{cases}}

For continuous

\mathrm{\text { at } x \rightarrow 0^{+}}

                      \mathrm{\lim _{x \rightarrow 0^{+}} f(x)=0}

\mathrm{\text { at } x \rightarrow 0^{-}}

                      \mathrm{\lim _{x \rightarrow 0^{-}} f(x)=0}

\because             \mathrm{\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)}

                                    \mathrm{=f(0)=0}

\therefore The function is continuous.
For differentiability
\mathrm{at \, \, \mathrm{x} \rightarrow \mathrm{O}^{-}}
                     \mathrm{ \left.\frac{d f}{d x}\right|_{x=0^{-}}=\frac{d}{d x}(-x)=-1 }
\mathrm{\text { at } x \rightarrow 0^{+}}

                    \mathrm{\left.\frac{\mathrm{df}}{\mathrm{dx}}\right|_{\mathrm{x}=0^{+}}=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})=1}
\mathrm{\because} The given function is not differentiable.

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Riya

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