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  • At the point x = 1, the function                              <img alt="\mathrm{ f(x)= \begin{cases}x^3-1 ; & 1

At the point x = 1, the function
                             \mathrm{ f(x)= \begin{cases}x^3-1 ; & 1<x<\infty \\ x-1 ; & -\infty<x \leq 1\end{cases} }

Option: 1

continuous and differentiable.


Option: 2

continuous and not differentiable.


Option: 3

discontinuous and differentiable.


Option: 4

discontinuous and not differentiable.


Answers (1)

best_answer

We have, \mathrm{\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}(x-1)=0}

\mathrm{\text { and, } \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}\left(x^3-1\right)=0 \text {. Also } f(1)=1-1=0 \text {. }}

So, f(x) is continuous at x=1. Clearly Lf '(1)=2 and Rf '(1)= 3. Therefore, f(x) is not differentiable at x=1.

 

Posted by

Divya Prakash Singh

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