Question

Let \[f\left( x \right) = \begin{cases}1 , & x \leq - 1 \\ \left| x \right|, & - 1 < x < 1 \\ 0 , & x \geq 1\end{cases}\] Then, f is 

पर्याय

• continuous at x = − 1

• differentiable at x = − 1

• everywhere continuous

• everywhere differentiable

Answer 1

(b) differentiable at x = − 1 

\[f\left( x \right) = \begin{cases}1 , & x \leq - 1 \\ \left| x \right|, & - 1 < x < 1 \\ 0 , & x \geq 1\end{cases}\]

Differentiabilty at x = − 1

(LHD x = − 1)

\[{lim}_{x \to - 1^-} \frac{f(x) - f( - 1)}{x + 1}\]
\[ = {lim}_{x \to - 1} \frac{f(x) - f( - 1)}{x + 1}\]
\[ = {lim}_{x \to - 1} \frac{1 - 1}{- 1 + 1}\]
\[ = 0\]

(RHD x = − 1)

\[= {lim}_{x \to - 1^+} \frac{f (x) - f( - 1)}{x + 1} \]
\[ = {lim}_{x \to - 1} \frac{f (x) - f( - 1)}{x + 1} \]
\[ = {lim}_{x \to - 1} \frac{f (x) - f( - 1)}{x + 1}\]
\[ = {lim}_{x \to - 1} \frac{|x| - | - 1|}{x + 1}\]
\[ = \frac{1 - 1|}{- 1 + 1}\]
\[ = 0 \]