#### Please solve RD Sharma class 12 chapter 9 Differentiability exercise Fill in the blanks question  1 maths textbook solution

Answer: $x=-1$

Hint: If f is differentiable at all $x\in R$, we must show $f^{'}(x)$exists at all $x\in R$

Given: The function $f(x)=|x+1|$

Solution:

$f(x)=|x+1|$

We know that

\begin{aligned} &|x+1|=\left\{\begin{array}{l} x+1, x+1 \geq 0 \\ -(x+1), x+1<0 \end{array}\right\} \\\\ &f(x)=\left\{\begin{array}{l} x+1, x \geq-1 \\ -x-1, x<-1 \end{array}\right\} \end{aligned}

\begin{aligned} &\lim _{x \rightarrow-1^{+}}(x+1)=-1+1=0 \\\\ &\lim _{x \rightarrow-1^{-}}(-x-1)=1-1=0 \end{aligned}

Thus, the function is continuous.

\begin{aligned} &x \geq-1, x<-1 \\\\ &f(x)=x+1, f(x)=-x-1 \\\\ &f^{\prime}(x)=1, f^{\prime}(x)=-1 \end{aligned}

\begin{aligned} &f^{\prime}\left(-1^{+}\right)=1, f^{\prime}\left(-1^{-}\right)=-1 \\\\ &f^{\prime}\left(-1^{+}\right)=1 \neq f^{\prime}\left(-1^{-}\right)=-1 \end{aligned}

Therefore, the function $f(x)=|x+1|$ is not differentiable at $x=-1$