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Provide solution for RD Sharma maths class 12 chapter 9 Differentiability exercise Fill in the blanks question  2

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Answer: x=\pm 1

Hint: If f is differentiable at all x\in R, we must showf'(x) exists at all x\in R

Given: g(x)=|x-1|+|x+1|

Solution:  we know that

\begin{aligned} g(x) &=\left\{\begin{array}{l} -x-1+1-x, x<-1 \\ x+1+1-x,-1 \leq x<1 \\ x-1+x+1, x \geq 1 \end{array}\right\} \\ &=\left\{\begin{array}{l} -2 x, x<-1 \\ 2,-1 \leq x<1 \\ 2 x, x \geq 1 \end{array}\right\} \end{aligned}

\begin{aligned} &g^{\prime}(x)=\left\{\begin{array}{l} -2, x<-1 \\ 0,-1 \leq x<1 \\ 2, x \geq 1 \end{array}\right\} \\\\ &g^{\prime}\left(-1^{-}\right)=-2, g^{\prime}\left(-1^{+}\right)=0 \end{aligned}

g^{\prime}\left(1^{-}\right)=0, g^{\prime}\left(1^{+}\right)=2

The above derivative calculations can be done from the first principle also.

Since LHS and RHS derivatives are not equal at the points 1 and -1.

 Hence the function is not differentiable at x=\pm 1.


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