The function g(x) = |x – 1| + |x + 1| is not differentiable at x = ____________.
$|x-1|= \begin{cases}x-1, & x \geq 1 \\ -(x-1), & x<1\end{cases}$
$|x+1|= \begin{cases}x+1, & x \geq-1 \\ -(x+1), & x<-1\end{cases}$
$\therefore g(x)=|x-1|+|x+1|= \begin{cases}-(x-1)-(x+1), & x<-1 \\ -(x-1)+x+1, & -1 \leq x<1 \\ x-1+x+1, & x \geq 1\end{cases}$
$\Rightarrow g(x)=|x-1|+|x+1|= \begin{cases}-2 x, & x<-1 \\ 2, & -1 \leq x<1 \\ 2 x, & x \geq 1\end{cases}$
When $x<-1, g(x)=-2 x$ which being a polynomial function is continuous and differentiable.
When $-1 \leq x<1, g(x)=2$ which being a constant function is continuous and differentiable.
When $x \geq 1, g(x)=2 x$ which being a polynomial function is continuous and differentiable.
Thus, the possible points of non-differentiability of $g(x)$ are $x=-1$ and $x=1$.
Now,
$L g^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{g(-1-h)-g(-1)}{-h}$
$\Rightarrow L g^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{-2(-1-h)-2}{-h}$
$\Rightarrow L g^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{2 h}{-h}$
$\Rightarrow L g^{\prime}(-1)=-2$
And
$R g^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{g(-1+h)-g(-1)}{h}$
$\Rightarrow R g^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{2-2}{-h}$
$\Rightarrow R g^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{0}{-h}$
$\Rightarrow R g^{\prime}(-1)=0$
$\therefore L g^{\prime}(-1) \neq R g^{\prime}(-1)$
So, g(x) is not differentiable at x = −1.
Also,
$L g^{\prime}(1)=\lim _{h \rightarrow 0} \frac{g(1-h)-g(1)}{-h}$
$\Rightarrow L g^{\prime}(1)=\lim _{h \rightarrow 0} \frac{2-2 \times 1}{-h}$
$\Rightarrow L g^{\prime}(1)=\lim _{h \rightarrow 0} \frac{0}{-h}$
$\Rightarrow L g^{\prime}(1)=0$
And
$R g^{\prime}(1)=\lim _{h \rightarrow 0} \frac{g(1+h)-g(1)}{h}$
$\Rightarrow R g^{\prime}(1)=\lim _{h \rightarrow 0} \frac{2(1+h)-2 \times 1}{h}$
$\Rightarrow R g^{\prime}(1)=\lim _{h \rightarrow 0} \frac{2 h}{h}$
$\Rightarrow R g^{\prime}(1)=2$
$\therefore L g^{\prime}(1) \neq R g^{\prime}(1)$
So, $g(x)$ is not differentiable at $x=1$.
Thus, the function $g(x)=|x-1|+|x+1|$ is not differentiable at $x=-1$ and $x=1$.
The function g(x) = |x – 1| + |x + 1| is not differentiable at x = ___±1___.