Discuss whether the function f(x) = |x + 1| + |x – 1| is differen… - Vidyadip

Question

Discuss whether the function f(x) = |x + 1| + |x – 1| is differentiable ∀ x ∈ R.

✓

Answer

$
\begin{array}{rlrl}
f (x) & =|x+1|+|x-1| & \\
& =(1-x)-(1+x), & & x<-1 \\
& =1+x+1-x, & & -1 \leq x<1 \\
& =x+1+x-1, & & x \geq 1 \\
\text { i.e., } & f (x)=-2 x, & & x<-1 \\
& =2, & & -1 \leq x<1 \\
& =2 x, & & x \geq 1
\end{array}
$
Differentiability at $x=-1$ :
$
\begin{aligned}
\text { Lf }^{\prime}(-1) & =\lim _{h \rightarrow 0^{-}} \frac{f(-1+h)-f(-1)}{h} \\
& =\lim _{h \rightarrow 0^{-}} \frac{-2(-1+h)-(2)}{h} \\
& =\lim _{h \rightarrow 0^{-}}\left(\frac{-2 h}{h}\right)_ \quad \ldots[\because h \rightarrow 0, \therefore h \neq 0] \\
& =-2 \quad \cdots h(-2) \\
R^{\prime}(-1) & =\lim _{h \rightarrow 0^{+}} \frac{f(-1+h)-f(-1)}{h} \\
& =\lim _{h \rightarrow 0^{+}} \frac{2-2}{h}=0
\end{aligned}
$
Here, Lf ' $(-1) \neq Rf ^{\prime}(-1)$
$f$ is not differentiable at $x=-1$.
Differentiability at $x=1$ :
$
\begin{aligned}
L f^{\prime}(1)= & \lim _{h \rightarrow 0^{-}} \frac{f(1+h)-f(1)}{h} \\
& =\lim _{h \rightarrow 0^{-}} \frac{2-2}{h}=0
\end{aligned}
$
$
\begin{aligned}
R^{\prime}(1) & =\lim _{h \rightarrow 0^{+}} \frac{f(1+h)-f(1)}{h} \\
& =\lim _{h \rightarrow 0^{+}} \frac{2(1+h)-(2)}{h} \\
& =\lim _{h \rightarrow 0^{-}}\left(\frac{2 h}{h}\right)\\
& =2 \quad \ldots[\because h \rightarrow 0, \therefore h \neq 0]
\end{aligned}
$
Here, $L f^{\prime}(1) \neq R f^{\prime}(1)$
$\therefore f$ is not differentiable at $x=1$.
$\therefore f$ is not differentiable at $x=-1$ and $x=1$.
$\therefore f$ is not differentiable $\forall x \in R$.

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