Question

Prove that $f\left(x\right)=|x-1|+|x-1|+|x-3|$ is continuous on R but not differentiable at $x=1,2$ and $3$ only.

Answer 1

$f\left(x\right)=|x-1|+|x-2|+|x-3|$

Also, $f\left(a^{-}\right)={lim}_{x\to a^{-}}f\left(x\right)$ and $f\left(a^{+}\right)={lim}_{x\to a^{+}}f\left(x\right)$

$f\left(1^{-}\right)=-(x-1)+1+2=3$, $f\left(1^{+}\right)=(x-1)+1+2=3$ Hence, continuous at $x=1$

$f\left(2^{-}\right)=1-(x-2)+1=2$, $f\left(2^{+}\right)=1+(x-2)+1=2$ Hence, continuous at $x=2$

$f\left(3^{-}\right)=2+1-(x-3)=3$, $f\left(3^{+}\right)=2+1+(x-3)=3$ Hence, continuous at $x=3$

Hence, the function is continuous on $R$ as these 3 points were the ones of change.

But, apart from continuity differentiability also needs to be checked.

At $x=1$,

$L{f}^{\prime }\left(1\right)={lim}_{h\to 0^{-}}\frac{f(1+h)-f\left(1\right)}{h}={lim}_{h\to 0^{-}}\frac{|1+h-1|+1+2-3}{h}={lim}_{h\to 0^{-}}\frac{-h}{h}=-1$

$R{f}^{\prime }\left(1\right)={lim}_{h\to 0^{+}}\frac{f(1+h)-f\left(1\right)}{h}={lim}_{h\to 0^{+}}\frac{|1+h-1|+1+2-3}{h}={lim}_{h\to 0^{+}}\frac{h}{h}=1$

As $L{f}^{\prime }\left(1\right)\ne R{f}^{\prime }\left(1\right)$, function is non-derivable at $x=1$

At $x=2$,

$L{f}^{\prime }\left(2\right)={lim}_{h\to 0^{-}}\frac{f(2+h)-f\left(2\right)}{h}={lim}_{h\to 0^{-}}\frac{1+|2+h-2|+1-2}{h}={lim}_{h\to 0^{-}}\frac{-h}{h}=-1$

$R{f}^{\prime }\left(2\right)={lim}_{h\to 0^{+}}\frac{f(2+h)-f\left(2\right)}{h}={lim}_{h\to 0^{+}}\frac{1+|2+h-2|+1-2}{h}={lim}_{h\to 0^{+}}\frac{h}{h}=1$

As $L{f}^{\prime }\left(2\right)\ne R{f}^{\prime }\left(2\right)$, function is non-derivable at $x=2$

At $x=3$,

$L{f}^{\prime }\left(3\right)={lim}_{h\to 0^{-}}\frac{f(3+h)-f\left(3\right)}{h}={lim}_{h\to 0^{-}}\frac{|3+h-3|+1+2-3}{h}={lim}_{h\to 0^{-}}\frac{-h}{h}=-1$

$R{f}^{\prime }\left(3\right)={lim}_{h\to 0^{+}}\frac{f(3+h)-f\left(3\right)}{h}={lim}_{h\to 0^{+}}\frac{|3+h-3|+1+2-3}{h}={lim}_{h\to 0^{+}}\frac{h}{h}=1$

As $L{f}^{\prime }\left(3\right)\ne R{f}^{\prime }\left(3\right)$, function is non-derivable at $x=3$

At $x=k$ where $k\ne \{1,2,\}$,

$L{f}^{\prime }\left(k\right)={lim}_{h\to 0^{-}}\frac{f(k+h)-f\left(k\right)}{h}$

$={lim}_{h\to 0^{-}}\frac{|k+h-1|+|k+h-2|+|k+h-3|-(|k-1|+|k-2|+|k-3|)}{h}={lim}_{h\to 0^{-}}\frac{0}{h}=0$

$R{f}^{\prime }\left(1\right)={lim}_{h\to 0^{+}}\frac{f(k+h)-f\left(k\right)}{h}$

$={lim}_{h\to 0^{+}}\frac{|k+h-1|+|k+h-2|+|k+h-3|-(|k-1|+|k-2|+|k-3|)}{h}={lim}_{h\to 0^{+}}\frac{0}{h}=0$

As $L{f}^{\prime }\left(k\right)=R{f}^{\prime }\left(k\right)$, function is derivable at $x=k$.

Hence, proved.