Question

Show that $f\left(x\right)=|3x+2|$ is differentiable at $x=2/3$.

Answer 1

We have,

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

L.H.D.

${lim}_{x\to \frac{2}{3}-}f\left(x\right)={lim}_{h\to 0}\frac{f\left(x\right)-f(x-h)}{h}$

$={lim}_{h\to 0}\frac{f\left(\frac{2}{3}\right)-f(\frac{2}{3}-h)}{h}$

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

$={lim}_{h\to 0}\frac{\left|4\right|-|(2-3h)+2|}{h}$

$={lim}_{h\to 0}\frac{4-|4-3h|}{h}$

$={lim}_{h\to 0}\frac{3h}{h}$

$=3$

R.H.D.

${lim}_{x\to \frac{2}{3}+}f\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

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

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

$={lim}_{h\to 0}\frac{|(2+3h)+2|-4}{h}$

$={lim}_{h\to 0}\frac{|4+3h|-4}{h}$

$={lim}_{h\to 0}\frac{3h}{h}$

$=3$

$L.H.D.=R.H.D.$

Therefore, differentiable at $\frac{2}{3}$

Hence, this is the answer.