Let $f\left(x\right)=\left\{\begin{array}{cc}\frac{\sin \left|x^2-5x+6\right|}{x^2-5x+6}& x\ne 2,3\\ 1& x=2 or 3\end{array}\right.$

the set of all points where$f$ is differentiable is

 

Solution:

The function is clearly differentiable except 

possibly at $x=2,3$

$\begin{array}{r}{f}^{\mathrm{\prime }}(2+)=\underset{h\to 0+}{lim} \frac{f(2+h)-f\left(2\right)}{h}\\ =\underset{h\to 0+}{lim} \frac{\sin h(1-h)+h(1-h)}{h^2(-1+h)}\\ =-\underset{h\to 0+}{lim} \left(\frac{\sin h(1-h)}{h(1-h)}+1\right)\frac{1}{h}\end{array}$

The last limit doesn’t exist. If this limit then $\underset{h\to 0+}{lim} \frac{1}{h}$ exist, 

which is not true

Thus f is not differentiable at $x=2$ . Similarly f is not dif-

ferentiable   at $x = 3$. Hence the set of all points where f is 

$(-\mathrm{\infty },\mathrm{\infty })~\{2,3\}$