Find the number of points where the given function f(x)=|x-1| |x-2| |x-3| sin πx is
1. Continous
2. Discontinuous
3. Derivable
4. Not Derivable

Dear student
{Using the result that f(x) g(x) is continuos if f(x) and g(x) are continuos.}

Since |x-1|, |x-2|, |x-3| and Sin $\pi x$ are all continuos for all real x, therefore f(x) = |x-1||x-2||x-3| Sin $\pi x$  will be continuous
f(x) is not discontinuous for any value of x.

If f(x) and g(x) are derivable at x=a then f(x) g(x) are derivable at x=a
Sin $\mathrm{\pi x}$ is differentiable for all real x, however |x-1|, |x-2| and |x-3| are not differentiable at x=1, x=2 and x=3 respectively.

These are the points at which we need to check continuity.
At x=1
Right hand derivative-

$$\begin{gathered} \underset{h\to 0^{+}}{\lim } \frac{f(1+h)-f\left(1\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{h(1-h)(2-h)Sin (\pi +\pi h)}{h} \\ =\underset{h\to 0^{+}}{\lim }-h(1-h)(2-h) \times \underset{h\to 0^{+}}{\lim } \frac{Sin \pi h}{\pi h} \times \pi \\ =\underset{h\to 0^{+}}{\lim }-h(1-h)(2-h) \times \pi \\ =0 \end{gathered}$$

Left hand derivative

$$\begin{gathered} \underset{h\to 0^{-}}{\lim } \frac{f(1-h)-f\left(1\right)}{-h}=\underset{h\to 0^{-}}{\lim }\frac{h(1+h)(2+h)Sin (\pi -\pi h)}{-h} \\ =\underset{h\to 0^{-}}{\lim }-h(1+h)(2+h) \times \underset{h\to 0^{-}}{\lim } \frac{Sin \pi h}{\pi h} \times \pi \\ =\underset{h\to 0^{-}}{\lim }-h(1+h)(2+h) \times \pi \\ =0 \end{gathered}$$
Left hand derivative=right hand derivative

therefore f(x) is differentiable at x=1


At x=2
Right hand derivative-

$$\begin{gathered} \underset{h\to 0^{+}}{\lim } \frac{f(2+h)-f\left(2\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{(1+h)h(1-h)Sin (\pi +\pi h)}{h} \\ =\underset{h\to 0^{+}}{\lim }-(1+h)h(1-h) \times \underset{h\to 0^{+}}{\lim } \frac{Sin \pi h}{\pi h} \times \pi \\ =\underset{h\to 0^{+}}{\lim }-(1+h)h(1-h) \times \pi \\ =0 \end{gathered}$$

Left hand derivative

$$\begin{gathered} \underset{h\to 0^{-}}{\lim } \frac{f(2-h)-f\left(2\right)}{-h}=\underset{h\to 0^{-}}{\lim }\frac{(1-h)\left(h\right)(h+1)Sin (2\pi -\pi h)}{-h} \\ =\underset{h\to 0^{-}}{\lim }h(1+h)(1-h) \times \underset{h\to 0^{-}}{\lim } \frac{Sin \pi h}{\pi h} \times \pi \\ =\underset{h\to 0^{-}}{\lim }h(1-h)(1+h) \times \pi \\ =0 \end{gathered}$$
Left hand derivative=right hand derivative

therefore f(x) is differentiable at x=2


At x=3
Right hand derivative-

$$\begin{gathered} \underset{h\to 0^{+}}{\lim } \frac{f(3+h)-f\left(3\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{(2+h)(1+h)\left(h\right)Sin (3\pi +\pi h)}{h} \\ =\underset{h\to 0^{+}}{\lim }-(2+h)(1+h)\left(h\right) \times \underset{h\to 0^{+}}{\lim } \frac{Sin \pi h}{\pi h} \times \pi \\ =\underset{h\to 0^{+}}{\lim }-(2+h)(1+h)\left(h\right) \times \pi \\ =0 \end{gathered}$$

Left hand derivative

$$\begin{gathered} \underset{h\to 0^{-}}{\lim } \frac{f(3-h)-f\left(3\right)}{-h}=\underset{h\to 0^{-}}{\lim }\frac{(2-h)(1-h)\left(h\right)Sin (3\pi -\pi h)}{-h} \\ =\underset{h\to 0^{-}}{\lim }-(2-h)(1-h)\left(h\right) \times \underset{h\to 0^{-}}{\lim } \frac{Sin \pi h}{\pi h} \times \pi \\ =\underset{h\to 0^{-}}{\lim }-(2-h)(1-h)\left(h\right) \times \pi \\ =0 \end{gathered}$$
Left hand derivative=right hand derivative

therefore f(x) is differentiable at x=3


Hence f(x) is derivable at all x