Question

The function $$f\left(x\right)=\{\begin{array}{rl}& |x-3|,x\ge 1\\ & {\displaystyle \frac{x^2}{4}}-{\displaystyle \frac{3x}{2}}+{\displaystyle \frac{13}{4}},x<1\end{array}$$ is
(a) continuous at $$x=1$$
(b) differentiable at $$x=1$$
(c) continuous at $$x=3$$
(d) differentiable at $$x=3$$

Answer 1

Hint: If the value of limit of the function at a point $$x=a$$ is equal to the value of the function at $$x=a$$ , the function is said to be continuous at $$x=a$$. A function is differentiable at $$x=a$$ , if the left-hand derivative of the function is equal to the right hand derivative of the function at $$x=a$$.

Complete step-by-step answer:
At $$x=3$$, $$f\left(x\right)=|x-3|$$.
We know $$|x-a|$$ is a function that is continuous everywhere but it is not differentiable at vertex i.e. at $$x=a$$.
So , $$f\left(x\right)$$is continuous at $$x=3$$ but not differentiable at $$x=3$$.
Now , we will check if the function is continuous or differentiable at critical points of the function .
A function is differentiable at $$x=a$$ , if the left-hand derivative of the function is equal to the right hand derivative of the function at $$x=a$$.
We know , the left hand derivative of $$f\left(x\right)$$ at $$x=a$$ is given as $$L^{{}^{\prime }}=\underset{h\to 0^{-}}{lim}{\displaystyle \frac{f(a-h)-f\left(a\right)}{-h}}$$ and the right hand derivative of $$f\left(x\right)$$at $$x=a$$ is given as $$R^{{}^{\prime }}=\underset{h\to 0^{+}}{lim}{\displaystyle \frac{f(a+h)-f\left(a\right)}{h}}$$.
We will consider the critical point $$x=1$$. To determine differentiability of $$f\left(x\right)$$ at $$x=1$$, we will evaluate its left-hand derivative.
$$L^{{}^{\prime }}=\underset{h\to 0^{-}}{lim}{\displaystyle \frac{f(1-h)-f\left(1\right)}{-h}}$$
$$=\underset{h\to 0^{-}}{lim}\left[{\displaystyle \frac{({\displaystyle \frac{{(1-h)}^2}{4}}-{\displaystyle \frac{3(1-h)}{2}}+{\displaystyle \frac{13}{4}})-({\displaystyle \frac{1^2}{4}}-{\displaystyle \frac{3\left(1\right)}{2}}+{\displaystyle \frac{13}{4}})}{-h}}\right]$$
$$=\underset{h\to 0^{-}}{lim}\left[{\displaystyle \frac{{\displaystyle \frac{h^2-2h+1}{4}}-{\displaystyle \frac{3-3h}{2}}+{\displaystyle \frac{13}{4}}-{\displaystyle \frac{1}{4}}+{\displaystyle \frac{3}{2}}-{\displaystyle \frac{13}{4}}}{-h}}\right]$$
$$=\underset{h\to 0^{-}}{lim}\left[{\displaystyle \frac{h^2+4h}{-4h}}\right]$$
$$=\underset{h\to 0^{-}}{lim}\left[{\displaystyle \frac{h+4}{-4}}\right]$$
$$=-1$$
Now , we will evaluate the right-hand derivative of $$f\left(x\right)$$ at $$x=1$$
$$R^{{}^{\prime }}=\underset{h\to 0^{+}}{lim}{\displaystyle \frac{f(1+h)-f\left(1\right)}{h}}$$
$$R^{{}^{\prime }}=\underset{h\to 0^{+}}{lim}{\displaystyle \frac{|1+h-3|-|1-3|}{h}}$$
Now, we know$$1+h<3$$,
So $$|1+h-3|=|h-2|=2-h$$
So, $$R^{{}^{\prime }}=\underset{h\to 0^{+}}{lim}{\displaystyle \frac{2-h-2}{h}}$$
$$=-1$$
Clearly, the left hand derivative is equal to the right hand derivative.
Hence $$f\left(x\right)$$ is differentiable at $$x=1$$.
Now, we also know that if a function is differentiable at a point then it must be continuous at that point.
So, $$f\left(x\right)$$ is continuous at $$x=1$$.
Hence , the function is differentiable and continuous at $$x=1$$ and the function is continuous but not differentiable at $$x=3$$.
Answer is (a) , (b) , (c)

Note: A function which is differentiable at a point is necessarily continuous at that point. But a function which is continuous at a point, may or may not be differentiable at that point.