Question

Let f(x)=\left\\{ \begin{aligned} & -x\text{ for }x\le 0 \\\ & {{\text{x}}^{2}}\text{ for } 0 < x\le 1 \\\ & {{x}^{3}}\text{-x+1 for } x > 1 \\\ \end{aligned} \right.
The number of points at which $f$ is continuous but not differentiable is:

Answer 1

If the left derivative of a function at a point is not equal to the right derivative of the function at the same point , then the function is said to be not differentiable at that point.

Complete step-by-step answer:
We will consider the following cases to check continuity and differentiability of the function.
Case 1: $x<0$
At $x<0$ , $f\left( x \right)=-x$ , i.e. it is a polynomial. Hence, it is continuous as well as differentiable.
Case 2: $x=0$
At $x=0$ ,
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( -x \right)=0$
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{x}^{2}}=0$
$f\left( 0 \right)=0$
$\because \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)={{\left[ f\left( x \right) \right]}_{x=0}}$
Hence, function is continuous at $x=0$ .
Now, we will check the differentiability of the function at $x=0$ .
The right-hand derivative of $f\left( x \right)$ at $x=0$ is given by:
$f_{+}^{'}\left( 0 \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 0+h \right)-f\left( 0 \right)}{h}$
$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( h \right)-\left( 0 \right)}{h}$
$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{h}^{2}}+0}{h}=0$
Now, the left-hand derivative of the function at $x=0$ is given by
$f_{-}^{'}\left( 0 \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 0-h \right)-f\left( 0 \right)}{-h}$
$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( -h \right)-\left( -0 \right)}{-h}$
$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{h+0}{-h}=-1$
$\because f_{+}^{'}\left( 0 \right)\ne f_{-}^{'}\left( 0 \right)$
Hence, $f\left( x \right)$ is not differentiable at $x=0$.
Case 3: $0\le x\le 1$
At $0\le x\le 1$ , $f\left( x \right)={{x}^{2}}$ i.e. it is a polynomial .
Hence, it is continuous as well as differentiable .
Case 4: $x=1$
We will check the continuity of the function at $x=1$ .
$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,{{x}^{2}}$
$\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,{{x}^{3}}-x+1$
$=1-1+1$
$=1$
Also, $f\left( 1 \right)={{\left( 1 \right)}^{2}}=1$ .
$\because \underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)={{\left[ f\left( x \right) \right]}_{x=1}}$
Hence, the function is continuous at $x=1$ .
Now, we will check the differentiability of $f\left( x \right)$ at $x=1$ .
The right-hand derivative of the function is given by:
$f_{+}^{'}\left( 1 \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+h \right)-f\left( 1 \right)}{h}$
$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{(1+h)}^{3}}-(1+h)+1-{{\left( 1 \right)}^{2}}}{h}$
$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{h}^{3}}+1+3{{h}^{2}}+3h-1-h+1-1}{h}$
$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{h}^{3}}+3{{h}^{2}}+2h}{h}=2$
The left-hand derivative of $f\left( x \right)$ at $x=1$ is given by:
$f_{-}^{'}\left( 1 \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1-h \right)-f\left( 1 \right)}{-h}$
$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\left( 1-h \right)}^{2}}-\left( 1 \right)}{-h}$
$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{h}^{2}}-2h+1-1}{-h}$
$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{h}^{2}}-2h}{h}$
$\underset{h\to 0}{\mathop{\lim }}\,h-2$
$=-2$
Since $f_{+}^{'}\left( 1 \right)\ne f_{-}^{'}\left( 1 \right)$ , hence, the function is not differentiable at $x=1$ .
Case 5: $x>1$
For $x>1$ , $f\left( x \right)={{x}^{3}}-x+1$ which is a polynomial. Hence, it is continuous as well as differentiable.
So, the number of points at which the function is continuous but not differentiable is $2$ i.e. at $x=0$ and at $x=1$ .

Note: A polynomial function is continuous as well as differentiable at all points as every polynomial function consists of some combination of basic mathematical operations, like addition, multiplication, etc. of constant function and identity function, i.e. $f(x)=x$ . Now, every constant and identity function is differentiable and continuous everywhere. So, a polynomial function is also differentiable and continuous everywhere.