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Question: Let \[f(x)=\left\\{ \begin{aligned} & {{x}^{2}}\text{ for }x\le 0 \\\ & 1\text{ for } 0 < x\...

Let f(x)=\left\\{ \begin{aligned} & {{x}^{2}}\text{ for }x\le 0 \\\ & 1\text{ for } 0 < x\le 1 \\\ & \dfrac{1}{x}\text{ for } x > 1 \\\ \end{aligned} \right.
The number of points at which ff is not differentiable is

Explanation

Solution

If the left-hand derivative off(x)f(x) is not equal to right-hand derivative of f(x)f(x) at x=ax=a, then the function is said to be not differentiable at x=ax=a.

Complete step-by-step answer:
The given function is f(x)=\left\\{ \begin{aligned} & {{x}^{2}}\text{ for }x\le 0 \\\ & 1\text{ for } 0 < x\le 1 \\\ & \dfrac{1}{x}\text{ for } x > 1 \\\ \end{aligned} \right.
For a function to be differentiable at x=ax=a , the right hand derivative of the function at x=ax=ashould be equal to the left hand derivative at x=ax=a.
Now ,we will check the differentiability of the function at critical points i.e. at the point x=0x=0 and at the point x=1x=1.
We know , the left-hand derivative of a function f(x)f(x) at x=ax=a is given as f−′(a)=lim⁡h→0 f(a−h)−f(a)−hf_{-}^{'}\left( a \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( a-h \right)-f\left( a \right)}{-h}
and the right-hand derivative of a function f(x)f(x) at x=ax=a is given as f+′(a)=lim⁡h→0 f(a+h)−f(a)hf_{+}^{'}\left( a \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( a+h \right)-f\left( a \right)}{h}.
At x=0x=0, the left-hand derivative of the function f(x)f(x) is given as
f−′(0)=lim⁡h→0 f(0−h)−f(0)−hf_{-}^{'}\left( 0 \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 0-h \right)-f\left( 0 \right)}{-h}
=lim⁡h→0 f(−h)−f(0)−h=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( -h \right)-f(0)}{-h}
=lim⁡h→0 (−h)2−0−h=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{(-h)}^{2}}-0}{-h}
=lim⁡h→0 (−h)=0=\underset{h\to 0}{\mathop{\lim }}\,(-h)=0
And , the right- hand derivative of the function f(x)f(x) is given as
f+′(0)=lim⁡h→0 f(0+h)−f(0)hf_{+}^{'}\left( 0 \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 0+h \right)-f\left( 0 \right)}{h}
=lim⁡h→0 f(h)−(0)h=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( h \right)-\left( 0 \right)}{h}
=lim⁡h→0+ (1−0h)=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\left( \dfrac{1-0}{h} \right)
=lim⁡h→0+ (1h)=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\left( \dfrac{1}{h} \right)
=∞=\infty
Now , we can clearly see that the left hand derivative of the function at x=0x=0 is not equal to the right hand derivative of the function at x=0x=0.
Hence , function is not differentiable at x=0x=0.
Now, we will check differentiability of the function at x=1x=1.
Atx=1x=1, the left-hand derivative of the function f(x)f(x) is given as
f−′(1)=lim⁡h→0 f(1−h)−f(1)−hf_{-}^{'}\left( 1 \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1-h \right)-f\left( 1 \right)}{-h}
=lim⁡h→0 1−1−h=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1-1}{-h}
=0=0
Now , the right-hand derivative of the function f(x)f(x) is given as
f+′(1)=lim⁡h→0 f(1+h)−f(1)hf_{+}^{'}\left( 1 \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( 1+h \right)-f\left( 1 \right)}{h}
=lim⁡h→0 1h−1h=\underset{h\to 0}{\mathop{\lim }}\,{{\dfrac{\dfrac{1}{h}-1}{h}}_{{}}}
=lim⁡h→0 1−hh2=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1-h}{{{h}^{2}}}
=1−00=∞=\dfrac{1-0}{0}=\infty
Now , we can clearly see that the left-hand derivative of the function at x=1x=1 is not equal to the right hand derivative of the function at x=1x=1.
Hence , function is not differentiable at x=1x=1.
Hence the number of points at which f(x)f(x) is not differentiable =2=2i.e. at x=0x=0 and at x=1x=1.

Note: A function is said to be differentiable at a point if the limit exists at the point and the function is continuous. Also , the left-hand derivative of the function at the point should be equal to the right-hand derivative of the function at the same point .