Question

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

Answer 1

Hint: If the left-hand derivative of\[f(x)\] is not equal to right-hand derivative of \[f(x)\] at \[x=a\], then the function is said to be not differentiable at \[x=a\].

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

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 .