Question

Find whether the following function is differentiable at \[x = 1,2\] .
 \[f\left( x \right) = \left\{
  x \\
  2 - x \\
   - 2 + 3x - {x^2} \\
  \right.\] \[
  , \\
  , \\
  , \\
 \] \[
  x \leqslant 1 \\
  1 \leqslant x \leqslant 2 \\
  x > 2 \\
 \]

Answer 1

Hint: First of all, the given function is continuous for all \[x < 1\] , \[1 < x < 2\] and \[x > 2\] , because it is a polynomial function. Now, for a function to be differentiable at any value of x, the L.H.D. (Left Hand side Derivative) must be equal to the R.H.D. (Right Hand side Derivative). So, find the L.H.D. and R.H.D. individually at the values of \[x = 1,2\] , and see whether L.H.D = R.H.D.

Complete step-by-step answer:
We have
 \[f\left( x \right) = \left\{
  x \\
  2 - x \\
   - 2 + 3x - {x^2} \\
  \right.\] \[
  , \\
  , \\
  , \\
 \] \[
  x \leqslant 1 \\
  1 \leqslant x \leqslant 2 \\
  x > 2 \\
 \]
It is clear that, given function is continuous for all \[x < 1\] , \[1 < x < 2\] and \[x > 2\] , because it is a polynomial function. So, the points where the function may be non-differentiable are at the values of \[x = 1\] and \[x = 2\] .
For a function to be differentiable at any value of \[x\], the Left Hand side Derivative (L.H.D.) must be equal to the Right Hand side Derivative (R.H.D.). So, we will check L.H.D. and R.H.D. individually at the values of \[x = 1\] and \[x = 2\] .
Now,
 \[f\left( 1 \right) = 2 - 1 = 1\] (at \[x = 1\] )
And
 \[f\left( 2 \right) = 2 - 2 = 0\] (at \[x = 2\] )
At \[x = 1\] ,
L.H.D. \[ = \mathop {\lim }\limits_{x \to {1^ - }} \dfrac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}}\]
             \[
   = \mathop {\lim }\limits_{x \to 1} \dfrac{{x - 1}}{{x - 1}} \\
   = 1 \\
 \]
R.H.D. \[ = \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}}\]
              \[
   = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2 - x} \right) - 1}}{{x - 1}} \\
   = \mathop {\lim }\limits_{x \to 1} \dfrac{{ - \left( {x - 1} \right)}}{{x - 1}} \\
   = - 1 \\
 \]
\[\therefore \]L.H.D \[ \ne \] R.H.D.
Thus, the given function is not differentiable at \[x = 1\] .
At \[x = 2\] ,
L.H.D. \[ = \mathop {\lim }\limits_{x \to {2^ - }} \dfrac{{f\left( x \right) - f\left( 2 \right)}}{{x - 2}}\]
             \[
   = \mathop {\lim }\limits_{x \to 2} \dfrac{{2 - x - 0}}{{x - 2}} \\
   = \mathop {\lim }\limits_{x \to 2} \dfrac{{ - \left( {x - 2} \right)}}{{x - 2}} \\
   = - 1 \\
 \]
R.H.D. \[ = \mathop {\lim }\limits_{x \to {2^ + }} \dfrac{{f\left( x \right) - f\left( 2 \right)}}{{x - 2}}\]
             \[
   = \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( { - 2 + 3x - {x^2}} \right) - 0}}{{x - 2}} \\
   = \mathop {\lim }\limits_{x \to 2} \dfrac{{ - \left( {{x^2} - 3x + 2} \right) - 0}}{{x - 2}} \\
   = \mathop {\lim }\limits_{x \to 2} \dfrac{{ - \left( {x - 1} \right)\left( {x - 2} \right)}}{{x - 2}} \\
   = \mathop {\lim }\limits_{x \to 2} - \left( {x - 1} \right) \\
   = - 1 \\
 \]
\[\therefore \]L.H.D. \[ = \] R.H.D.
Thus, the given function is differentiable at \[x = 2\] .

Note: Every polynomial equation is always continuous everywhere on \[\left( { - \infty ,\infty } \right)\] .
A function can be differentiable at a particular given point if and only if the Left Hand side Derivative (L.H.D.) and the Right Hand side Derivative (R.H.D.) co-exist and are equal.