Question

Discuss the continuity of the function f, where f is defined by \[f(x) = \left\{ {\begin{array}{*{20}{c}}
  { - 2;if{\text{ }}x \leqslant - 1} \\
  {2x;if{\text{ }} - 1 < x \leqslant 1} \\
  {2;if{\text{ }}x > 1}
\end{array}} \right\}\].

Answer 1

Hint: In this question, we need to check the continuity of the given function at x=1 and x=-1. For this, we will use the relation of the left hand limit and the right hand limit for the given function.

Complete step-by-step answer:
For a function to be continuous, the functions should have the same value at the left-hand limit, right-hand limit as well as on the limits.
Here in the question, we need to determine the continuity at the point x=1 and x=-1 as the function given is associated with the same.
Now, the value of the function at x=-1 is defined as $ f\left( x \right) = - 2 $ so, the value of the function at x=-1 is given by
 $ f( - 1) = - 2 $
The value of the left-hand limit of the function at x=-1 is defined as $ f\left( {{x^ - }} \right) = - 2 $ so, the value of the function at $ x = - 1 - h $ where h is infinitesimally small and is tending to zero is given by
 $
  f\left( {{x^ - }} \right) = \mathop {\lim }\limits_{x \to - {1^ - }} \left( { - 2} \right) \\
   = \left( { - 2} \right) \\
   = - 2 - - - - (i) \\
  $
The value of the right-hand limit of the function at x=-1 is defined as $ f\left( {{x^ + }} \right) = 2x $ so, the value of the function at $ x = - 1 + h $ where h is infinitesimally small and is tending to zero is given by
\[
  f\left( {{x^ + }} \right) = \mathop {\lim }\limits_{x \to - {1^ + }} 2x \\
   = \mathop {\lim }\limits_{h \to 0} \left( {2( - 1 + h)} \right) \\
   = \mathop {\lim }\limits_{h \to 0} \left( { - 2 + 2h} \right) \\
   = - 2 - - - - (ii) \\
 \]
From the equations (i) and (ii) we can see that the value of the functions at the left-hand limit is equal to the value of the function at the right-hand limits, so the given function is not continuous at x=-1.
Now, the value of the function at x=1 is defined as $ f\left( x \right) = 2x $ so, the value of the function at x=1 is given by
 $
  f(1) = 2x \\
   = 2 \\
  $
The value of the left-hand limit of the function at x=1 is defined as $ f\left( {{x^ - }} \right) = 2x $ so, the value of the function at $ x = 1 - h $ where h is infinitesimally small and is tending to zero is given by
 $
  f\left( {{x^ - }} \right) = \mathop {\lim }\limits_{x \to {1^ - }} 2x \\
   = \mathop {\lim }\limits_{h \to 0} \left( {2(1 - h)} \right) \\
   = \mathop {\lim }\limits_{h \to 0} \left( {2 - 2h} \right) \\
   = 2 - - - - (iii) \\
  $
The value of the right-hand limit of the function at x=1 is defined as $ f\left( {{x^ + }} \right) = 2 $ so, the value of the function at $ x = 1 + h $ where h is infinitesimally small and is tending to zero is given by
 $
  f\left( {{x^ + }} \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( 2 \right) \\
   = 2 - - - - (iv) \\
  $
From the equations (iii) and (iv) we can see that the value of the functions at the left-hand limit is equal to the value of the function at the right-hand limits, so the given function is not continuous at x=1.
Hence, the given function is continuous at x=-1 and x=1.

Note: If a function is differentiable, then, the function must be continuous at the point of the investigation while it is not necessary that if it is continuous then, it must be differentiable as well.