Question

Discuss the continuity of the function  \[f\left( x \right) = \begin{cases}2x - 1 , & \text { if }  x < 2 \\ \frac{3x}{2} , & \text{ if  } x \geq 2\end{cases}\]

Answer 1

When x < 2, we have  

\[f\left( x \right) = 2x - 1\]

We know that a polynomial function is everywhere continuous.

So,

\[f\left( x \right)\]  is continuous for each x < 2.

When  \[x > 2\] , we have 

\[f\left( x \right) = \frac{3x}{2}\]

The functions 3x and 2 are continuous being the polynomial and constant function, respectively.

Thus, the quotient function 

\[\frac{3x}{2}\]  is continuous at each x > 2.

Now,
Let us consider the point x = 2.
Given:

\[f\left( x \right) = \binom{2x - 1, \text{ if  } x < 2}{\frac{3x}{2}, \text{ if } x \geq 2}\]

We have
(LHL at x = 2) =  \[\lim_{x \to 2^-} f\left( x \right) = \lim_{h \to 0} f\left( 2 - h \right) = \lim_{h \to 0} \left( 2\left( 2 - h \right) - 1 \right) = 4 - 1 = 3\]

(RHL at x = 2) =   \[\lim_{x \to 2^+} f\left( x \right) = \lim_{h \to 0} f\left( 2 + h \right) = \lim_{h \to 0} \frac{3\left( h + 2 \right)}{2} = 3\]

Also,

\[f\left( 2 \right) = \frac{3\left( 2 \right)}{2} = 3\]

∴  \[\lim_{x \to 2^-} f\left( x \right) = \lim_{x \to 2^+} f\left( x \right) = f\left( 2 \right)\]

Thus, 

\[f\left( x \right)\]  is continuous at x = 2.

Hence, 

\[f\left( x \right)\]  is continuous at x = 2.