Question

Find the points of discontinuity, if any, of the following functions:  \[f\left( x \right) = \begin{cases}\left| x - 3 \right|, & \text{ if }  x \geq 1 \\ \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4}, & \text{ if }  x < 1\end{cases}\]

Answer 1

When x > 1, then 

\[f\left( x \right) = \left| x - 3 \right|\]

Since modulus function is a continuous function,

\[f\left( x \right)\]  is continuous for each x > 1.

When x < 1, then

\[f\left( x \right) = \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4}\]

Since,

\[x^2 \text{ and } 3x\]  are continuous being polynomial functions,

\[\frac{x^2}{4} \text{and} \frac{3x}{2}\] will also be continuous.
Also,

 \[\frac{13}{4}\]  is continuous being a polynomial function.

\[\Rightarrow \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4}\]  is continuous for each \[x < 1\]

\[\Rightarrow f\left( x \right)\]  is continuous for each x < 1.

At x = 1, we have
(LHL at x=1) =  \[\lim_{x \to 1^-} f\left( x \right) = \lim_{h \to 0} f\left( 1 - h \right) = \lim_{h \to 0} \left[ \frac{\left( 1 - h \right)^2}{4} - \frac{3\left( 1 - h \right)}{2} + \frac{13}{4} \right] = \frac{1}{4} - \frac{3}{2} + \frac{13}{4} = 2\]

(RHL at x=1) =  \[\lim_{x \to 1^+} f\left( x \right) = \lim_{h \to 0} f\left( 1 + h \right) = \lim_{h \to 0} \left[ \left| 1 + h - 3 \right| \right] = \left| - 2 \right| = 2\]

\[f\left( 1 \right) = \left| 1 - 3 \right| = \left| - 2 \right| = 2\]

 Thus, 

\[\lim_{x \to 1^-} f\left( x \right) = \lim_{x \to 1^+} f\left( x \right) = f\left( 1 \right)\]

Hence , 

\[f\left( x \right)\] is continuous at x= 1.

Thus, the given function is nowhere discontinuous.