Question

Find the points of discontinuity, if any, of the following functions: 

\[f\left( x \right) = \begin{cases}x^3 - x^2 + 2x - 2, & \text{ if }x \neq 1 \\ 4 , & \text{ if } x = 1\end{cases}\]

 

Answer 1

When x 

\[\neq\] 1 then 

\[f\left( x \right) = x^3 - x^2 + 2x - 2\]

We know that a polynomial function is everywhere continuous.
So,

\[f\left( x \right) = x^3 - x^2 + 2x - 2\] is continuous at each x

\[\neq\] 1

At x = 1, we have

(LHL at x = 1) =

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

(RHL at x = 1) =

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

Also,

\[f\left( 1 \right) = 4\]

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

Thus,

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

Hence, the only point of discontinuity for 

\[f\left( x \right)\] is x = 1.