Q. 44 - Calculus

Question

Use Theorem 1.16 and left and right limits to determine whether each function f is continuous at its break point(s). For each discontinuity of f , describe the type of discontinuity and any one-sided discontinuity. 

$f\left(x\right)=\left\{\begin{array}{r}{x}^3,\text{ if }x\le 0\\ 1-x,\text{ if }0<x<3\\ x-5,\text{ if }x\ge 3\end{array}\right.$

Step-by-Step Solution

Answer

The function has a jump discontinuity at break point x = 0.

The function is continuous at break point x = 3.

Step 1. Given information.

We have been given a function:

$f\left(x\right)=\left\{\begin{array}{r}{x}^3,\text{ if }x\le 0\\ 1-x,\text{ if }0<x<3\\ x-5,\text{ if }x\ge 3\end{array}\right.$

We have to determine whether this function f is continuous at its break point(s). For each discontinuity of f , we have to describe the type of discontinuity and any one-sided discontinuity.  

Step 2. Check continuity at break point x = 0.

$$\begin{gathered} \underset{x\to 0^{-}}{lim} f\left(x\right) \\ =\underset{x\to 0^{-}}{lim} \left(x^3\right) \\ =0^3 \\ =0 \\ \underset{x\to 0^{+}}{lim} f\left(x\right) \\ =\underset{x\to 0^{+}}{lim} (1-x) \\ =1-0 \\ =1 \\ \underset{x\to 0}{lim} f\left(x\right) \\ =\underset{x\to 0}{lim} \left(x^3\right) \\ =0^3 \\ =0 \end{gathered}$$

Since both side limits are not equal, the function has jump discontinuity.

This function is left continuous at x = 0.

Step 3. Check continuity at break point x = 3.

$$\begin{gathered} \underset{x\to 3^{-}}{lim} f\left(x\right) \\ =\underset{x\to 3^{-}}{lim} (1-x) \\ =1-3 \\ =-2 \\ \underset{x\to 3^{+}}{lim} f\left(x\right) \\ =\underset{x\to 3^{+}}{lim} (x-5) \\ =3-5 \\ =-2 \\ \underset{x\to 3}{lim} f^{\mathrm{\prime }}\left(x\right) \\ =\underset{x\to 3}{lim} (x-5) \\ =3-5 \\ =-2 \end{gathered}$$

Since all the values are equal, the function is continuous at x = 3.