Question

graph f, locate all points of discontinuity, and discuss the behavior of f at these points.
f(x)=\begin{cases}-1 + x & \text{if }x< - 1\\5 - x & \text{if }xgeq - 1end{cases}
locate all points of discontinuity. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the point(s) of discontinuity is/are x = - 1. (use a comma to separate answers as needed.)
b. there are no points of discontinuity.
discuss the behavior of f at its point(s) of discontinuity.
find f(x) at any points of discontinuity. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. at the point(s) of discontinuity, the value(s) of the function f(x) is/are
(use a comma to separate answers as needed.)
b. there are no points of discontinuity.

Explanation:

Step1: Calculate left - hand limit

For \(x\to - 1^{-}\), \(f(x)=-1 + x\). So \(\lim_{x\to - 1^{-}}f(x)=\lim_{x\to - 1^{-}}(-1 + x)=-1+( - 1)=-2\).

Step2: Calculate right - hand limit

For \(x\to - 1^{+}\), \(f(x)=5 - x\). So \(\lim_{x\to - 1^{+}}f(x)=\lim_{x\to - 1^{+}}(5 - x)=5-( - 1)=6\).

Step3: Calculate function value at \(x=-1\)

For \(x = - 1\), \(f(x)=5 - x\), so \(f(-1)=5-( - 1)=6\).
Since \(\lim_{x\to - 1^{-}}f(x)=-2
eq\lim_{x\to - 1^{+}}f(x)=6\), there is a discontinuity at \(x = - 1\).

Answer:

A. The point(s) of discontinuity is/are \(x=-1\).
At the point of discontinuity \(x = - 1\), the left - hand limit is \(-2\) and the right - hand limit is \(6\) and the function value \(f(-1) = 6\). The function has a jump discontinuity at \(x=-1\). So for the second part:
A. At the point(s) of discontinuity, the value(s) of the function \(f(x)\) is/are \(6\) (the value of the function at \(x=-1\) from the right - hand side definition).