Question

graph f, locate all points of discontinuity, and discuss the behavior of f at these points.
f(x)=\begin{cases}2 + x & \text{if } xleq1\\4 - x & \text{if } x > 1end{cases}
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. the point(s) of discontinuity defined by f(x) have the value(s) . all other points of discontinuity are not defined by f(x). (use a comma to separate answers as needed.)
b. the point(s) of discontinuity are not defined by f(x).
c. there are no points of discontinuity.
find (lim_{x
ightarrow c}f(x)), where c is/are the point(s) at which f(x) is discontinuous. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. (lim_{x
ightarrow c}f(x)=). all other points of discontinuity have a limit that does not exist. (use a comma to separate answers as needed.)
b. the limit(s) of the point(s) of discontinuity do(es) not exist.
c. there are no points of discontinuity.

Explanation:

Step1: Find left - hand limit

For \(x\to1^{-}\), \(f(x)=2 + x\). So, \(\lim_{x\to1^{-}}f(x)=\lim_{x\to1^{-}}(2 + x)=2+1 = 3\).

Step2: Find right - hand limit

For \(x\to1^{+}\), \(f(x)=4 - x\). So, \(\lim_{x\to1^{+}}f(x)=\lim_{x\to1^{+}}(4 - x)=4 - 1=3\).

Step3: Find function value at \(x = 1\)

When \(x = 1\), \(f(1)=2+1 = 3\) (using the first - part of the piece - wise function since \(x = 1\) satisfies \(x\leq1\)).

Step4: Check continuity

Since \(\lim_{x\to1^{-}}f(x)=\lim_{x\to1^{+}}f(x)=f(1)=3\), the function \(f(x)\) is continuous at \(x = 1\). And since \(f(x)\) is a linear function on each of its sub - domains (\(x\leq1\) and \(x>1\)), it is continuous everywhere.

Answer:

C. There are no points of discontinuity.
C. There are no points of discontinuity.