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}
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=)
(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. the point(s) of discontinuity defined by f(x) have the value(s)
(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.

Explanation:

Step1: Calculate left - hand limit

We find $\lim_{x
ightarrow1^{-}}f(x)$. Since $x
ightarrow1^{-}$ means $x < 1$, $f(x)=2 + x$. So, $\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{-}}(2 + x)=2+1 = 3$.

Step2: Calculate right - hand limit

We find $\lim_{x
ightarrow1^{+}}f(x)$. Since $x
ightarrow1^{+}$ means $x>1$, $f(x)=4 - x$. So, $\lim_{x
ightarrow1^{+}}f(x)=\lim_{x
ightarrow1^{+}}(4 - x)=4 - 1=3$.

Step3: Calculate function value at $x = 1$

When $x = 1$, $f(1)=2 + 1=3$ (because for $x\leq1$, $f(x)=2 + x$).

Step4: Check continuity

A function $y = f(x)$ is continuous at $x=a$ if $\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)$. Here, $\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{+}}f(x)=f(1) = 3$, so the function is continuous everywhere.

Answer:

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