Question

suppose that the function f is defined, for all real numbers, as follows.
f(x)=\begin{cases}-x + 2&\text{if }xleq1\\5x - 2&\text{if }x>1end{cases}
graph the function f. then determine whether or not the function is continuous.
is the function continuous?
yes
no

Explanation:

Step1: Find the left - hand limit as x approaches 1

We use the part of the function for \(x\leq1\), so \(\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{-}}(-x + 2)\). Substitute \(x = 1\) into \(-x+2\), we get \(-1 + 2=1\).

Step2: Find the right - hand limit as x approaches 1

We use the part of the function for \(x>1\), so \(\lim_{x
ightarrow1^{+}}f(x)=\lim_{x
ightarrow1^{+}}(5x - 2)\). Substitute \(x = 1\) into \(5x-2\), we get \(5\times1-2 = 3\).

Step3: Evaluate the function at x = 1

We use the part of the function for \(x\leq1\), so \(f(1)=-1 + 2=1\).

Step4: Check the continuity condition

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, \(\lim_{x
ightarrow1^{-}}f(x)=1\), \(\lim_{x
ightarrow1^{+}}f(x)=3\), and since \(1
eq3\), the function is not continuous.

Answer:

No