Question

suppose that the function (f) is defined, for all real numbers, as follows.
f(x)=\begin{cases}-3x - 1&\text{if }x < - 1\\-x + 1&\text{if }xgeq - 1end{cases}
graph the function (f). then determine whether or not the function is continuous.
is the function continuous?
yes
no

Explanation:

Step1: Analyze \(f(x)= - 3x - 1\) for \(x < - 1\)

When \(x=-1\), \(y=-3\times(-1)-1 = 2\). The slope is \(-3\) and the \(y\) - intercept is \(-1\) for \(x < - 1\). We draw a dashed - line (since \(x
eq - 1\) for this part) with slope \(-3\) for \(x < - 1\).

Step2: Analyze \(f(x)=-x + 1\) for \(x\geq - 1\)

When \(x=-1\), \(y=-(-1)+1=2\). The slope is \(-1\) and the \(y\) - intercept is \(1\). We draw a solid - line (since \(x=-1\) is included) with slope \(-1\) for \(x\geq - 1\).

Step3: Check continuity

A function \(y = f(x)\) is continuous at \(x = a\) if \(\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)\).
For \(a=-1\), \(\lim_{x
ightarrow - 1^{-}}f(x)=\lim_{x
ightarrow - 1^{-}}(-3x - 1)=-3\times(-1)-1 = 2\).
\(\lim_{x
ightarrow - 1^{+}}f(x)=\lim_{x
ightarrow - 1^{+}}(-x + 1)=-(-1)+1 = 2\).
And \(f(-1)=-(-1)+1 = 2\).

Answer:

Yes