Question

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

Explanation:

Step1: Analyze left - hand limit

For \(x < - 1\), \(f(x)=-x + 3\). Calculate \(\lim_{x
ightarrow - 1^{-}}f(x)\). Substitute \(x=-1\) into \(-x + 3\), we get \(\lim_{x
ightarrow - 1^{-}}(-x + 3)=-(-1)+3=4\).

Step2: Analyze right - hand limit

For \(x\geq - 1\), \(f(x)=-3x + 1\). Calculate \(\lim_{x
ightarrow - 1^{+}}f(x)\). Substitute \(x = - 1\) into \(-3x+1\), we get \(\lim_{x
ightarrow - 1^{+}}(-3x + 1)=-3\times(-1)+1=4\).

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

When \(x=-1\), using \(f(x)=-3x + 1\) (since \(x=-1\) satisfies \(x\geq - 1\)), \(f(-1)=-3\times(-1)+1=4\).

Step4: Determine continuity

Since \(\lim_{x
ightarrow - 1^{-}}f(x)=\lim_{x
ightarrow - 1^{+}}f(x)=f(-1) = 4\), and the two linear - piece functions \(y=-x + 3\) for \(x < - 1\) and \(y=-3x + 1\) for \(x\geq - 1\) are continuous on their respective domains, the function \(f(x)\) is continuous.

Answer:

Yes