Question

suppose that the function f is defined, for all real numbers, as follows. f(x) = { 3x + 1 if x < -2; x - 3 if x ≥ -2 } graph the function f. then determine whether or not the function is continuous.

Explanation:

Step1: Find left - hand limit as x approaches - 2

We use the part of the function for $x < - 2$, which is $f(x)=3x + 1$.
$\lim_{x
ightarrow - 2^{-}}f(x)=\lim_{x
ightarrow - 2^{-}}(3x + 1)=3\times(-2)+1=-6 + 1=-5$

Step2: Find right - hand limit as x approaches - 2

We use the part of the function for $x\geq - 2$, which is $f(x)=x - 3$.
$\lim_{x
ightarrow - 2^{+}}f(x)=\lim_{x
ightarrow - 2^{+}}(x - 3)=-2-3=-5$

Step3: Find the value of the function at x = - 2

We use the part of the function for $x\geq - 2$, so $f(-2)=-2-3=-5$

Step4: Check continuity condition

Since $\lim_{x
ightarrow - 2^{-}}f(x)=\lim_{x
ightarrow - 2^{+}}f(x)=f(-2)=-5$, the function is continuous.
To graph:

• For $y = 3x+1$ when $x < - 2$: The $y$-intercept of $y = 3x + 1$ is 1 and the slope is 3. We draw a ray with an open - circle at $x=-2$ (because $x < - 2$). When $x=-2$, $y=3\times(-2)+1=-5$.
• For $y=x - 3$ when $x\geq - 2$: The $y$-intercept of $y=x - 3$ is - 3 and the slope is 1. We draw a ray with a closed - circle at $x = - 2$ (because $x\geq - 2$), and when $x=-2$, $y=-2-3=-5$.

Answer:

The function $f(x)$ is continuous.