Question

problems 15 - 16, write an equation of a rational function with the following limit properties. 15. $lim_{x \to 2^{-}}f(x)=3$, $lim_{x \to 2^{+}}f(x)=3$, $lim_{x \to 4^{-}}f(x)=+infty$, $lim_{x \to 4^{+}}f(x)=-infty$ 16. $lim_{x \to - 3^{-}}f(x)=6$, $lim_{x \to - 3^{+}}f(x)=6$, $lim_{x \to - 1^{-}}f(x)=+infty$, $lim_{x \to - 1^{+}}f(x)=+infty$

Explanation:

Step1: Analyze vertical - asymptotes

The vertical asymptotes occur where the limit is $\pm\infty$. For problem 15, $x = 4$ is a vertical asymptote. A factor of $(x - 4)$ in the denominator will cause this behavior. For problem 16, $x=-1$ is a vertical asymptote, so a factor of $(x + 1)$ in the denominator is needed.

Step2: Analyze horizontal - like behavior

For problem 15, as $x\to2^{\pm}$, $\lim_{x\to2}f(x)=3$. This suggests that the function has a hole or a horizontal - like behavior at $x = 2$. We can introduce a factor of $(x - 2)$ in both the numerator and denominator. For problem 16, as $x\to - 3^{\pm}$, $\lim_{x\to - 3}f(x)=6$, so we introduce a factor of $(x + 3)$ in both the numerator and denominator.

Step3: Construct the rational function

For problem 15:
Let $f(x)=\frac{3(x - 2)}{(x - 2)(x - 4)}=\frac{3}{x - 4}$ (the $(x - 2)$ terms cancel out for $x
eq2$).
For problem 16:
Let $f(x)=\frac{6(x + 3)}{(x + 3)(x + 1)}=\frac{6}{x + 1}$ (the $(x + 3)$ terms cancel out for $x
eq - 3$).

Answer:

1. $f(x)=\frac{3}{x - 4}$
2. $f(x)=\frac{6}{x + 1}$