Question

problems 15 - 16, write an equation of a rational function with the following limit properties. 15. $lim_{x
ightarrow2^{-}}f(x)=3$, $lim_{x
ightarrow2^{+}}f(x)=3$, $lim_{x
ightarrow - 3^{-}}f(x)=6$, $lim_{x
ightarrow - 3^{+}}f(x)=6$. 16. $lim_{x
ightarrow - 1^{-}}f(x)=+infty$, $lim_{x
ightarrow - 1^{+}}f(x)=+infty$, $lim_{x
ightarrow4^{-}}f(x)=+infty$, $lim_{x
ightarrow4^{+}}f(x)=-infty$. problems 17 - 18, sketch the graph of a rational function that has the following properties. 17. $f(x)$ has a hole at $x = 2$

Explanation:

Step1: Analyze vertical - asymptote conditions

For $\lim_{x
ightarrow a^{-}}f(x)=\infty$ and $\lim_{x
ightarrow a^{+}}f(x)=\infty$, there is a vertical asymptote at $x = a$. From $\lim_{x
ightarrow - 1^{-}}f(x)=\infty$ and $\lim_{x
ightarrow - 1^{+}}f(x)=\infty$, we know $x=-1$ is a vertical asymptote. Also, from $\lim_{x
ightarrow4^{-}}f(x)=\infty$ and $\lim_{x
ightarrow4^{+}}f(x)=-\infty$, $x = 4$ is a vertical asymptote.

Step2: Analyze hole - related conditions

Since there is a hole at $x = 2$, the function has a common factor in the numerator and denominator that cancels out at $x = 2$.

Step3: Analyze non - asymptote limits

For $\lim_{x
ightarrow2^{-}}f(x)=3$ and $\lim_{x
ightarrow2^{+}}f(x)=3$, and $\lim_{x
ightarrow - 3^{-}}f(x)=6$ and $\lim_{x
ightarrow - 3^{+}}f(x)=6$, the function approaches these non - infinite values at these points.

Step4: Construct the rational function

A rational function of the form $f(x)=\frac{(x - 2)(x + 3)(x + k)}{(x - 2)(x + 1)(x - 4)}$. To satisfy $\lim_{x
ightarrow - 3}f(x)=6$, we substitute $x=-3$ into $\frac{(x + k)}{(x + 1)(x - 4)}$ and set it equal to $1$ (because when $x=-3$, $(x - 2)$ and $(x + 3)$ cancel out). Substituting $x=-3$ into $\frac{x + k}{(x + 1)(x - 4)}$, we have $\frac{-3 + k}{(-3 + 1)(-3 - 4)}=1$. Solving $\frac{-3 + k}{(-2)\times(-7)}=1$, we get $\frac{-3 + k}{14}=1$, then $-3 + k=14$, and $k = 17$. So a possible rational function is $f(x)=\frac{(x - 2)(x + 3)(x + 17)}{(x - 2)(x + 1)(x - 4)}$.

Answer:

$f(x)=\frac{(x - 2)(x + 3)(x + 17)}{(x - 2)(x + 1)(x - 4)}$