# Answer the following questions for the function, r ( x ) = 3 x − 18/x^2+5 x+6 ? a)What is the domain of r ( x ) ? Give your answer using interval notation. b)r ( − 4 ) = c)For what value(s) of x does r ( x ) = 0 ? x= d)r ( x ) = − 3 5 . What is x ? x =

Jun 23, 2018

It is given

$r \left(x\right) = 3 x - \frac{18}{x} ^ 2 + 5 x + 6 = 8 x + 6 - \frac{18}{x} ^ 2$

(a) The maximum domain of a function represents all the values it can take without becoming undefined/undetermined.

In our case, we see that

$r \left(0\right) = 6 - \frac{18}{0} \to \text{ Undefined}$

As such, the domain of $r$ is the real numbers without $0$.

$r : {\mathbb{R}}^{\text{*}}$

${\mathbb{R}}^{\text{*"=(-oo,+oo) "\}} 0 = \left(- \infty , 0\right) \cup \left(0 , + \infty\right)$

(b) $r \left(- 4\right) = - 32 + 6 - \frac{18}{16} = - \frac{217}{8}$

(c) We wish to find the roots of $r$:

$r \left(x\right) = 0$
$8 x + 6 - \frac{18}{x} ^ 2 = 0$

This is going to be a bit tricky. First of all, multiply both sides by ${x}^{2}$, as $x$ cannot be $0$:

$8 {x}^{3} + 6 {x}^{2} - 18 = 0$
$4 {x}^{3} + 3 {x}^{2} - 9 = 0$

See this link on how to solve this cubic equation.

Finally, we find out the only root of $r$ to be

$x = \frac{1}{4} \left(- 1 + \sqrt[3]{71 - 12 \sqrt{35}} + \sqrt[3]{71 + 12 \sqrt{35}}\right) \approx 1.1022$

(d) $r \left(x\right) = - 35$

$8 x + 6 - \frac{18}{x} ^ 2 = - 35$
$8 {x}^{3} + 6 {x}^{2} - 18 = - 35 {x}^{2}$

I won't even try to solve this. By trial and error, you can approximate the roots of our new formed equation; for a fact, it has $3$ real roots. I highly suggest you check out this Wikipedia page. All of the roots' exact forms contain complex numbers and are pretty "ugly".

Anyway, here are the three values of $x$:

${x}_{1} \approx - 5.0363$
${x}_{2} \approx - 0.71422$
${x}_{3} \approx 0.62552$