# Question e46e8

Feb 15, 2018

$\left(- \infty , 0\right) \cup \left[\frac{5}{3} , + \infty\right)$

#### Explanation:

$\text{rearrange making x the subject}$

$\Rightarrow y \left(3 - {x}^{2}\right) = 5$

$\Rightarrow 3 y - y {x}^{2} = 5$

$\Rightarrow {x}^{2} = \frac{3 y - 5}{y} \Rightarrow x = \sqrt{\frac{3 y - 5}{y}}$

$\text{now } \sqrt{\frac{3 y - 5}{y}} \ge 0$

$3 y - 5 = 0 \Rightarrow y = \frac{5}{3} \text{ and } y = 0$

$\text{these values split the range into 3 intervals}$

(-oo,0),(0,5/3],([5/3,+oo)

$\text{choose a "color(blue)"test point in each interval}$

$x = - 10 \to \textcolor{red}{\text{positive}}$

$x = 1 \to \textcolor{b l u e}{\text{negative}}$

$x = 10 \to \textcolor{red}{\text{positive}}$

$\left(- \infty , 0\right) \cup \left[\frac{5}{3} , + \infty\right)$
graph{5/(3-x^2) [-10, 10, -5, 5]}

Feb 15, 2018

The range is $y \in \left(- \infty , 0\right) \cup \left[\frac{5}{3} , + \infty\right)$

#### Explanation:

To find the range, proceed as follows

$y = \frac{5}{3 - {x}^{2}}$

Therefore,

$3 - {x}^{2} = \frac{5}{y}$

${x}^{2} = 3 - \frac{5}{y}$

${x}^{2} = \frac{3 y - 5}{y}$

$x = \sqrt{\frac{3 y - 5}{y}}$

What's under the sqrt sign $\ge 0$

So,

$f \left(y\right) = \frac{3 y - 5}{y} \ge 0$

Solve this inequality with a sign chart,

$\textcolor{w h i t e}{a a a a}$$y$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a a}$$0$$\textcolor{w h i t e}{a a a a a a a a}$$\frac{5}{3}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$y$$\textcolor{w h i t e}{a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$3 y - 5$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$color(white)(aaaaa)-#$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(y\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a}$$+$

Therefore,

$f \left(y\right) \ge 0$, $y \in \left(- \infty , 0\right) \cup \left[\frac{5}{3} , + \infty\right)$

graph{5/(3-x^2) [-7.9, 7.9, -3.95, 3.95]}