Question #e46e8

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Question #e46e8

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Answer 1 · 2018-02-15T08:24:43

$(- \infty, 0) \cup [\frac{5}{3}, + \infty)$ $(-oo,0)uu[5/3,+oo)$

Explanation:

$rearrange making x the subject$ $"rearrange making x the subject"$

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

$\Rightarrow 3 y - y x^{2} = 5$ $rArr3y-yx^2=5$

$\Rightarrow x^{2} = \frac{3 y - 5}{y} \Rightarrow x = \sqrt{\frac{3 y - 5}{y}}$ $rArrx^2=(3y-5)/yrArrx=sqrt((3y-5)/y)$

$now \sqrt{\frac{3 y - 5}{y}} \geq 0$ $"now "sqrt((3y-5)/y)>=0$

$3 y - 5 = 0 \Rightarrow y = \frac{5}{3} and y = 0$ $3y-5=0rArry=5/3" and "y=0$

$these values split the range into 3 intervals$ $"these values split the range into 3 intervals"$

$(- \infty, 0), (0, \frac{5}{3}], ([\frac{5}{3}, + \infty)$ $(-oo,0),(0,5/3],([5/3,+oo)$

$choose a test point in each interval$ $"choose a "color(blue)"test point in each interval"$

$x = - 10 \to positive$ $x=-10tocolor(red)"positive"$

$x = 1 \to negative$ $x=1tocolor(blue)"negative"$

$x = 10 \to positive$ $x=10tocolor(red)"positive"$

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

Answer 2 · 2018-02-15T08:52:30

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

Explanation:

To find the range, proceed as follows

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

Therefore,

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

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

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

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

What's under the $\sqrt{}$ $sqrt$ sign $\geq 0$ $>=0$

So,

$f (y) = \frac{3 y - 5}{y} \geq 0$ $f(y)=(3y-5)/y>=0$

Solve this inequality with a sign chart,

$a a a a$ $color(white)(aaaa)$ $y$ $y$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a a a a$ $color(white)(aaaaaaa)$ $0$ $0$ $a a a a a a a a$ $color(white)(aaaaaaaa)$ $\frac{5}{3}$ $5/3$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $y$ $y$ $a a a a a a a a$ $color(white)(aaaaaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $∣ ∣$ $||$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a a$ $color(white)(aaaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $3 y - 5$ $3y-5$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ #color(white)(aaaaa)-# $a a$ $color(white)(aa)$ $0$ $0$ $a a$ $color(white)(aa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (y)$ $f(y)$ $a a a a a a$ $color(white)(aaaaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $∣ ∣$ $||$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a$ $color(white)(aa)$ $0$ $0$ $a a$ $color(white)(aa)$ $+$ $+$

Therefore,

$f (y) \geq 0$ $f(y)>=0$ , $y \in (- \infty, 0) \cup [\frac{5}{3}, + \infty)$ $y in (-oo,0) uu[5/3, +oo)$

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

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Question #e46e8

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