Question #e46e8

2 Answers
Feb 15, 2018

Answer:

#(-oo,0)uu[5/3,+oo)#

Explanation:

#"rearrange making x the subject"#

#rArry(3-x^2)=5#

#rArr3y-yx^2=5#

#rArrx^2=(3y-5)/yrArrx=sqrt((3y-5)/y)#

#"now "sqrt((3y-5)/y)>=0#

#3y-5=0rArry=5/3" and "y=0#

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

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

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

#x=-10tocolor(red)"positive"#

#x=1tocolor(blue)"negative"#

#x=10tocolor(red)"positive"#

#(-oo,0)uu[5/3,+oo)#
graph{5/(3-x^2) [-10, 10, -5, 5]}

Feb 15, 2018

Answer:

The range is #y in (-oo,0) uu[5/3, +oo)#

Explanation:

To find the range, proceed as follows

#y=5/(3-x^2)#

Therefore,

#3-x^2=5/y#

#x^2=3-5/y#

#x^2=(3y-5)/y#

#x=sqrt((3y-5)/y)#

What's under the #sqrt# sign #>=0#

So,

#f(y)=(3y-5)/y>=0#

Solve this inequality with a sign chart,

#color(white)(aaaa)##y##color(white)(aaaa)##-oo##color(white)(aaaaaaa)##0##color(white)(aaaaaaaa)##5/3##color(white)(aaaa)##+oo#

#color(white)(aaaa)##y##color(white)(aaaaaaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aaaaa)##+#

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

#color(white)(aaaa)##f(y)##color(white)(aaaaaa)##+##color(white)(aaaa)##||##color(white)(aaaa)##-##color(white)(aa)##0##color(white)(aa)##+#

Therefore,

#f(y)>=0#, #y in (-oo,0) uu[5/3, +oo)#

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