How do you find the domain and range of #t/sqrt(t^2-25) #?

1 Answer
Jim G.
Jan 22, 2018

#"see explanation"#

Explanation:

#f(t)=t/(sqrt(t^2-25))#

#"the denominator cannot equal zero as this would make"#
#"f(t) undefined"#

#rArrt^2-25!=0rArrt!=+-5#

#"also "t^2-25>0#

#rArr(t-5)(t+5)>0#

#rArrt<-5" or "t>5#

#rArr"domain is "(-oo,-5)uu(5,+oo)#

#f(t)=t/(sqrt(t^2(1-25/t^2)))=t/(tsqrt(1-25/t^2)#

#color(white)(f(t))=cancel(t)^1/(cancel(t)^1sqrt(1-25/t^2))#

#"as "t to+-oo,f(t)to1/sqrt(1-0)#

#rArry=-1,y=1larrcolor(blue)"excluded values"#

#rArr"range is "(-1,-oo)uu(1,+oo)#
graph{x/(sqrt(x^2-25)) [-10, 10, -5, 5]}

Related questions
See all questions in Domain and Range of a Function
Impact of this question
2203 views around the world
You can reuse this answer
Creative Commons License