How do you find the domain and range of #g(t) = ( 1 + 8t ) / ( 1 + t^2 )#?

1 Answer
Nov 1, 2017

Answer:

The domain is #t in RR#
The range is #g(t) in [-3.531,4.531]#

Explanation:

As #AA t in RR,(1+t^2)>0#

The domain is #t in RR#

Let #y=(1+8t)/(1+t^2)#

#y(1+t^2)=1+8t#

#y+yt^2=1+8t#

#yt^2-8t+y-1=0#

This is a quadratic equation in #t^2# and in order for this equation to have solutions, the discriminant #Delta>=0#

#Delta=b^2-4ac=(-8)^2-(4)(y)(y-1)>=0#

#64-4y^2+4y>=0#

#y^2-y-16<=0#

#y=(1+-sqrt(1+64))/(2)=(1+-sqrt65)/2#

#y_1=(1+sqrt65)/2=4.531#

#y_2=(1-sqrt65)/2=-3.531#

Therefore,

the range is #g(t) in [-3.531,4.531]#

graph{(1+8x)/(1+x^2) [-12.66, 12.65, -6.33, 6.33]}