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

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Answer 1 · 2017-11-01T09:32:46

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

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]}

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