What is the domain and range of $f (x) = \frac{3 x - 1}{x^{2} + 9}$ $f(x)=(3x-1)/(x^2+9)$ ?

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Answer 1 · 2017-10-13T16:33:03

The domain is $x in RR$
The range is $f(x) in [-0.559,0.448]$

The function is $f(x)=(3x-1)/(x^2+9)$

$AA x in RR$ , the denominator is $x^2+9>0$

Therefore,

The domain is $x in RR$

To find the range, proceed as follows

Let $y=(3x-1)/(x^2+9)$

Rearranging,

$yx^2+9y=3x-1$

$yx^2-3x+9y+1=0$

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

$Delta=b^2-4ac=(-3)^2-(4)*(y)(9y+1)>=0$

$9-36y^2-4y>=0$

$36y^2+4y-9<=0$

Solving this inequality,

$y=(-4+-sqrt(4^2+4*9*36))/(2*36)=(-4+-sqrt1312)/(72)$

$y_1=(-4-36.22)/(72)=-0.559$

$y_2=(-4+36.22)/(72)=0.448$

We can make a sign chart.

The range is $y in [-0.559,0.448]$

graph{(3x-1)/(x^2+9) [-10, 10, -5, 5]}

What is the domain and range of f(x)=(3x-1)/(x^2+9)f(x)=3x−1x2+9 f(x)=(3x-1)/(x^2+9)?