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

1 Answer
Oct 13, 2017

Answer:

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

Explanation:

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]}