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

2 Answers
Oct 25, 2017

Answer:

#-oo < x < oo#

#-1 <= y <= 1#

Explanation:

The domain is the set of real values that #x# can take to give a real value.

The range is the set of real values you can get out of the equation.

With fractions you often have to make sure that the denominator is not #0#, because you can't divide by #0#. However, here the denominator cannot equal #0#, because if

#x^2 + 9 = 0#

#x^2 = -9#

#x = sqrt(-9)#, which doesn't exist as a real number.

Therefore, we know we can put pretty much anything into the equation.

The domain is #-oo < x < oo#.

The range is found by recognising that #abs(x^2 + 9) >= abs(x + 3)# for any real value of #x#, which means that #abs((x+3)/(x^2+9)) <= 1#

This means that the range is

#-1 <= y <= 1#

Oct 25, 2017

Answer:

The domain is #x in RR# and the range is #y in [-0.069, 0.402]#

Explanation:

The domain is #x in RR# as the denominator is

#(x^2+9)>0, AA x in RR#

For the range, proceed as follows,

Let #y=(x+3)/(x^2+9)#

Then,

#yx^2+9y=x+3#

#yx^2-x+9y-3=0#

This is a quadratic equation in #x#

In order for this equation to have solutions, the discriminant #Delta>=0#

Therefore,

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

#1-36y^2+12y>=0#

#-36y^2+12y+1>=0#

#y=(-12+-sqrt(12^2-4(-36)(1)))/(2*-36)#

#y=(-12+-sqrt288)/(-72)=-((-1+-sqrt2)/(6))#

#y_1=(1+sqrt2)/6=0.402#

#y_2=(1-sqrt2)/6=-0.069#

Therefore,

The range is #y in [-0.069, 0.402]#

You can cofirm this with a sign chart and a graph

graph{(x+3)/(x^2+9) [-7.9, 7.9, -3.95, 3.95]}