What is the domain and range of #1 / (x^2 + 5x + 6)#?

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Answer 1 · 2018-05-24T12:39:05

The domain is # x in (-oo,-3)uu(-3, -2)uu(-2, +oo)#. The range is #y in (-oo,-4]uu [0, +oo)#

The denominator is

#x^2+5x+6=(x+2)(x+3)#

As the denominator must be #!=0#

Therefore,

#x!=-2# and #x!=-3#

The domain is # x in (-oo,-3)uu(-3, -2)uu(-2, +oo)#

To find the range, proceed as follows :

Let #y=1/(x^2+5x+6)#

#y(x^2+5x+6)=1#

#yx^2+5yx+6y-1=0#

This is a quadratic equation in #x# and the solutions are real only if the discriminant is #>=0#

#Delta=b^2-4ac=(5y)^2-4(y)(6y-1) >=0#

#25y^2-24y^2+4y>=0#

#y^2+4y>=0#

#y(y+4)>=0#

The solutions of this inequality is obtained with a sign chart.

The range is #y in (-oo,-4]uu [0, +oo)#

graph{1/(x^2+5x+6) [-16.26, 12.21, -9.17, 5.07]}