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]}

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