How do you find the domain of #f(x) = (4x^2 - 9) /(x^2 + 5x + 6)#?

2 Answers
Michael · Jacobi J.
May 23, 2018

#x#= All real numbers except #-2# and #-3#

Explanation:

We know that the denominator cannot be zero because you cannot divide by zero. So we must find what x has to equal to make it zero so we know what it can't be.

#x^2+5x+6=0#

Factor

#(x+3)(x+2)=0#

Now we can set bot of these equal to zero

#x+3=0#

#x=-3#

#x+2=0#

#x=-2#

Therefore x is equal to all real numbers except -2 and -3.

Jacobi J.
May 23, 2018

#x inRR, x!=-2, -3#

Explanation:

Our numerator is a polynomial, so the domain is defined for all real numbers. The denominator, however, cannot be equal to zero, otherwise it would be undefined.

We can set the denominator equal to zero to find what #x# cannot be.

We have the following:

#x^2+5x+6=0#

To factor this, we can think of two numbers that have a sum of #5# and a product of #6#. After some trial and error, we come up with #2# and #3#, thus we have

#(x+2)(x+3)=0#

Setting both factors equal to zero, we arrive at

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

These are the only values for which the function is undefined, because they will make the denominator equal to zero, which is undefined.

#x inRR, x!=-2, -3#

Hope this helps!

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