Finding Domain of f(x)=\frac{1}{x^{2}-x-6}

[badatmath]

I'm supposed to find the domain of this function, I know I'm supposed to set the denominator equal to 0 but I don't know what to do after that, please help!

$$f(x)=\frac{1}{x^{2}-x-6}$$

I can't figure out how to make this look like a real fraction but it's supposed to be 1 over x squared minus x minus 6.

 

[MarkFL]

Okay, to find the values that need to be excluded from the domain because they cause division by zero, we set:

$$x^2-x-6=0$$

Can the quadratic on the left perhaps be factored?

 

[badatmath]

So I add 6. That makes x^2-x=6, and then what?

 

[MarkFL]

So I add 6. That makes x^2-x=6, and then what?

No, you want to ask yourself, "Are there two factors of -6 whose sum is -1?"

Look at the following:

$$(x+a)(x+b)=x^2+(a+b)x+ab$$

So, you need to find an $a$ and a $b$ such that $ab=-6$ and $a+b=-1$...what do you find?

 

[badatmath]

I figured it out!

$$x^2-x-6=0$$

$$(x-3)(x+2)=0$$

$$x\in\{-2,3\}$$

And so the domain, in interval notation, is:

$$(-\infty,-2)\,\cup\,(-2,3)\,\cup\,(3,\infty)$$

 

[Deveno]

The proper way to factor a quadratic is "to complete the square".

In general, if we have:

$x^2 + px + q = 0$, we can through some algebraic trickery, make it be:

$x^2 + 2\left(\dfrac{p}{2}\right)x + \dfrac{p^2}{4} - \dfrac{p^2}{4} + q = 0$

or:

$\left(x + \dfrac{p}{2}\right)^2 = \dfrac{p^2 - 4q}{4}$.

In your case, we have $p = -1$, and $q = -6$, and the "ugly" expression on the right becomes:

$\dfrac{1 - 4(-6)}{4} = \dfrac{25}{4} = \left(\dfrac{5}{2}\right)^2$.

This is most convenient, since we can now "just take square roots", either:

$x - \dfrac{1}{2} = \dfrac{5}{2}$, or:

$x - \dfrac{1}{2} = -\dfrac{5}{2}$

Although MarkFL's approach is valid, it only works on "easily factorable" quadratics, and you might not know ahead of time if you have one.

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I figured it out!

$$x^2-x-6=0$$

$$(x-3)(x+2)=0$$

$$x\in\{-2,3\}$$

And so the domain, in interval notation, is:

$$(-\infty,-2)\,\cup\,(-2,3)\,\cup\,(3,\infty)$$

Good job!