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

  • Thread starter Thread starter badatmath
  • Start date Start date
  • Tags Tags
    Domain
AI Thread Summary
To find the domain of the function f(x)=1/(x^2-x-6), the first step is to set the denominator equal to zero, resulting in the equation x^2-x-6=0. This quadratic can be factored into (x-3)(x+2)=0, leading to the solutions x=-2 and x=3, which are the values that must be excluded from the domain. Consequently, the domain in interval notation is (-∞,-2)∪(-2,3)∪(3,∞). The discussion highlights the importance of factoring and provides an alternative method of completing the square for solving quadratics.
badatmath
Messages
3
Reaction score
0
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.
 
Last edited by a moderator:
Mathematics news on Phys.org
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?
 
So I add 6. That makes x^2-x=6, and then what?
 
badatmath said:
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?
 
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:

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

- - - Updated - - -

badatmath said:
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!
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Back
Top