Domain of Quadratic Equations

[thornluke]

Hi guys,

I'm really confused in finding the domain of quadratic equations.

For example: when finding a suitable domain so that an inverse exists, why is the domain of x²-4 x>0
whilst,
the domain of 2x²+3 is x≥0

Can the domain of x²-4 be x≥0?

Furthermore, what is the largest domain and how do I know whether or not it is the largest?

Thanks,
Thorn.

 

[HallsofIvy]

Hi guys,

I'm really confused in finding the domain of quadratic equations.

For example: when finding a suitable domain so that an inverse exists, why is the domain of x²-4 x>0
whilst,
the domain of 2x²+3 is x≥0

Can the domain of x²-4 be x≥0?

Furthermore, what is the largest domain and how do I know whether or not it is the largest?

Thanks,
Thorn.

The assumptions you make for your first question are wrong. The domains of the two functions $x^2- 4$ and $2x^2+ 3$ are not different. There exist an infinite number of domains on which $x^2- 4$ or $2x^2+ 3$ are "one-to-one" and so have inverses. $x\ge 0$ and any subset of that are such domains. So are $x\le 0$ and any subsets of that.

As for finding the largest domain (which is what is normally meant by "domain" without other condition), look at the graph. For these quadratics, the simplest thing to do is to find the vertex of each parabola. For $x^2- 4$, the vertex is (0, -4) and for $2x^2+ 3$, it is (0, 3).

Each parabola "turns back" at its vertex so you have two different x values, on either side of the vertex, with the same y value- the function is not "one to one". As long as your set lies on one side of the vertex, it is a domain on which the function is invertible. A largest possible such domain is all points on one side of the vertex together with the vertex itself. Since the two functions, $x^2- 4$ and $2x^2+ 3$ both have vertex at x= 0, there are two such "largest domains", $x\ge 0$ and $x\le 0$, for both.

Also, since this question has nothing to do with "Abstract and Linear Algebra", I am moving it to "General Math".