   SEARCH HOME Math Central Quandaries & Queries  Question from Genius, a student: State the domain and range g(x)=x(x-1) Hi,

In looking for the domain of a function $f(x)$ it is often easier to look for values of $x$ that are not in the domain of $f(x).$ That is for functions from the real numbers to the real numbers you look for values of $x$ so that $f(x)$ does not produce a real number. Thus for example if $f(x) = \sqrt{x}$ then $x$ cannot be negative. For all other values of $x,\; f(x)$ is a real number so the domain of $f(x) = \sqrt{x}$ is all nonnegative real numbers.

For your function $g(x) = x(x-1)$ are there any real numbers $x$ so that $g(x)$ does not produce a real number?

Again for the range of a function $f(x)$ it is often easier to look for numbers $y$ which you never get as an outcome of the function $f(x).$ Thus for example if $f(x) = x^2$ then you never get a negative value of $f(x)$ but you can get all nonnegative numbers. Hence the range of $f(x)$ is all nonnegative numbers.

For the range of $g(x)$ I would prefer to write the expression $y = x(x - 1).$ I am going to show you two ways to approach this, one algebraically because I think that is what is expected and one graphically because I think it is easier, especially if you have software or a graphing calculator that will graph the function.

Geometric approach.

Plot the graph of $y = x(x-1).$ can you read off the graph the range of this expression?

Algebraic approach.

Rewrite the expression $y = x(x-1)$ as $x^2 - x - y = 0$ and solve for $x$ using the general quadratic. For what values of $y$ is there no $x$ so that $x^2 - x - y = 0?$

I suggest you solve the problem using both approach to see if your answers agree.

I hope this helps,
Penny      Math Central is supported by the University of Regina and the Imperial Oil Foundation.