Domains (graphs and equations)

Function Notation and domains

When we talk about, say, the function $f\left(x\right)=x^2+1$f(x)=x2+1, what we mean is, in four parts, that:

1. There is a function $f$f (something that does "work")
2. ...which takes a value $x$x (generally called $x$x but stands for some number)
3. ...that belongs to a set of allowable values (the set is called the domain) 
4. ...and maps it to another number (calculated in this case as $1$1 more than the square of $x$x). 

So $f\left(1\right)=1^2+1=2$f(1)=12+1=2, and $f\left(0\right)=0^2+1=1$f(0)=02+1=1, and $f\left(-7\right)=-7^2+1=50$f(−7)=−72+1=50 etc.

By allowable values, we mean one of two things.

Firstly there are sometimes natural restrictions on what the function can handle. Just as we can't put milk into a petrol tank and not expect problems, so we can't put certain numbers into certain functions and not expect mathematical problems.

For example the function $f\left(x\right)=\sqrt{x}$f(x)=√x, defined for real numbers, cannot handle negative real numbers. The function $f\left(x\right)=\frac{1}{x}$f(x)=1x​ cannot handle zero. These exclusions define what is called a natural domain. In our first example, $f\left(x\right)=x^2+1$f(x)=x2+1 has no exclusions and so its natural domain includes all real numbers.   

Secondly, there might be what could be called user-defined restrictions where certain restrictions are put onto the domain so that the function makes sense in the real world. For example, we might decide to define our function $f\left(x\right)=3.5x$f(x)=3.5x with a domain given by the set of positive integers, because we are actually using the function to determine the revenue on selling $x$x apples. Again we might define the function $f\left(t\right)=200t-4.9t^2$f(t)=200t−4.9t2 where $t$t is the elapsed time in seconds after the stroke of midnight on 31 December 2015. Negative time makes no sense, so we restrict the input to the time domain  $t\ge0$t≥0. 

An example:

Suppose we consider the function given by $f\left(x\right)=\sqrt{x-x^2}$f(x)=√x−x2, defined over the reals. 

If we try $x=2$x=2, we immediately find a problem, since $f\left(2\right)=\sqrt{2-2^2}=\sqrt{-2}$f(2)=√2−22=√−2. Clearly the natural domain is some subset of the real numbers that doesn't include $2$2. But how do we find it?

We know that we need $x-x^2\ge0$x−x2≥0, or after factorising, $x\left(1-x\right)\ge0$x(1−x)≥0. In words this says that the product of the two numbers $x$x and $1-x$1−x must be positive. A little thought leads to the realisation that $x$x must be confined to the interval $0\le x\le1$0≤x≤1.

Any number outside that interval will cause either $x$x or $1-x$1−x (but never both) to become negative, and hence the product will become negative as well. So the natural domain is given by $0\le x\le1$0≤x≤1.

This severe restriction on $x$x has, in this case, also restricted the size of the numbers the function can produce. Think about the possible values of  $x$x and $1-x$1−x. If $x=0.1$x=0.1, then $1-x=0.9$1−x=0.9 and so their product is $0.09$0.09. If $x=0.9$x=0.9, then $1-x=0.1$1−x=0.1 and the product is still $0.09$0.09. Perhaps there is a maximum size of the product when $x$x and $1-x$1−x are equal. If $x=0.5$x=0.5, then $1-x=0.5$1−x=0.5 and the product is $0.25$0.25. Try different values of $x$x to satisfy yourself that this is indeed the case.

If $x=0.5$x=0.5, then $f\left(x\right)=\sqrt{0.5-0.5^2}=1$f(x)=√0.5−0.52=1, and this seems to be the largest function value possible.

To progress further, we might call $y=f\left(x\right)$y=f(x), so that $y=\sqrt{x-x^2}$y=√x−x2. Now let's re-arrange this expression:

$y$y	$=$=	$\sqrt{x-x^2}$√x−x2
$y^2$y2	$=$=	$\left(x-x^2\right)$(x−x2)
$y^2$y2	$=$=	$x-x^2$x−x2
$x^2-x+y^2$x2−x+y2	$=$=	$0$0
$\left(x^2-x+\frac{1}{4}\right)+y^2$(x2−x+14​)+y2	$=$=	$\frac{1}{4}$14​
$\left(x-\frac{1}{2}\right)^2+y^2$(x−12​)2+y2	$=$=	$\left(\frac{1}{2}\right)^2$(12​)2

This is a circle, centre $\left(\frac{1}{2},0\right)$(12​,0), radius $\frac{1}{2}$12​. However, because our function can only produce non-negative values in the range $0\le y\le\frac{1}{2}$0≤y≤12​, then our function is the semicircle in the first quadrant as shown in the graph below.

 

Worked Examples

Question 1

Question 2

Question 3