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:
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 gas 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.
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 factoring, $x\left(1-x\right)\ge0$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.
Consider $f\left(x\right)=x+3$f(x)=x+3 for the domain {$-5$−5, $-4$−4, $0$0, $1$1}. Complete the table of values. Plot the ordered pairs.
$x$x
$f\left(x\right)$f(x)
$\left(x,f\left(x\right)\right)$(x,f(x))
$-5$−5
$\editable{}$
$($($\editable{}$, $\editable{}$$)$)
$-4$−4
$\editable{}$
$($($\editable{}$, $\editable{}$$)$)
$0$0
$\editable{}$
$($($\editable{}$, $\editable{}$$)$)
$1$1
$\editable{}$
$($($\editable{}$, $\editable{}$$)$)
The function $f$f is used to determine the area of a square given its side length.
Which of the following values is not part of the domain of the function?
$-8$−8
$6.5$6.5
$9$9$\frac{1}{3}$13
$\sqrt{78}$√78
For $n\ge0$n≥0, state the area function for a side length of $n$n.
Plot the graph of the function $f$f.
The function $f\left(x\right)=\sqrt{x+1}$f(x)=√x+1 has been graphed.
State the domain of the function. Express as an inequality.
Is there a value in the domain that can produce a function value of $-2$−2?
Yes
No