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Grade 12

Domains (graphs and equations)

Lesson

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 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)=200t4.9t2 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  $t\ge0$t0

An example:

Suppose we consider the function given by $f\left(x\right)=\sqrt{x-x^2}$f(x)=xx2, 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)=222=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 $x-x^2\ge0$xx20, or after factoring, $x\left(1-x\right)\ge0$x(1x)0. In words this says that the product of the two numbers $x$x and $1-x$1x must be positive. A little thought leads to the realisation that $x$x must be confined to the interval $0\le x\le1$0x1.

Any number outside that interval will cause either $x$x or $1-x$1x (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$0x1.

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$1x. If $x=0.1$x=0.1, then $1-x=0.9$1x=0.9 and so their product is $0.09$0.09. If $x=0.9$x=0.9, then $1-x=0.1$1x=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$1x are equal. If $x=0.5$x=0.5, then $1-x=0.5$1x=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)=0.50.52=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=xx2. Now let's re-arrange this expression:

$y$y $=$= $\sqrt{x-x^2}$xx2
$y^2$y2 $=$= $\left(x-x^2\right)$(xx2)
$y^2$y2 $=$= $x-x^2$xx2
$x^2-x+y^2$x2x+y2 $=$= $0$0
$\left(x^2-x+\frac{1}{4}\right)+y^2$(x2x+14)+y2 $=$= $\frac{1}{4}$14
$\left(x-\frac{1}{2}\right)^2+y^2$(x12)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}$0y12, then our function is the semicircle in the first quadrant as shown in the graph below.

 

Worked Examples

Question 1

Consider $f\left(x\right)=x+3$f(x)=x+3 for the domain {$-5$5, $-4$4, $0$0, $1$1}.

  1. Complete the table of values.

    $x$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{}$$)$)
    $1$1 $\editable{}$ $($($\editable{}$, $\editable{}$$)$)
  2. Plot the ordered pairs.

    Loading Graph...

Question 2

The function $f$f is used to determine the area of a square given its side length.

  1. Which of the following values is not part of the domain of the function?

    $-8$8

    A

    $6.5$6.5

    B

    $9$9$\frac{1}{3}$13

    C

    $\sqrt{78}$78

    D
  2. For $n\ge0$n0, state the area function for a side length of $n$n.

  3. Plot the graph of the function $f$f.

    Loading Graph...

Question 3

The function $f\left(x\right)=\sqrt{x+1}$f(x)=x+1 has been graphed.

Loading Graph...

  1. State the domain of the function. Express as an inequality.

  2. Is there a value in the domain that can produce a function value of $-2$2?

    Yes

    A

    No

    B

 

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