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India
Class XI

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 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)=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 factorising, $x\left(1-x\right)\ge0$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

 

Outcomes

11.SF.RF.2

Definition of relation, pictorial diagrams, domain, co-domain and range of a relation. Function as a special kind of relation from one set to another. Pictorial representation of a function, domain, co-domain and range of a function. Real valued function of the real variable, domain and range of these functions, constant, identity, polynomial, rational, modulus, signum and greatest integer functions with their graphs. Sum, difference, product and quotients of functions.

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