# Representing a Function

There are three possible ways to represent function:

• Analytically (by formula)
• Numerically (by table of values)
• Visually (by graph)

Analytical representation of the function.

This is the most common way to represent function. We define formula that contains arithmetic operations on constant values and variable $x$ that we need to perform to obtain value of $y$.

For example, $y=1/(1+x^2)$, $y=3x^2-5$ etc.

Now, from function $y=1/(1+x^2)$ we can find that $y(2)=1/(1+2^2)=1/5$.

However, this is not the only way to represent function. We can represent it analytically without a formula.

For example, consider function $y=[x]$ (floor function). Clearly $[1]=1,[2.5]=2,[sqrt(13)]=3,[-pi]=-4$ etc. but there is no formula that expresses $[x]$.

Another example is $f(x)=x! =1*2*3*...*x$. Domain of this function is set of natural numbers because $x$ can take only natural values. So, $f(3)=3! =6$, but again thereis no formulas that expresses $x!$.

Now, let's talk about domain of the function.

We can state domain of the function explicitly, for example, $f(x)=x^2,3<=x<=5$. This means that $x$ can change only in interval $[3,5]$.

But often domain is given implicitly, i.e. domain is all values of $x$ where $f(x)$ is defined.

For example, for function $f(x)=x^2+1$ domain is set of all real numbers, i.e. interval $(-oo,oo)$, because function is defined for any value of $x$.

Domain of the function $h(x)=1/(x-2)$ is all real numbers except $x=2$, i.e. $(-oo,2)uu(2,oo)$. $x=2$ is not in domain because $h(2)$ is not defined (denominator equals 0).

Domain of the function $y(x)=sqrt(1-x^2)$ is $[-1,1]$ because for this values expression under root is non-negative and, so, function $y(x)$ is defined.

Moreover, in real world application we also need to make sure that function makes sense. For example, consider law of free fall $s=1/2 g t^2$. This function is defined for all $t$ but it doesn't make sense when $t<0$ (time can't be negative). Therefore, domain of the function in given situation is $(0,oo)$.

Example 1. Suppose perimeter of rectangle is 24 cm. Find its area as function of one side. Find domain of this function.

Let one side is $x$ and second side is $u$. Then $2x+2u=24$ or $u=12-x$.

Therefore, area $A=xu=x(12-x)$. Thus, $A(x)=x(12-x)$.

Since length should be positive then $x>0$. Also area should be positive, so $12-x>0$ or $x<12$. Thus, domain is interval $(0,12)$.

Example 2. State the domain of $f(x)=1/(x^2-3x)$ .

Since $x^2-3x=x(x-3)$ and denominator can't equal 0 then domain of the function is all real numbers except $x=0$ and $x=3$.

Example 3. Find the domain of the function $f(t)=sqrt(t)+root(3)(t)$ .

Since square root should be non-negative then domain of the function is $t>=0$ .

Example 4. Find $f(x+h)$ if $f(x)=sqrt(x)/(x^2+3x+4)$.

$f(x+h)=sqrt(x+h)/((x+h)^2+3(x+h)+4)$ .

Numerical (tabular) representation of the function.

In real life dependence between variables is formed based on experiments or observations. For example, at any given time $t$ we can calculate world population $N$. We say that $N$ is a function of $t$ but can't find explicit formula to express this. All we have is table of values as below:

 $t$ (in years) 1970 1980 1990 2000 2010 $N$ (in millions of people) 3835 4100 4545 5600 7300

However, in further topics of calculus we will be able to find function that will approximate this values.

Graphical Representation of the function.

Though in calculus functions are not defined graphically, but graphical representation is very helpful in studying of functions.

Suppose that we are given function $y=f(x)$ with domain $X$. Imagine on plane two perpendicular axes: x-axis and y-axis. Consider pair of corresponding values $x$ and $y$, where $x$ is taken from set $X$, and $y=f(x)$. Image of this pair is point $A(x,y)$ with x-coordinate $x$ and y-coordinate $y$. When variable $x$ is changing within interval $X$, this point describes some curve. This curve is graph of the function $y=f(x)$.

So, to draw graph of the function take from interval $X$ points that are close to each other: $x_1,x_2,x_3,...x_n$. Now find corresponding y-values: $y_1=f(x_1),y_2=f(x_2),...,y_n=f(x_n)$.

Draw points $(x_1,y_1),(x_2,y_2),...(x_n,y_n)$. Connect these points by smooth curve. We obtained graph of the function. Note, the more points you take, the more accurarte graph you will obtain.

Example 5. Draw graph of the function $y=x^2$ on interval $[0,1]$.

Let's take following points: $x_1=0,x_2=0.2,x_3=0.4,x_4=0.6,x_5=0.8,x_6=1$.

Now find corresponding y-values: $y_1=(0)^2=0$, $y_2=(0.2)^2=0.04$, $y_3=(0.4)^2=0.16$, $y_4=(0.6)^2=0.36$, $y_5=(0.8)^2=0.64$, $y_6=(1)^2=1$.

Draw points $(0,0),(0.2,0.04),(0.4,0.16),(0.6,0.36),(0.8,0.64),(1,1)$ on the xy-plane and connect them by smooth curve.

Example 6. Consider function whose graph is shown to the left. Find $f(0)$, $f(2)$ and state domain and range of $f$.

We take point $x=0$ and move up until intersection of curve. Intersection is at point 3, so $f(0)=3$. Similarly $f(2)=1$.

Function is defined when $-1<=x<=3$, therefore, domain of the function is interval $[-1,3]$, range is $[1,5.5]$.

The graph of the function is a curve in the xy-plane. But the question arises: which curves in the xy-plane are graphs of functions? This is answered by the following test.

Vertical Line Test. A curve in the xy-plane is the graph of a function of $x$ if and only if no vertical line intersects the curve more than once.

Graphical representation allows us easily to determine whether curve is function.

Consider graph of the function $y^2+x^2=4$. It is a cirlcle with radius 2. Clearly it is not a function.

To see why it is so express $y$ in terms of $x$: $y^2=4-x^2$ or $y=+-sqrt(4-x^2)$. Because of $+-$ we have two values of $y$ for every value of $x$. Also as can be seen from graph function fails vertical line test.

Parametric representation of the function.

Parametric representation is slightly changed form of representing function by formula.

When we define function by formula we use following record $y=f(x)$ where $f$ is some formula.

But often it is more convenient to represent function using two functions and parameter.

If we define function as $x=u(t),y=v(t)$ then by changing value of $t$ (parameter) we can find corresponding pairs of $x$ and $y$.

In general we set some restrictions on $t$. If we require $a<=t<=b$ then point $(u(a),v(a))$ is called initial, point $(u(b),v(b))$ is called terminal.

In some cases (but not always) it is possible to eliminate paremeter $t$ and obtain standard representation of $y$ as function of $x$ (or $x$ as a function of $y$).

Also, note that parametric equations can generate curve that are not functions.

Example 7. Draw graph of the function $x=t^3+1,y=t^2-1$, $-2<=t<=2$.

Let's find a couple of points that correspond to different value of $t$.

 $t$ $x$ $y$ -2 $(-2)^3+1=-7$ $(-2)^2-1=3$ -1 $(-1)^3+1=0$ $(-1)^2-1=0$ 0 $0^3+1=1$ $0^2-1=-1$ 1 $1^3+1=2$ $1^2-1=0$ 2 $2^3+1=9$ $2^2-1=3$

Graph of this function is shown to the right.

Note, that in this case we can eliminate parameter $t$: from first equation $t=(x-1)^(1/3)$, plugging this equation for $t$ in second equation gives $y=(x-1)^(2/3)-1$.

Example 8. Draw function $x=t^2+t,y=t^3-t^2$, $-2<=t<=2$.

Let's find a couple of points that correspond to different values of $t$.

 $t$ $x$ $y$ -2 $(-2)^2+(-2)=2$ $(-2)^3-(-2)^2=-12$ -1 $(-1)^2+(-1)=0$ $(-1)^3-(-1)^2=-2$ 0 $0^2+0=0$ $0^3-0^2=0$ 1 $1^2+1=2$ $1^3-1^2=0$ 2 $2^2+2=6$ $2^3-2^2=4$

Note, that this is not a function because it fails vertical line test. Also it is very hard to eliminate parameter $t$.