# Representing a Function

There are three possible ways to represent a function:

• analytically (by formula);
• numerically (by table of values);
• visually (by graph).

Analytical representation of a function.

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

For example, ${y}=\frac{{1}}{{{1}+{{x}}^{{2}}}}$, ${y}={3}{{x}}^{{2}}-{5}$, etc.

Now, from the function ${y}=\frac{{1}}{{{1}+{{x}}^{{2}}}}$, we can find that ${y}{\left({2}\right)}=\frac{{1}}{{{1}+{{2}}^{{2}}}}=\frac{{1}}{{5}}$.

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

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

Another example is ${f{{\left({x}\right)}}}={x}!={1}\cdot{2}\cdot{3}\cdot\ldots\cdot{x}$. The domain of this function is a set of natural numbers, because ${x}$ can take only natural values. So, ${f{{\left({3}\right)}}}={3}!={6}$, but again, there is no formula that expresses ${x}!$.

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

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

But often, the domain is given implicitly, i.e. the domain is all values of ${x}$ where ${f{{\left({x}\right)}}}$ is defined.

For example, for the function ${f{{\left({x}\right)}}}={{x}}^{{2}}+{1}$, the domain is a set of all real numbers, i.e. the interval ${\left(-\infty,\infty\right)}$, because the function is defined for any value of ${x}$.

The domain of the function ${h}{\left({x}\right)}=\frac{{1}}{{{x}-{2}}}$ is all real numbers except ${x}={2}$, i.e. ${\left(-\infty,{2}\right)}\cup{\left({2},\infty\right)}$. ${x}={2}$ is not in the domain, because ${h}{\left({2}\right)}$ is not defined (the denominator equals ${0}$).

The domain of the function ${y}{\left({x}\right)}=\sqrt{{{1}-{{x}}^{{2}}}}$ is ${\left[-{1},{1}\right]}$, because for these values the expression under the root is non-negative, and, thus, the function ${y}{\left({x}\right)}$ is defined.

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

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

Let's suppose that one side is ${x}$ and the second side is ${u}$. Then, ${2}{x}+{2}{u}={24}$, or ${u}={12}-{x}$.

Therefore, the area is ${A}={x}{u}={x}{\left({12}-{x}\right)}$. Thus, ${A}{\left({x}\right)}={x}{\left({12}-{x}\right)}$.

Since length should be positive, ${x}>{0}$. Also, area should be positive; so, ${12}-{x}>{0}$, or ${x}<{12}$. Thus, the domain is the interval ${\left({0},{12}\right)}$.

Let's work another quick example.

Example 2. State the domain of ${f{{\left({x}\right)}}}=\frac{{1}}{{{{x}}^{{2}}-{3}{x}}}$.

Since ${{x}}^{{2}}-{3}{x}={x}{\left({x}-{3}\right)}$ and the denominator can't equal ${0}$, the domain of the function is all real numbers except ${x}={0}$ and ${x}={3}$.

One more quick example for practice.

Example 3. Find the domain of the function ${f{{\left({t}\right)}}}=\sqrt{{{t}}}+{\sqrt[{{3}}]{{{t}}}}$.

Since the square root should be non-negative, the domain of the function is ${t}\ge{0}$.

And let's do another short example before passing on.

Example 4. Find ${f{{\left({x}+{h}\right)}}}$, if ${f{{\left({x}\right)}}}=\frac{\sqrt{{{x}}}}{{{{x}}^{{2}}+{3}{x}+{4}}}$.

${f{{\left({x}+{h}\right)}}}=\frac{\sqrt{{{x}+{h}}}}{{{{\left({x}+{h}\right)}}^{{2}}+{3}{\left({x}+{h}\right)}+{4}}}$.

Numerical (tabular) representation of a function.

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

Graphical representation of a function.

Although in calculus functions are not defined graphically, graphical representation is very helpful in studying functions.

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

So, to draw the graph of the function, take from the interval ${X}$ points that are close to each other: ${x}_{{1}},{x}_{{2}},{x}_{{3}},\ldots{x}_{{n}}$. Now, find the corresponding y-values: ${y}_{{1}}={f{{\left({x}_{{1}}\right)}}}$, ${y}_{{2}}={f{{\left({x}_{{2}}\right)}}}$,..., ${y}_{{n}}={f{{\left({x}_{{n}}\right)}}}$.

Draw the points ${\left({x}_{{1}},{y}_{{1}}\right)}$, ${\left({x}_{{2}},{y}_{{2}}\right)}$,..., ${\left({x}_{{n}},{y}_{{n}}\right)}$. Connect these points by a smooth curve. We've obtained the graph of the function. Note: the more points you take, the more accurate graph you will obtain.

Let's move on to practical examples.

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

Let's take the 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 the corresponding y-values: ${y}_{{1}}={{\left({0}\right)}}^{{2}}={0}$, ${y}_{{2}}={{\left({0.2}\right)}}^{{2}}={0.04}$, ${y}_{{3}}={{\left({0.4}\right)}}^{{2}}={0.16}$, ${y}_{{4}}={{\left({0.6}\right)}}^{{2}}={0.36}$, ${y}_{{5}}={{\left({0.8}\right)}}^{{2}}={0.64}$, ${y}_{{6}}={{\left({1}\right)}}^{{2}}={1}$.

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

And a bit more work to memorize it better.

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

We take the point ${x}={0}$ and move up until the intersection of the curve. The intersection is at the point ${3}$; so, ${f{{\left({0}\right)}}}={3}$. Similarly, ${f{{\left({2}\right)}}}={1}$.

The function is defined when $-{1}\le{x}\le{3}$; therefore, the domain of the function is the interval ${\left[-{1},{3}\right]}$, and the range is ${\left[{1},{5.5}\right]}$.

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 to determine easily whether the curve is a function.

Consider the graph of the function ${{y}}^{{2}}+{{x}}^{{2}}={4}$. It is a cirlcle with a radius of 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 ${y}=\pm\sqrt{{{4}-{{x}}^{{2}}}}$. Because of $\pm$, we have two values of ${y}$ for every value of ${x}$. Also, as can be seen from the graph, the function fails the vertical line test.

Parametric representation of a function.

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

When we define a function by a formula, we use the following record: ${y}={f{{\left({x}\right)}}}$, where ${f{}}$ is some formula.

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

If we define a function as ${x}={u}{\left({t}\right)},{y}={v}{\left({t}\right)}$, by changing the value of ${t}$ (parameter), we can find the corresponding pairs of ${x}$ and ${y}$.

In general, we set some restrictions on ${t}$. If we require ${a}\le{t}\le{b}$, the point ${\left({u}{\left({a}\right)},{v}{\left({a}\right)}\right)}$ is called initial, and the point ${\left({u}{\left({b}\right)},{v}{\left({b}\right)}\right)}$ is called terminal.

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

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

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

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

 ${t}$ ${x}$ ${y}$ $-{2}$ ${{\left(-{2}\right)}}^{{3}}+{1}=-{7}$ ${{\left(-{2}\right)}}^{{2}}-{1}={3}$ $-{1}$ ${{\left(-{1}\right)}}^{{3}}+{1}={0}$ ${{\left(-{1}\right)}}^{{2}}-{1}={0}$ ${0}$ ${{0}}^{{3}}+{1}={1}$ ${{0}}^{{2}}-{1}=-{1}$ ${1}$ ${{1}}^{{3}}+{1}={2}$ ${{1}}^{{2}}-{1}={0}$ ${2}$ ${{2}}^{{3}}+{1}={9}$ ${{2}}^{{2}}-{1}={3}$

The graph of this function is shown to the right.

Note that in this case we can eliminate the parameter ${t}$: from the first equation, ${t}={{\left({x}-{1}\right)}}^{{\frac{{1}}{{3}}}}$; plugging this equation for ${t}$ into the second equation gives ${y}={{\left({x}-{1}\right)}}^{{\frac{{2}}{{3}}}}-{1}$.

And one more final example.

Example 8. Draw the function ${x}={{t}}^{{2}}+{t},{y}={{t}}^{{3}}-{{t}}^{{2}}$, $-{2}\le{t}\le{2}$.

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

 ${t}$ ${x}$ ${y}$ $-{2}$ ${{\left(-{2}\right)}}^{{2}}+{\left(-{2}\right)}={2}$ ${{\left(-{2}\right)}}^{{3}}-{{\left(-{2}\right)}}^{{2}}=-{12}$ $-{1}$ ${{\left(-{1}\right)}}^{{2}}+{\left(-{1}\right)}={0}$ ${{\left(-{1}\right)}}^{{3}}-{{\left(-{1}\right)}}^{{2}}=-{2}$ ${0}$ ${{0}}^{{2}}+{0}={0}$ ${{0}}^{{3}}-{{0}}^{{2}}={0}$ ${1}$ ${{1}}^{{2}}+{1}={2}$ ${{1}}^{{3}}-{{1}}^{{2}}={0}$ ${2}$ ${{2}}^{{2}}+{2}={6}$ ${{2}}^{{3}}-{{2}}^{{2}}={4}$

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