### Functions and graphs: Relations and functions

### Isolation of a variable

Rewriting an implicit relationship between variables in a form in which one of the variables, say \(v\), is present on its own on the left-hand side of an equation, i.e., creating an equation of the form \(v = \mathrm{formula\;without\;}v\), is called the **isolation of the variable** \(v\text.\) The example below shows how it works.

Is \(x\) also a function of \(y\)? If so, what is the function definition?

In other words, can you express \(x\) in \(y\), in the form \(x=\mathrm{formula\;in\;} y\).

You can also get to the solution by entering intermediate equations:

Then you see if you are still on track, but in the end you must get to the equation in the form \(x=\ldots\)

Multiply both sides of the equation with \(5x-3\) and simplify:

\[\begin{aligned}

(5x-3)y &= (5x-3)\cdot \frac{9x}{(5x-3)}\\ \\

5xy-3y&=9x

\end{aligned}\] Now you have obtained an equation without denominators and subsequently only have to manipulate polynomial equations.

Move all terms with \(x\) to the left-hand side and move all terms without \(x\) to the right-hand side, and factorise:

\[\begin{aligned}

5xy-9x&= 3y\\ \\

x(5y-9)&=3y

\end{aligned}\] Division of the last equation by \(5y-9\) leads to the requested form: \[x=\frac{3y}{5y-9}\] This can be read as a definiton of a function \(x\) of \(y.\)

The above example may seem artificial, but it has a direct application in cell biology.

Suppose that an experiment in which a small molecule binds to a membrane protein shows that the concentration \(b\) of bound molecules is as follows related to the concentration \(c\) of unbound molecules \[b=\frac{E_0\cdot c}{K+c}\] where \(E_0\) is the total protein concentration and \(K\) is a dissociation constant. Suppose that the parameters \(E_0, K\) and the concentration \(b\) are known, then we can use the formula to compute \(c\). Instead of doing this for various values of \(b\) again and again, it is more convenient to write \(c\) as a function of \(b\). This can be done as follows: \[\begin{aligned} b=\frac{E_0\cdot c}{K+c} &\implies b\cdot (K+c)=E_0\cdot c &\blue{\mathrm{removal\;of\;fraction}} \\ \\ &\implies b\cdot K+b\cdot c=E_0\cdot c &\blue{\mathrm{expansion\;of\;brackets}}\\ \\ &\implies b\cdot K = E_0\cdot c - b\cdot c &\blue{\mathrm{everything\;with\;}c\mathrm{\;to\;the\;right}}\\ \\ &\implies b\cdot K = (E_0-b)\cdot c &\blue{\mathrm{factorisation}}\\ \\ &\implies \frac{b\cdot K}{(E_0-b)}=c &\blue{\mathrm{isolation\;of\;}c}\\ \\ &\implies c=\frac{K\cdot b}{(E_0-b)}&\blue{\mathrm{swapping\;left\;and\;right}}\end{aligned}\]

We will discuss in the remainder of this chapter the function concept and related concepts. In the following three chapters we discuss various standard functions.

Mathcentre video

Transposition or Rearrangement of Formulae (38:34)