Functions

A function is simply a rule that assigns a unique value of a dependent variable (e.g. \(f(x)\)) to each value of an independent variable (e.g. \(x\)): \(x \rightarrow f(x)\)

Something is not a function if it assigns multiple values of \(y\) for the same value of \(x\) (e.g. on a graph, a vertical line)
We can relate any independent variable (e.g. \(x\)) to any dependent variable (e.g. \(y\)) so get comfortable using variables other than \(x\) and \(y\)!

In its general form a function can be written as:

\[q = q(p)\]

“Quantity (\(q\)) is a function of price (\(p\))”
This expresses that there is a relationship between \(q\) and \(p\), it doesn’t tell us the specific form of that relationship
- \(q\) is the dependent or “endogenous” variable, its value is determined by \(p\)
- \(p\) is the independent or “exogenous” variable, its value is given and not dependent on other variables
- The specific form of this function might be:

\[q = 100-6p\]

The numbers 100 and 6 are known as parameters, they are parts of the quantitative relationship between quantity and price (the variables) that do not change
If we have values of \(p\), we can find the value of \(q(p)\):
- When \(p=10\) \[\begin{align*} q(p)&=100-6p\\ q(10)&=100-6(10)\\ q(10)&=100-60\\ q(10)&=40\\ \end{align*}\]
- When \(p=5\): \[\begin{align*} q(p)&=100-6p\\ q(5)&=100-6(5)\\ q(5)&=100-30\\ q(5)&=70\\ \end{align*}\]

Multivariate functions have multiple independent variables, such as:

\[q=f(k,l)\]

“Output (\(q\)) is a function of both capital (\(k\)) and labor (\(l\))”
In economics, we often restrict the domain and range of functions to positive real numbers, \(\mathbb{R}_+\), since prices and quantities are never negative in the real world
- Domain: the set of \(x\)-values
- Range: the set of \(y\)-values determined by the function

Inverse Functions

Many functions have a useful inverse, where we switch the independent variable and dependent variable
- For example, if we have the demand function:

\[q=100-6p\]

we may want find the inverse demand function, an equation where \(p\) is the dependent variable, rather than \(q\) (this is how we normally graph Supply and Demand functions!) - To do this, we need to solve the above equation for \(p\):

\[ \begin{align*} q&=100-6p &&\text{The original equation}\\ q+6p&=100 && \text{Add } 6p \text{ to both sides}\\ 6p&=100-q && \text{Subtract } q \text{ from both sides}\\ p&=\frac{100}{6}-\frac{1}{6}q && \text{Divide both sides by 6}\\ \end{align*} \]

Functions with Fractions

Many people are rusty on a few useful algebra rules we will need, one being how to deal with fractions in equations
To get rid of a fraction, multiply both sides of the equation by the fraction’s reciprocal (swap the numerator and denominator), which will yield just 1

\[\begin{align*} 100&=\frac{1}{4}x & & \text{The equation to be solved for} x\\ \frac{4}{1} \big(\frac{100}{1}\big) &=\frac{4}{1}\bigg(\frac{1}{4}x\bigg) && \text{Multiplying by the reciprocal of } \frac{1}{4} \text{, which is } \frac{4}{1} \\ \frac{400}{1}&=\frac{4}{4}x & & \text{Cross multiplying fractions}\\ 400 & = x & & \text{Simplifying}\\ \end{align*}\]

Alternatively (if possible), re-imagining the fraction as a decimal may help:

\[\begin{align*} 100&=\frac{1}{4}x && \text{The original equation}\\ 100&=0.25x && \text{Converting to a decimal}\\ 400 & = x && \text{Dividing both sides by }0.25\\ \end{align*}\]

Add fractions by finding a common denominator:

\[\begin{align*} \frac{4}{3} &+ \frac{2}{5}\\ \bigg(\frac{4 \times 5}{3 \times 5}\bigg) &+ \bigg(\frac{2 \times 3}{5 \times 3}\bigg)\\ \frac{20}{15}&+\frac{6}{15}\\ &= \frac{26}{15}\\ \end{align*}\]

Multiply fractions straight across the numerator and denominator

\[ \frac{4}{3} \times \frac{2}{5}=\frac{4 \times 2}{3 \times 5}=\frac{8}{15} \]