## 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

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