Functions

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. $x$): $x\to f(x)$

• 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 $p=10$ $$\begin{array}{rl}q(p)& =100-6p\\ q(10)& =100-6(10)\\ q(10)& =100-60\\ q(10)& =40\end{array}$$
  • When $p=5$: $$\begin{array}{rl}q(p)& =100-6p\\ q(5)& =100-6(5)\\ q(5)& =100-30\\ q(5)& =70\end{array}$$

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 $p$: $$\begin{array}{rlrl}q& =100-6p& & \text{The original equation}\\ q+6p& =100& & \text{Add }6p\text{ to both sides}\\ 6p& =100-q& & \text{Subtract }q\text{ from both sides}\\ p& =\frac{100}{6}-\frac{1}{6}q& & \text{Divide both sides by 6}\end{array}$$

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

$$\begin{array}{rlrl}100& =\frac{1}{4}x& & \text{The equation to be solved for}x\\ \frac{4}{1}(\frac{100}{1})& =\frac{4}{1}(\frac{1}{4}x)& & \text{Multiplying by the reciprocal of }\frac{1}{4}\text{, which is }\frac{4}{1}\\ \frac{400}{1}& =\frac{4}{4}x& & \text{Cross multiplying fractions}\\ 400& =x& & \text{Simplifying}\end{array}$$

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

$$\begin{array}{rlrl}100& =\frac{1}{4}x& & \text{The original equation}\\ 100& =0.25x& & \text{Converting to a decimal}\\ 400& =x& & \text{Dividing both sides by }0.25\end{array}$$

• Add fractions by finding a common denominator:

$$\begin{array}{rl}\frac{4}{3}& +\frac{2}{5}\\ (\frac{4\times 5}{3\times 5})& +(\frac{2\times 3}{5\times 3})\\ \frac{20}{15}& +\frac{6}{15}\\ & =\frac{26}{15}\end{array}$$

• Multiply fractions straight across the numerator and denominator $$\frac{4}{3}\times \frac{2}{5}=\frac{4\times 2}{3\times 5}=\frac{8}{15}$$