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

1.2 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 $$\begin{array}{cccc}100& =\frac{1}{4}x& & \text{The equation to be solved for}x\\ \frac{4}{1}(\frac{100}{1})& =\frac{4}{1}(\frac{1}{4}x)& & \text{Multiplying by the reciprocal of}\frac{1}{4}\text{, which is}\frac{4}{1}\\ \frac{400}{1}& =\frac{4}{4}x& & \text{Cross multiplying fractions}\\ 400& =x& & \text{Simplifying}\end{array}$$
• Alternatively (if possible), re-imagining the fraction as a decimal may help: $$\begin{array}{cccc}100& =\frac{1}{4}x& & \text{The original equation}\\ 100& =0.25x& & \text{Converting to a decimal}\\ 400& =x& & \text{Dividing both sides by}0.25\end{array}$$
• Add fractions by finding a common denominator: $$\begin{array}{cc}\frac{4}{3}& +\frac{2}{5}\\ (\frac{4\times 5}{3\times 5})& +(\frac{2\times 3}{5\times 3})\\ \frac{20}{15}& +\frac{6}{15}\\ & =\frac{26}{15}\end{array}$$
• Multiply fractions straight across the numerator and denominator $$\frac{4}{3}\times \frac{2}{5}=\frac{4\times 2}{3\times 5}=\frac{8}{15}$$

2.1 Slope-Intercept Form

• A linear function of two variables can be written in slope-intercept form: $$y=ax+b$$
  • $y$ is the dependent variable (on the vertical axis)
  • $x$ is the independent variable (on the horizontal axis)
  • $a$ is the slope of the line
    • $a=\frac{\text{rise}}{\text{run}}=\frac{\text{change in}y}{\text{change in}x}=\frac{\Delta x}{\Delta y}$
    • you may have been taught the slope as “$m$”, this is just personal taste, but again, get used to using different letters!
  • $b$ is the vertical-intercept, a constant number where the line crosses the vertical axis
    • if $y$ is the dependent variable, this is the “$y$-intercept”, $y$ where $x=0$
• Any point on the line has an $x$-coordinate and $y$-coordinate $(x_i,y_i)$

2.2 Other Forms

• A linear function can equivalently be expressed in the following form: $$a_1{x}_1+a_2{x}_2=c$$
  • $x_2$ is the dependent variable (on the vertical axis)
  • $x_1$ is the independent variable (on the horizontal axis)
  • $c$ is a constant
• This is a valid equation, but is difficult to visualize in the traditional graph as above. Simply solve for the dependent variable on the vertical axis $(x_2$ as if you were solving for $y)$:

$$\begin{array}{ccc}{a}_1{x}_1+a_2{x}_2=c& & \text{Original}\\ a_2{x}_2=c-a_1{x}_1& & \text{Subtracting}{x}_1\text{term}\\ x_2=\frac{c}{a_2}-\frac{a_1}{a_2}{x}_1& & \text{Dividing by}{a}_2\end{array}$$

• The vertical intercept is $\frac{c}{a_2}$

• The horizontal intercept is $\frac{c}{a_1}$

• The slope is $-\frac{a_1}{a_2}$

• This is extremely useful for dealing with constraints in constrained optimization problems: budget constraints and isocost lines

2.3 Drawing a Graph From an Equation

If we already have a linear equation that we would like to graph, we can follow these steps:

1. Take the equation and plug in two values, e.g. if we have:

$$p=\frac{1}{2}q+4$$

2. We can find two points on the graph. The easiest one to find is the vertical-intercept, where the line crosses the vertical axis, where $q=0$, so plug in $q=0$: $$\begin{array}{cc}p& =\frac{1}{2}\left(0\right)+4\\ p& =4\end{array}$$ Thus, one point is $(0,4)$. Note that the constant in the function itself is the $p-intercept$! So one valid point will always be $(0,b)$!

3. For our second point, let’s plug in $q=2$: $$\begin{array}{cc}p& =\frac{1}{2}\left(2\right)+4\\ p& =5\end{array}$$ Thus, another point is $(2,5)$

4. Now, plot the two points on the graph, and connect them with a line

Note: A quick shortcut to plot a line is to find the vertical intercept and plot that, and then find the next point using the slope. Here, start our line at 4 on the vertical axis, and then, as the slope is $\frac{1}{2}$, for every one unit increase in $q$, $p$ increases by $\frac{1}{2}$. Our second point, (2,5), is a 2 unit increase in $q$ resulting in a $1$ unit increase in $p$.

2.4 Finding an Equation from a Graph

In order to find the equation of an existing line, we follow these steps:

1. First, take two points on the line and find the slope, $a$, between them. Let’s pick $(1,6)$ and $(3,2)$. $$\begin{array}{cc}\text{Slope}& =\frac{rise}{run}\\ & =\frac{(p_2-p_1)}{(q_2-q_1)}\\ & =\frac{(2-6)}{(3-1)}\\ & =\frac{-4}{2}\\ & =-2\end{array}$$

There is a shortcut that we can use to find the slope faster by eye-balling the graph: When $q$ changes by 1, how many units does $p$ change? If we move from (1,6) to (2,5), $q$ increases by 1, but $p$ falls by 2. So the slope is $-2$. For every one unit increase in $q$, $p$ changes by -2.

2. Now with the slope, we need to find the vertical intercept, or $b$, we solve this by plugging in the slope and any point on the graph, we will use (1,6): $$\begin{array}{cc}p& =aq+b\\ \left(6\right)& =-2\left(1\right)+b\\ 6& =-2+b\\ 8& =b\end{array}$$ Note, there is another easy way to eye-ball what this value is. It is simply that $p$ value where $q=0$, or at what $p$ value the graph crosses the vertical axis. We can see it is at 8.

3. Thus, we have the slope and the intercept, so our equation is: $$p=-2q+8$$

3.1 Elasticity

Using logs and percentage changes helps us talk about elasticity, an extremely useful concept with vast applications all over economics. Elasticity measures the percentage change in one variable ($y$) as a response to a 1% change in another ($x$) at a particular value of $x$ and $y$. $${ϵ}_{yx}=\frac{\%\Delta y}{\%\Delta x}=\frac{\left(\frac{\Delta y}{y}\right)}{\left(\frac{\Delta x}{x}\right)}=\frac{\Delta y}{\Delta x}\ast \frac{x}{y}$$

• Interpretation: A 1% change in $x$ will lead to a ${ϵ}_{yx}$% change in $y$

For example, the price elasticity of demand measures the percentage change in quantity demanded to a 1% change in price (at a particular price point), note here: $x=P$ and $y=q$: $${ϵ}_D=\frac{\%\Delta q}{\%\Delta p}=\frac{\frac{\Delta q}{q}}{\frac{\Delta p}{p}}=\frac{\Delta q}{\Delta p}\ast \frac{p}{q}$$

• Note that $\frac{\Delta q}{\Delta p}$ is $\frac{1}{slope}$ of the demand curve (which is $\frac{\Delta p}{\Delta q}$)
• Note though we would technically multiply by $\frac{100}{100}$ to get percentage change, this term obviously is just 1. Elasticity is unitless.
• Note also that on a graph we usually express $q$ as our independent variable and $p$ as our dependent variable

3.2 Derivatives (Calculus)

Often, $\Delta y$ refers to a very small change in $y$, a marginal change in $y$. A rate of change is the ratio of two changes, such as the change between $x$ and $y=f\left(x\right)$: $$\frac{\Delta f\left(x\right)}{\Delta x}=\frac{f(x+\Delta x)-f\left(x\right)}{\Delta x}$$

• This measures how $f\left(x\right)$ changes as $x$ changes
• If $\Delta$ is infinitesimally small, then we have expressed the (first) derivative of $f\left(x\right)$ with respect to $x$, written variously as $f^{\prime }\left(x\right)$ or $\frac{df\left(x\right)}{dx}$ $$\frac{df\left(x\right)}{dx}=\underset{\Delta x\to 0}{lim}\frac{f(x+\Delta x)-f\left(x\right)}{\Delta x}$$

The derivative of a linear function $(y=ax+b)$ is a constant (i.e. the slope), $a$ $$\frac{df\left(x\right)}{dx}=a$$

The derivative of the first derivative is the second derivative of a function $f\left(x\right)$ with respect to $x$, denoted $f^{\prime \prime }\left(x\right)$ or $\frac{d^{2}f\left(x\right)}{d{x}^2}$

• The second derivative measures the curvature of a function
• It used for proving when a function has reached a maximum or minimum, or is concave or convex (often used in #Nonlinear-Functions-&-Optimization)

4.1 Optimization

For most curves, we often want to find the value where the function reaches its maximum or minimum (in general, these are types of “extrema”) along some interval

Formally, a function reaches a maximum at $x^{\ast }$ if $f\left(x^{\ast }\right)\ge f\left(x\right)$ for all $x$; or a minimum at $x^{\ast }$ if $f\left(x^{\ast }\right)\le f\left(x\right)$ for all $x$ - In the graph above, the function reaches a maximum at $x_1$ and a minimum at $x_2$

The maximum or minimum of a function occurs where the slope (first derivative) is zero, known in calculus as the “first-order condition” $$\frac{df\left(x^{\ast }\right)}{dx}=0$$

• To distinguish between maxima and minima, we have the “second-order condition”
  • A minimum occurs when the second derivative of the function is positive, and the curve is convex $$\frac{d^{2}f\left(x^{\ast }\right)}{d{x}^2}>0$$
  • A maximum occurs when the second derivative of the function is negative, and the curve is concave $$\frac{d^{2}f\left(x^{\ast }\right)}{d{x}^2}<0$$
  • An inflection point occurs where the second derivative of the function is zero
  • All three are known as “critical points”

This is often useful for unconstrained optimization, e.g. finding the quantity of output that maximizes profits

Note, if we have a multivariate function $y=f(x_1,x_2)$ and want to find the maximum or minimum ($x_1^{\star },x_2^{\star }$), the first order conditions (FOC, plural) are where all the partial derivatives (derivative with respect to $x_1$ and derivative with respect to $x_2$) are zero $$\begin{array}{cc}\frac{\partial f(x_1^{\ast },x_2^{\ast })}{\partial x_1}& =0\\ \frac{\partial f(x_1^{\ast },x_2^{\ast })}{\partial x_2}& =0\end{array}$$ There are second order conditions as well, to demonstrate whether an extremum is a maximum or minimum, but they are too complex to discuss here.

Often we want to find the maximum or minimum of a function over some restricted values of $(x_1,x_2)$, known as constrained optimization. This is one of the most important modeling tools in microeconomics, and will show up in many contexts.

• We want to find the maximum of some function: $$\begin{array}{c}\underset{x_1,x_2}{max}f(x_1,x_2)\\ \text{subject to}g(x_1,x_2)=c\end{array}$$
  • $f(x_1,x_2)$ is the “objective function” we wish to maximize (or minimize)
  • $g(x_1,x_2)=c$ is the “constraint” that limits us within some specified set of $x_1$ and $x_2$ values

Much of microeconomic modeling is about figuring out what an agent’s objective is (e.g. maximize profits, maximize utility, minimize costs) and what their constraints are (e.g. budget, time, output).

There are several ways to solve a constrained optimization problem (see Appendix to Ch. 5 in textbook), the most frequent (but requiring calculus) is Lagrangian multiplier method.

Graphically, the solution to a constrained optimization problem is the point where a curve (objective function) and a line (constraint) are tangent to one another: they just touch, but do not intersect (e.g. at point A below).

At the point of tangency (A), the slope of the curve (objective function) is equal to the slope of the line (constraint) - This is extremely useful and is always the solution to simple constrained optimization problems, e.g. - e.g. maximizing utility subject to income - e.g. minimizing cost subject to a certain level of output - We can find the equation of the tangent line using point slope form

$$y-y_1=m(x-x_1)$$ - We need to know the slope $m$, which we would know from the slope of the function at that point - We know $(x_1,y_1$) is the point of tangency

5.1 Substitution

First, we take one equation and solve for one of the variables. Here, we take the first, and solve for $x$:

$$\begin{array}{cc}12x+4y& =36\\ 12x& =36-4y\\ x& =3-\frac{1}{3}y\end{array}$$

Now we take this value and plug it into the other equation:

$$\begin{array}{cc}6x-3y& =3\\ 6(3-\frac{1}{3}y)-3y& =3\\ 18-2y-3y& =3\\ 18-5y& =3\\ -5y& =-15\\ y& =3\end{array}$$

Now that we know the value of one variable, plug it into either of the original equations to solve for the value of the other variable:

$$\begin{array}{cc}12x+4y& =36\\ 12x+4\left(3\right)& =36\\ 12x+12& =36\\ 12x& =24\\ x& =2\end{array}$$

We should verify that our variables are correct, so plug $x$ and $y$ into each equation and make sure it is true. Let’s start with the first equation:

$$\begin{array}{cc}12x+4y& =36\\ 12\left(2\right)+4\left(3\right)& =36\\ 24+12& =36✓\end{array}$$

We can do the same with the other equation:

$$\begin{array}{cc}6x-3y& =3\\ 6\left(2\right)-3\left(3\right)& =3\\ 12-9& =3✓\end{array}$$

5.2 Elimination

We will multiply the equations by constants to make the coefficients of one variable equal. Here, let us try to make the coefficients in front of each equation’s $x$ equal.

$$\begin{array}{cc}[12x+4y& =36]\ast 6\\ [6x-3y& =3]\ast 12\end{array}$$

To do so, we will multiply the first equation by 6, and the second equation by 12. (We multiply each equation by the coefficient in front of the other equation’s $x$ variable):

$$\begin{array}{cc}72x+24y& =216\\ 72x-36y& =36\end{array}$$

Now we subtract the second equation from the first, which should get rid of $x$:

$$\begin{array}{cc}72x+24y& =216\\ -[72x-36y& =36]\end{array}$$

Be careful to distribute the minus sign carefully:

$$\begin{array}{cc}[72x-72x]+[24y-(-36y\left)\right]& =[216-36]\\ 60y& =180\\ y& =3\end{array}$$

Now that we have the value of one variable, plug it in to either equation.

$$\begin{array}{cc}12x+4y& =36\\ 12x+4\left(3\right)& =36\\ 12x+12& =36\\ 12x& =24\\ x& =2\end{array}$$

Now that we have both variables, we can plug them in to each equation to double check them, same as before.