Suppose you want to know the probability that you draw a jack or a red card from a standard 52-card deck.  You know there are four jacks in the deck, one of each of the four suits (hearts, diamonds, clubs and spades). There are 13 cards of each color, so there are 26 red cards. 

So one might think the probability is 4/52 for the jacks plus 26/52 for the red cards to get 30/52. But that isn't correct because if you do that you care basically counting the jack of hearts and jack of diamonds twice, since they are in both sets of outcomes.

To fix that, you have to subtract 2 from the total to get 28/52, simplified to 7/13.

The basic formula for two events that can both occur simultaneously, you take P(First event) + P(Second event) - P(intersection of the events).

Some students have difficulty determining the domain and range of a function. If you can graph the function using a graphing utility or making an x,y table and plotting points to create the graph, then seeing the domain and range is much simpler.

The domain is the set of x values, or the input of the function. The range is the output or the y-value of the function.

For example, if the function is f(x) = x^2 + 3x and the input is x = 1, the output is f(1), which is the same as y = (1)^2 + 3(1) = 4.

Looking at a graph, the amount the graph stretches horizontally is the domain and the amount the graph stretches vertically is the range.

If the function is just represented as an (x,y) table, the domain is all the x values and the range is all the y values.

Examples of using the chain rule, product rule and quotient rule for differentiation

f(x) = cos^3 (2x)

-3sin^2(2x)(2)

-6sin^2(2x)


f(x) = 2x(5x^2 + 3)^4

Use product rule and chain rule

2x(4(5x^2 + 3)^3)(10x)

80x^2(5x^2+3)

f(x) = 6x^2/(3x^2 + 4)^3

[(3x^2 + 4)^3(12x) - (6x^2)(3(3x^2 + 4)^2(6x))]/(3x^2 + 4)^6


Product rule is first term times derivative of the second plus second term times derivative of the first.

Quotient rule is denominator times derivative of numerator times numerator times derivative of denominator, all divided by denominator squared.

How do you find the inverse of a function? Here's some simple steps.

Suppose f(x) = 3x + 8

1. Substitute f(x) with y
2. Interchange y and x
3. Solve equation for y


In this example f(x) = 3x + 8 becomes y = 3x + 8.

Next interchange x and y to get x = 3y + 8.

Now solve equation for y to get (x  - 8)/3 = y

If you have two functions and you want to determine if they are inverses of each other, find the inverse of each. If the inverse matches the opposite function, then the functions are inverses.

For example,  f(x) = x^2 and g(x) = square root(x).  Get the inverse of f(x) first, which is

y = x^2

x = y^2

Solve for y to get y = square root(x) which is g(x).

Now get the inverse of g(x)

y = square root(x)

x = square root(y)

solve for y to get y = x^2, which is f(x). Therefore, f(x) and g(x) are inverses.

Here's a sample SAT problem that uses the concept of exponents.

2^x + 2^x + 2^x + 2^x  = 2^7.  What does x equal?

It's temping to say 7/4 because 2^x appears 4 times and 7/4 times 4 = 7. But that isn't correct.

You need to calculate 2^7 first, which is 128.

Now since 2^x is added to itself 4 times, you can rewrite the equation as

4(x^2) = 128

Therefore, dividing by 4 gives x^2 = 32. 

The value of x is five since 2 * 2 * 2 * 2 * 2 = 32

Suppose you have the graphs of f(x) and g(x) and want to find the derivative of f(x)g(x) at x = 2.  What you do is use the product rule for differentiation to get

f'(x)g(x) + f(x)g'(x), where f'(x) and g'(x) are the derivatives of f(x) and g(x), respectively.

Then using the graph you can get the values of f(x) and g(x) at x = 2. To get the derivative at x = 2, you look at the slope of the graph at that point on the graph and substitute those values into the equation.

Sometimes students have problems remembering formulas for area or volume. One easy way to know the volume of a cylinder is to think of a cylinder as a bunch of very thin circles stacked on top of each other. Think of a CD and stacking them on top of each other to a certain height. Assuming there was no hole in the middle of the CD, it's basically a cylinder, but if you think of it as just a circle, you can get the area of the circle and multiply it by the height of the stack. That is the volume of a cylinder.

We know the area of a circle is Pi times radius squared or Pi(r^2). Therefore the volume of a cylinder is Pi times radius squared times height or Pi(r^2)h.