When you get into more complicated algebraic functions in algebra II, you may run across the inverse function. The inverse function is a tricky calculation, but it's not too difficult if you keep a few simple rules in mind.

Define what the function is. Set a rule where "y = some function of x." If you are working with a word problem, read the problem carefully to determine how you will set up this function and if there are limits to the function. Limits are what "x" is aloud to be equal to. An example of this is "y = 2x + 1" where "x = 0  20." This means between 0 and 20 on the xaxis, "y = 2 times x."

Replace x with y and y with x in the function. In the above example, this gives you something that looks like "x = 2y + 1." You are essentially switching the values of x and y.

Solve for y in your function. In the simple example above this would look like "y = (x  1)/2." This is the inverse function of "y = 2x + 1." Be sure to note that the limits, or the values that x is aloud to be, do not change, even though you have redefined what y is equal to. In the above example, x is still equal to 0 through 20.

Note some special examples in the trig functions. For example, the inverse function of sin is arc sin. This could be shown as "y = arc sin x" or "x = sin y," if you want to define the function in terms of y as a function of x.

Be sure you have the correct inverse notations. Do not notate your function as "1/y" as this is incorrect. If you take the above example and replace "y = 2x +1" with "y = 1/(2x +1)" you would not get the "y = (x  1)/2" answer.