 # INVERSE FUNCTIONS.

## Presentation on theme: "INVERSE FUNCTIONS."— Presentation transcript:

INVERSE FUNCTIONS

Remember we talked about functions---taking a set X and mapping into a Set Y
1 2 3 4 5 10 8 6 1 2 2 4 3 6 4 8 10 5 Set X Set Y An inverse function would reverse that process and map from SetY back into Set X

If we map what we get out of the function back, we won’t always have a function going back.
1 2 2 4 3 6 4 8 5 Since going back, 6 goes back to both 3 and 5, the mapping going back is NOT a function These functions are called many-to-one functions Only functions that pair the y value (value in the range) with only one x will be functions going back the other way. These functions are called one-to-one functions.

This would not be a one-to-one function because to be one-to-one, each y would only be used once with an x. 1 2 3 4 5 10 8 6 1 2 2 4 3 6 4 8 5 10 This is a function IS one-to-one. Each x is paired with only one y and each y is paired with only one x Only one-to-one functions will have inverse functions, meaning the mapping back to the original values is also a function.

Recall that to determine by the graph if an equation is a function, we have the vertical line test.
If a vertical line intersects the graph of an equation more than one time, the equation graphed is NOT a function. This is NOT a function This is a function This is a function

This is a many-to-one function This then IS a one-to-one function
To be a one-to-one function, each y value could only be paired with one x. Let’s look at a couple of graphs. Look at a y value (for example y = 3)and see if there is only one x value on the graph for it. For any y value, a horizontal line will only intersection the graph once so will only have one x value This is a many-to-one function This then IS a one-to-one function

This is NOT a one-to-one function This is NOT a one-to-one function
If a horizontal line intersects the graph of an equation more than one time, the equation graphed is NOT a one-to-one function and will NOT have an inverse function. This is NOT a one-to-one function This is NOT a one-to-one function This is a one-to-one function

Let’s consider the function and compute some values and graph them.
Notice that the x and y values traded places for the function and its inverse. These functions are reflections of each other about the line y = x Let’s consider the function and compute some values and graph them. This means “inverse function” x f (x) (2,8) (8,2) x f -1(x) Let’s take the values we got out of the function and put them into the inverse function and plot them (-8,-2) (-2,-8) Yes, so it will have an inverse function Is this a one-to-one function? What will “undo” a cube? A cube root

So geometrically if a function and its inverse are graphed, they are reflections about the line y = x and the x and y values have traded places. The domain of the function is the range of the inverse. The range of the function is the domain of the inverse. Also if we start with an x and put it in the function and put the result in the inverse function, we are back where we started from. Given two functions, we can then tell if they are inverses of each other if we plug one into the other and it “undoes” the function. Remember subbing one function in the other was the composition function. So if f and g are inverse functions, their composition would simply give x back. For inverse functions then:

Verify that the functions f and g are inverses of each other.
If we graph (x - 2)2 it is a parabola shifted right 2. Is this a one-to-one function? This would not be one-to-one but they restricted the domain and are only taking the function where x is greater than or equal to 2 so we will have a one-to-one function.

Verify that the functions f and g are inverses of each other.
Since both of these = x, if you start with x and apply the functions they “undo” each other and are inverses.

Steps for Finding the Inverse of a One-to-One Function
y = f -1(x) Solve for y Trade x and y places Replace f(x) with y

Let’s check this by doing
Find the inverse of y = f -1(x) or Solve for y Trade x and y places Yes! Replace f(x) with y Ensure f(x) is one to one first. Domain may need to be restricted.

Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint. Shawna has kindly given permission for this resource to be downloaded from and for it to be modified to suit the Western Australian Mathematics Curriculum. Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar