How to find the Inverse of a Function

The inverse of a function is the function that undoes its operation. The notation for the inverse of the function f is f^-1. In function notation this can be written like this.

f^-1(f(x))=x
f(f^-1(x))=x

We can also write it in terms of composition of functions. (See How to find Compositions of Functions .)

f^-1_°f(x)=x
f_°f^-1(x)=x

Not all functions have inverses. A function must be one to one in order to have one. A one to one function passes the horizontal line test as well as the vertical line test. In the graph of a function, no vertical line can pass through more than one point. In order to be a one to one function the same also has to be true for horizontal lines, no horizontal line can pass through more than one point of the graph. In terms of the function as an operation, this means that there can be no collapsing, two different inputs can't give you the same output, so for example f(x)=x² isn't one to one, because when you put in negative numbers you get out the same thing as when you put in positive numbers.

Now suppose we have a function that is one to one. How can we find its inverse? Even this is not always possible, but for a good number of simple functions it is not too difficult and here is how to do it.
 

1. Replace f(x) with y.
2. Reverse the roles of x and y, that is replace every occurrence of x with y and every occurrence of y with x.
3. Solve for y in terms of x.
4. Replace y with f^-1(x).

Examples

In the following problems the instruction is to find f^-1.

Example 1:

f(x)=3x+2

Solution:

1. y=3x+2

2.

x=3y+2

3.

3y=x-2   y = x-2    3

4.

f-1(x)=

x-2   3

Example 2:

f(x)=x³+1

Solution:

1. y=x³+1

2.

x=y³+1 3.

y3=x-1

You can do this with cube root because there is no plus or minus needed, because every number has only one real cube root. If this had been a square instead of a cube, though, the function would not have been one to one, so it wouldn't have had an inverse.

4.

Example 3:

f(x)=

1  x+1

Solution:

1.

y=

1  x+1

2.

x=

1  y+1

3.

(y+1)x=1 xy+x=1 xy=1-x   y= 1-x   x

Here I am multiplying both sides by y+1 at the beginning to clear the denominator. Then we have to get y alone, so we get everything without a y on the other side of the equation and then divide by the coefficient on y.

4.

f-1(x)=

1-x   x

Example 4:

ã

f(x)=

2x   x-3

Solution:

1.

y=

2x   x-3

2.

x=

2y   y-3

3.

(y-3)x=2y yx-3x=2y -3x=2y-xy 2y-xy=-3x (2-x)y=-3x   y=  -3x  2-x =   3x x-2

This one is a little bit harder because there is more than one occurrence of y. What we have to do is think of y as the only variable and pretend that x is a number and then do what we would normally do in such a situation, get all of the variables on one side of the equation and all of the numbers on the other side. The factoring out of the y in the 5th step is like combining like terms, just like you would do if it had been 7y-3y. The only difference is that you can do the subtraction in 2-x, because you don't know what x is, so you just have to write it. At the very end I multiplied top and bottom by -1 to make it look prettier. This isn't really required, but it is better form, so nice to do if you think of it.

4.

f-1(x)=

3x   x-2