Here we have the function f(x) = 2x+3, written as a flow diagram:
The Inverse Function goes the other way:
So the inverse of: 2x+3 is: (y3)/2
The inverse is usually shown by putting a little "1" after the function name, like this:
f^{1}(y)
We say "f inverse of y"
So, the inverse of f(x) = 2x+3 is written:
f^{1}(y) = (y3)/2
(I also used y instead of x to show that we are using a different value.)
The cool thing about the inverse is that it should give us back the original value:
When the function f turns the apple into a banana,
Then the inverse function f^{1} turns the banana back to the apple
Example: Using the formulas from above, we can start with x = 4:
f(4) = 2×4+3 = 11
We can then use the inverse on the 11:
f^{1}(11) = (113)/2 = 4
And we magically get 4 back again!
We can write that in one line:
f^{1}( f(4) ) = 4
"f inverse of f of 4 equals 4"
So applying a function f and then its inverse f^{1} gives us the original value back again:
f^{1}( f(x) ) = x
We could also have put the functions in the other order and it still works:
f( f1(x) ) = x
Example: Start with:
f^{1}(11) = (113)/2 = 4
And then:
f(4) = 2×4+3 = 11
So we can say:
f( f^{1}(11) ) = 11
"f of f inverse of 11 equals 11"
We can work out the inverse using Algebra. Put "y" for "f(x)" and solve for x:
This method works well for more difficult inverses.
A useful example is converting between Fahrenheit and Celsius:
To convert Fahrenheit to Celsius: f(F) = (F  32) × 5/9
The Inverse Function (Celsius back to Fahrenheit): f^{1}(C) = (C × 9/5) + 32
For you: see if you can do the steps to create that inverse!
It has been easy so far, because we know the inverse of Multiply is Divide, and the inverse of Add is Subtract, but what about other functions?
Here is a list to help you:
Did you see the "Careful!" column above? That is because some inverses work only with certain values.
Example: Square and Square Root
When we square a negative number, and then do the inverse, this happens:
Square: (−2)^{2 }= 4
Inverse (Square Root): √(4) = 2
But we didn't get the original value back! We got 2 instead of −2. Our fault for not being careful!
So the square function (as it stands) does not have an inverse
But we can fix that!
Restrict the Domain (the values that can go into a function).
Example: (continued)
Just make sure we don't use negative numbers.
In other words, restrict it to x ≥ 0 and then we can have an inverse.
So we have this situation:
No Inverse?
Let us see graphically what is going on here:
Just think ... if there are two or more xvalues for one yvalue, how do we know which one to choose when going back?
Imagine we came from x1 to a particular y value, where do we go back to? x_{1} or x_{2}?
In that case we can't have an inverse.
But if we can have exactly one x for every y we can have an inverse.
It is called a "onetoone correspondence" or Bijective, like this
So a bijective function follows stricter rules than a general function, which allows us to have an inverse.
So what is all this talk about "Restricting the Domain"?
In its simplest form the domain is all the values that go into a function (and the range is all the values that come out).
As it stands the function above does not have an inverse, because some yvalues will have more than one xvalue.
But we could restrict the domain so there is a unique x for every y ...
... and now we can have an inverse:
Note also:
Let's plot them both in terms of x ... so it is now f^{1}(x), not f^{1}(y):
f(x) and f^{1}(x) are like mirror images (flipped about the diagonal).
In other words:
The graph of f(x) and f^{1}(x) are symmetric across the line y = x
Example: Square and Square Root (continued)
First, we restrict the Domain to x ≥ 0:
And you can see they are "mirror images" of each other about the diagonal y=x.
Note: when we restrict the domain to x ≤ 0 (less than or equal to 0) the inverse is then f^{1}(x) = −√x:
Which are inverses, too.
It is sometimes not possible to find an Inverse of a Function.
Example: f(x) = x/2 + sin(x)
We cannot work out the inverse of this, because we cannot solve for "x":
y = x/2 + sin(x)
y ... ? = x
Even though we write f1(x), the "1" is not an exponent (or power):
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