CALCULUS OF INVERSE FUNCTIONS

What is an Inverse Function?

An Inverse Function \( f^{-1} \) is an 'Undo' Function, it undoes what the original function \( f \) did to the entered values.
Pretty simple. Let me give you a few examples:

Example 1: If \( f(x) = x+5 \) then \( f^{-1}(x) = x-5 \) .
The undo function of the 'add 5' function \( f(x) = x+5 \) is of course the 'subtract 5' function \( f^{-1}(x) = x - 5 \).
For instance: \( f(100) = 105 \) then \( f^{-1}(105) = 105 - 5 = 100 \).
We started and ended with 100.

Example 2: If you travel to Mexico and get 20 Pesos for 1 Dollar then \( f(x) = 20x \). After your travels you exchange your Pesos \( f^{-1}(x) = (1/20)x \) .
The undo function of the 'times 20' function \( f(x) = 20x \) is of course the 'divide by 20' function \( f^{-1}(x) = (1/20)x \) .
For instance: \( f(100) = 20*100 = 2000 \) then \( f^{-1}(2000) = 2000 / 20 = 100 \).
Again, we started and ended with 100.

Example 3: If \( f(x) = 20x+5 \) then \( f^{-1}(x) = (x-5)/20 \) .
Here, since we first multiply by 20 and then add 5 the Inverse Function first must subtract 5 then divide by 20. \( f^{-1}(x) = (x-5)/20 \) .
For instance: \( f(100) = 20*100 + 5 = 2005 \)
then \( f^{-1}(2005) = (2005 - 5)/20 = 100 \).
Again, we started and ended with 100.

Example 4: If \( f(x) = 5/9*(x-32) \) then \( f^{-1}(x) = 9/5*x+32 \), the Fahrenheit to Celsius conversion.
Here, since we first subtract by 32 and then multiply by 5/9 the Inverse Function, the Celsius to Fahrenheit conversion, we first must divide by 5/9 (or muliply by 9/5) then add 32. \( f^{-1}(x) = 9/5x+32 \) .
For instance: \( f(86) = 5/9*(86 - 32) = 5/9*54 = 30 \)
then \( f^{-1}(30) = 9/5*30 + 32 = 54+32 = 86 \).
Again, we started and ended with 86.

An Inverse Functions \(f^{-1} \) 'undoes' what a Function f does to an entered value a, so we can write:

\( f(a)=c\)   then   \(f^{-1}(c) = a \)

4 Conclusions on Inverse Functions

1) Inverse Functions are unique. Every Function has its dedicated Inverse Function.

2) The Output,c, or Range, of a Function f is the Input, or Domain, of its Inverse Function.

\( f(a)=c\)   then  \(f^{-1}(c) = a \)

3) The Input,a, or Domain, of a Function f is the Output, or Range, of its Inverse Function.

\( f(a)=c\)   then   \(f^{-1}(c) = a \)

We can summarize 2) and 3) as follows:




4) A Function f and its Inverse have swapped x and y coordinates: \( f(a)=c\)   then   \(f^{-1}(c) = a \) . A table may look you as follows:


We will use this property to find and also graph Inverse Functions below.

5) The composition of \( f \) and its Inverse \( f^{-1} \) :
Plugging the output of \( f \), \( f(a)\), into the Inverse \( f^{-1} (x) \) is written as \( f^{-1}(f(a)) \)
The Output of the Composition just yields the Input, we may write:

    \( f^{-1}(f(a)) = a \)   or   \( f(f^{-1}(c)) = c \)

We will use this property to find the Derivatve of Inverse Functions below.

How do I find Inverse Functions?

Since a Function \( f \) and its Inverse Function \( f^{-1} \) have swapped x and y coordinates we just swap x and y in the given equation.
Again, this is pretty simple. Let me give you a few examples:

Example 1: The Inverse Function of \( y = x+5 \) is found by swapping x and y : \( x = y+5 \) and solving for the new y by subtracting 5: \( x-5 = y \).
Thus, the Inverse Function is \( f^{-1}(x) = x - 5 \).

Example 2: The Inverse Function of \( f(x) = 20x+5 \) is found by swapping x and y : \( x = 20y+5 \) and
solving for the new y yields: \( 20x = (y-5) \) and \( x = (y-5)/20 \).
Thus, the Inverse Function is \( f^{-1}(x) = (x-5)/20 \).

Example 3: If \( f(x) = 5/9*(x-32) \) then \( f^{-1}(x) = 9/5*x+32 \), the Fahrenheit to Celsius conversion.
Since we first subtract by 32 and then multiply by 5/9 the Inverse Function,
to do the Celsius to Fahrenheit conversion, we first must divide by 5/9 (or muliply by 9/5) then add 32.
Thus, the Inverse is \( f^{-1}(x) = 9/5x+32 \) .

How do I Graph Inverse Functions?

Since a Function \( f \) and its Inverse Function \( f^{-1} \) have just swapped x and y coordinates we just swap x and y on our graphs. Again pretty simple. Let me give you a few examples:

Example 1: The Inverse Function of \( y = x+5 \) is \( f^{-1}(x) = x - 5 \) . To Graph the Inverse notice the swapped x and y coordinates.


Notice the swapped x and y-coordinates of the 4 intercepts.

Example 2: The Inverse Function of \( f(x) = 5/9*(x-32) \) is \( f^{-1}(x) = 9/5*x+32 \) .
To Graph the Inverse we just swap x and y coordinates.


Notice the swapped x and y-coordinates of the 4 intercepts.

Also, notice that both Inverse Functions in blue are reflections over the green y=x, the 45-degree line, as shown below:


Notice the 2 Functions never intersect since they have the same slope.

The Graphs in Example2 do intersect because of their different slopes 5/9 and 9/5.

Notice again the swapped x and y-coordinates of the 4 intercepts.
Corresponding points have the same distances to the y=x line.

Horizontal Line Test: Are Inverses Functions?

No, not always! Many Inverses do pass the Vertical Test and are thus Functions.
The Vertical Line Test applied to an Inverse \( f^{-1}(x) \) states: If any Vertical Line placed on the Graph of an Inverse intersects no more than once, then the Inverse is a Function (and not a Relation).
That is: Each x in the Domain of \( f^{-1}(x) \) corresponds to one y in its Range.
Mathematician's say that the function is injective.

Let us now apply both the known Vertical Line Test and the new Horizontal Line Test using the previous 2 examples:

Example 1: The Inverse \( f^{-1}(x) = x - 5 \) is a Function as any vertical line intersects only once, see below:


Instead of applying the Vertical Line Test on the Inverse \( f^{-1}(x) \) we can apply the Horizontal Line Test on \( f(x) \):
If any Horizontal Line placed on the Graph of a Function \( f(x) \) intersects no more than once, then the Inverse is a Function (and not a Relation).

Let us apply the Horizontal Line Test on \( f(x) = x + 5 \):



Since no horizontal line intersects the graph of \( f(x) = x + 5 \) more than once the Horizontal Line Test concludes that the Inverse \( f^{-1}(x) = x - 5 \) is indeed a Function.
In fact, the Inverse of any Linear Function as a Function itself simply because horizontal lines cannot intersect another lines more than once (unless they lie on top of each other).

Example 2: The Inverse of \( f(x) = 5/9*(x-32) \) is \( f^{-1}(x) = 9/5*x+32 \) which is a Function.
Applying the Vertical Line Test on \( f(x) = 5/9*(x-32) \) shows no more than one intersection between its Graph and the vertical lines, see below:


Alternatively, we may apply the Horizontal Line Test on \( f(x) = 5/9*(x-32) \) to see that its Inverse is a Function because no more than one intersection between its Graph and the horizontal lines occur, see below:




Example 3: The Inverse of \( f(x) = x^{2} \) is not a Function since the Horizontal Line Test on \( f(x) = x^2 \) shows two intersection between its Graph and the green dotted horizontal lines (which means 2 intersections between the vertical lines and the red Inverse). See below:



2 Notes on the Horizontal and Vertical Line Tests:
1) The advantage of applying the Horizontal Line Test on \( f(x) \) over the Vertical Line Test on \( f^{-1}(x) \) lies in the fact that we don't even need to find the Inverse to determine if \( f^{-1}(x) \) is a Function or a Relation.
2) The Horizontal Line Test on \( f(x) \) and the Vertical Line Test on \( f^{-1}(x) \) can be used interchangably. The reason lies in the fact that the Function \( f(x) \) along with vertical lines are to be reflected over the y=x (the 45 degree) line to obtain the graph of the Inverse \( f^{-1}(x) \) along with the horizontal lines.

Calculus of Inverse Functions: How do I find the Derivative of an Inverse?

We start by differentiating the composition of \( f \) and its Inverse \( f^{-1} \):


       \( f^{-1}(f(x)) = x \)


Applying the Chain Rule we obtain:


       \( d/dx(f^{-1}(f(x)))*f'(x) = 1 \)


Dividing by \( f'(x) \) yields the Derivative of the Inverse:


       \( d/dx(f^{-1}(f(x))) = 1/f'(x) \)


Since we will be finding the slope of the Inverse at a given point \( f^{-1}(c) \) we write f(a) = c to get the Slope of the Inverse at x=a:

This formula shows us that Inverses have reciprocal slopes at corresponding points which is a result of the reflection of the Inverse over the 45 degree line. The above examples suggested that a steeper Function \( f \) results in a flatter Inverse \( f^{-1} \).

When finding the slope at a point of the Inverse \( f^{-1} \), the input c of the Inverse \( f^{-1} \) equals output of the Function f(a). So we must do the following 3 steps:

  1) Find the value a that yields f( a ) = c .

  2) Find the Derivative of f .

  3) Find the Reciprocal .

Lets apply the above 3 steps to the following examples below:

Example 1: Find the Slope of the Inverse of \( f(x) = 3x+5 \) when c , that is \( f^{-1} ( \)8\() \).
1) The value a that yields f(a)=8 is a=1 since f(1)=8 .
2) The Derivative of 3x+5 is \( f'(x)=3 \) .
3) The Reciprocal of 3 is 1/3, thus \( f^{-1} ( \)8\()=1/3 \) .

Example 2: Find the Slope of the Inverse of \( f(x) = 5/9*(x-32) \) when c=0 , that is \( f^{-1} (\)0\( ) \).
1) The value a that yields f(a)=0 is a=32 since f(32)=0 .
2) The Derivative of 5/9*(x-32) is \( f'(x)=5/9 \) .
3) The Reciprocal of 5/9 is 9/5, thus \( f^{-1} ( \)8\( ) = 9/5 \) .
Both examples are Linear Functions with constant slopes that did not depend on the particular c values.

Let's study Functions other than Linear.

Example 3: Find the Slope of the Inverse of \( f(x) = x^{3} \) when c=8 , that is \( f^{-1} (\)8 \( ) \).
1) The value a that yields f(a)=0 is a=2 since f(2) = \( 2^3=8 \) .
2) The Derivative of \( x^{3} \) is \( f'(x)=3x^{2} \) then \( f'(2)=3*2^{2}=12 \) .
3) The Reciprocal of 12 is 1/12, thus \( f^{-1} ( \)8 \( ) = 1/12 \).

Example 4: Find the Slope of the Inverse of \( f(x) = x^{3} \) when c=0 , that is \( f^{-1} (\)0 \( ) \).
1) The value a that yields f(a)=0 is a=0 since f(0 ) = \( 0^3=0 \) .
2) The Derivative of \( x^{3} \) is \( f'(x)=3x^{2} \) then \( f'(0)=3*0^{2}=0 \) .
3) The Reciprocal of 0 is 1/0 which does not exist. Since \( f'(0)=0 \) which means f is horizontal when x=0, thus its inverse, when reflected over the y=x line, is vertical and thus has an undefined slope.

Example 5: Find the Slope of the Inverse (in red) of \( f(x) = x^{2} \) when c=4 , that is \( f^{-1} (\)4 \( ) \).
1) The value a that yields f(a)=0 is a=0 since f(2 and -2) = \( 2^2=4 \) .
2) The Derivative of \( x^{2} \) is \( f'(x)=2x \) then \( f'(2)=4 \) and \( f'(-2)=-4 \) .
3) The Reciprocals are 1/4 and -1/4. \( f(x) = x^{2} \) does not pass the horizontal line test implying that the Inverse is not a Function.

We found the two slopes at both points on the Inverse when \(x=4 \) as shown below: