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CALCULUS OF INVERSE FUNCTIONS
Table of Content (8 min. reading time)
1 – What is an Inverse Function?
2 – How do I find the Inverse of a Function?
3 – Inverse Functions SOLVER: Find your Inverse.
4 – How do I Graph Inverse Functions?
5 – Horizontal Line Test: Is the Inverse a Function?
6 – Calculus of Inverses: How do I find the Derivative of an Inverse?
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:
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 :
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 .
Inverse Function Solver
Enter the problem to be solved below, next click the Solve button at the right to get the solution.
Is there another APP that finds Inverse Functions?
Yes, Microsoft’s Math Solver is a free math problem scanner with step by step solutions – just like the Free Photomath Online version. It is available on the web at https://math.microsoft.com/en
Is there a Video that shows me how to find an Inverse Function?
Of course. Just watch this 1min Video. It shows you how easy it is:
Calculus Made Easy
- Step by Step - Integration
- Step by Step - Differentiation
- Step by Step - Limits (L'Hopital Rule)
- Step by Step - Differential Equations
- Step by Step - Implicit Differentiation
- Step by Step - Area Approximation
- Step by Step - Multivariable Calculus
- Step by Step - Inverse Slope
- Step by Step - Volume (by Revolution)
- Step by Step - Parametric Equations
- Step by Step - Polar Equations
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.
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.
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.
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)=x2 is not a Function since the Horizontal Line Test on f(x)=x2 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)=x3 when c=8 , that is f−1(8 ).
1) The value a that yields f(a)=0 is a=2 since f(2) = 23=8 .
2) The Derivative of x3 is f′(x)=3×2 then f′(2)=3∗22=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)=x3 when c=0 , that is f−1(0 ).
1) The value a that yields f(a)=0 is a=0 since f(0 ) = 03=0 .
2) The Derivative of x3 is f′(x)=3×2 then f′(0)=3∗02=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)=x2 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) = 22=4 .
2) The Derivative of x2 is f′(x)=2x then f′(2)=4 and f′(−2)=−4 .
3) The Reciprocals are 1/4 and -1/4. f(x)=x2 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:
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:
Applying 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:
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 for 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.
So let’s study Functions other than Linear.
Example 3: Find the Slope of the Inverse of f(x)=x3 when c=8 , that is f−1(8 ).
1) The value a that yields f(a)=0 is a=2 since f(2) = 23=8 .
2) The Derivative of x3 is f′(x)=3×2 then f′(2)=3∗22=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)=x3 when c=0 , that is f−1(0 ).
1) The value a that yields f(a)=0 is a=0 since f(0 ) = 03=0 .
2) The Derivative of x3 is f′(x)=3×2 then f′(0)=3∗02=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 of f(x)=x2 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) = 22=4 .
2) The Derivative of x2 is f′(x)=2x then f′(2)=4 and f′(−2)=−4 .
3) The Reciprocals are 1/4 and -1/4. f(x)=x2 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:
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