# 1.6.6E: Inverse Functions

section 1.6 exercises

Assume that the function f is a one-to-one function.

1. If (f(6)=7), find (f^{-1} (7))

2. If (f(3)=2), find (f^{-1} (2))

3. If (f^{-1} (-4)=-8), find (f(-8))

4. If (f^{-1} (-2)=-1), find (f(-1))

5. If (f(5)=2), find ((f(5))^{-1})

6. If (f(1)=4), find ((f(1))^{-1})

7. Using the graph of (f(x)) shown

a. Find (f(0))

b. Solve (f(x)=0)

c. Find (f^{-1} (0))

d. Solve (f^{-1} (x)=0)

8. Using the graph shown

a. Find (g(1))

b. Solve (g(x)=1)

c. Find (g^{-1} (1))

d. Solve (g^{-1} (x)=1)

9. Use the table below to find the indicated quantities.

 (x) 0 1 2 3 4 5 6 7 8 9 (f(x)) 8 0 7 4 2 6 5 3 9 1

a. Find (f(1))

b. Solve (f(x)=3)

c. Find (f^{-1}(0))

d. Solve (f^{-1}(x)=7)

10. Use the table below to fill in the missing values.

 (t) 0 1 2 3 4 5 6 7 8 (h(t)) 6 0 1 7 2 3 5 4 9

a. Find (h(6))

b. Solve (h(t)=0)

c. Find (h^{-1} (5))

d. Solve (h^{-1} (t)=1)

For each table below, create a table for (f^{-1} (x).)

11.

 (x) 3 6 9 13 14 (f(x)) 1 4 7 12 16

For each function below, find (f^{-1} (x))

13. (f(x)=x+3)

14. (f(x)=x+5)

15. (f(x)= 2 - x)

16. (f(x)=3-x)

17. (f(x)=11x+7)

18. (f(x)=9+10x)

For each function, find a domain on which (f) is one-to-one and non-decreasing, then find the inverse of (f) restricted to that domain.

19. (f(x)=(x +7)^{2})

20. (f(x)=(x-6)^{2})

21. (f(x)=x^{2} -5)

22. (f(x)=x^{2} +1)

23. If (f(x)=x^{3} -5) and (g(x)=sqrt[{3}]{x+5}), find

a. (f(g(x)))

b. (g(f(x)))

c. What does this tell us about the relationship between (f(x)) and (g(x))?

24. If (f(x)=dfrac{x}{2+x}) and (g(x)=dfrac{2x}{1-x}), find

a. What does this tell us about the relationship between (f(x)) and (g(x))?

1. 6

3. -4

5. 1/2

7a. 3
b. 2
c. 2
d. 2

11.

 (x) 1 4 7 12 16 (f^{-1}(x)) 3 6 9 13 14

13. (f^{-1}(x) = x -3)

15. (f^{-1}(x) = -x + 2)

17. (f^{-1}(x) = dfrac{x - 7}{11})

19. Restricted domain (x ge -7), (f^{-1}(x) = sqrt{x} - 7)

21. Restricted domain (x ge 0), (f^{-1}(x) = sqrt{x + 5})

23a. (f(g(x)) = (sqrt[3]{x + 5})^3 - 5 = x)
b. (g(f(x)) = sqrt[3]{x^3 - 5 + 5} = x)
c. This means that they are inverse functions (of each other)

## Stewart Calculus 7e Solutions Chapter 6 Inverse Functions Exercise 6.8

Stewart Calculus 7e Solutions Chapter 6 Inverse Functions Exercise 6.8

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## Contents

with Jacobian determinant:

As an important result, the inverse function theorem has been given numerous proofs. The proof most commonly seen in textbooks relies on the contraction mapping principle, also known as the Banach fixed-point theorem (which can also be used as the key step in the proof of existence and uniqueness of solutions to ordinary differential equations). [2] [3]

Since the fixed point theorem applies in infinite-dimensional (Banach space) settings, this proof generalizes immediately to the infinite-dimensional version of the inverse function theorem [4] (see Generalizations below).

An alternate proof in finite dimensions hinges on the extreme value theorem for functions on a compact set. [5]

Yet another proof uses Newton's method, which has the advantage of providing an effective version of the theorem: bounds on the derivative of the function imply an estimate of the size of the neighborhood on which the function is invertible. [6]

### Manifolds Edit

The inverse function theorem can be rephrased in terms of differentiable maps between differentiable manifolds. In this context the theorem states that for a differentiable map F : M → N (of class C 1 > ), if the differential of F ,

is a diffeomorphism. Note that this implies that the connected components of M and N containing p and F(p) have the same dimension, as is already directly implied from the assumption that dFp is an isomorphism. If the derivative of F is an isomorphism at all points p in M then the map F is a local diffeomorphism.

### Banach manifolds Edit

These two directions of generalization can be combined in the inverse function theorem for Banach manifolds. [10]

### Constant rank theorem Edit

The inverse function theorem (and the implicit function theorem) can be seen as a special case of the constant rank theorem, which states that a smooth map with constant rank near a point can be put in a particular normal form near that point. [11] Specifically, if F : M → N has constant rank near a point p ∈ M , then there are open neighborhoods U of p and V of F ( p ) and there are diffeomorphisms u : T p M → U M o U!> and v : T F ( p ) N → V N o V!> such that F ( U ) ⊆ V and such that the derivative d F p : T p M → T F ( p ) N :T_

M o T_N!> is equal to v − 1 ∘ F ∘ u circ Fcirc u!> . That is, F "looks like" its derivative near p . Semicontinuity of the rank function implies that there is an open dense subset of the domain of F on which the derivative has constant rank. Thus the constant rank theorem applies to a generic point of the domain.

When the derivative of F is injective (resp. surjective) at a point p , it is also injective (resp. surjective) in a neighborhood of p , and hence the rank of F is constant on that neighborhood, and the constant rank theorem applies.

### Holomorphic functions Edit

If a holomorphic function F is defined from an open set U of C n ^!> into C n ^!> , and the Jacobian matrix of complex derivatives is invertible at a point p , then F is an invertible function near p . This follows immediately from the real multivariable version of the theorem. One can also show that the inverse function is again holomorphic. [12]

### Polynomial functions Edit

If it would be true, the Jacobian conjecture would be a variant of the inverse function theorem for polynomials. It states that if a vector-valued polynomial function has a Jacobian determinant that is an invertible polynomial (that is a nonzero constant), then it has an inverse that is also a polynomial function. It is unknown whether this is true or false, even in the case of two variables. This is a major open problem in the theory of polynomials.

## Inverse-gamma distribution

In probability theory and statistics, the inverse gamma distribution is a two-parameter family of continuous probability distributions on the positive real line, which is the distribution of the reciprocal of a variable distributed according to the gamma distribution. Perhaps the chief use of the inverse gamma distribution is in Bayesian statistics, where the distribution arises as the marginal posterior distribution for the unknown variance of a normal distribution, if an uninformative prior is used, and as an analytically tractable conjugate prior, if an informative prior is required.

α + ln ⁡ ( β Γ ( α ) ) − ( 1 + α ) ψ ( α )

However, it is common among Bayesians to consider an alternative parametrization of the normal distribution in terms of the precision, defined as the reciprocal of the variance, which allows the gamma distribution to be used directly as a conjugate prior. Other Bayesians prefer to parametrize the inverse gamma distribution differently, as a scaled inverse chi-squared distribution.

## 1.6.6E: Inverse Functions

What are inverse functions?

Take a function f: draw its graph in the usual way interchange x and y axes, and you have the graph of the inverse function f − 1 .
y = f ( x ) means x = f − 1 ( y ) .

This can be accomplished with a drawing on a piece of paper by turning the paper over, orienting so that the old first quadrant appears in the upper right corner, and looking through the paper at the old graph.

Do not confuse the inverse with the reciprocal function they are completely different concepts. Careless people may use notation that is the same for both. This is a bad thing to do since it promotes confusion.

Notice that if you substitute x = f − 1 ( y ) into y = f ( x ) you get x = f − 1 ( f ( x ) ) . This last equation can be used as an alternate definition of the inverse function to f .

There is a problem with defining inverse functions. A function can have only one ordered pair for each argument, while the same value can occur many times. This means that interchanging arguments and values which is what we do in creating an inverse, will create a non-function, unless the original function takes on each value exactly once.

When a function takes a value more than once, we have to do extra work to define an inverse function for it. Namely, we have to select one of its values to be the new argument and throw away the others. This can be done in many different ways when f is not single valued, so that there is always some arbitrariness in the definition of f − 1 when f is not single valued.

The clearest example for this is the function x 2 . It takes on each positive value twice. Both 4 and -4 have the same square. The standard thing to do for this function is to define its inverse, x 1 / 2 , to be the positive square root, ignoring the negative one. (The negative square root is then denoted by − x 1 / 2 .) This definition has two virtues: one is that positive numbers are more positive than negative ones. The other is, that with this definition (and not had we chosen the negative root as inverse) the square root of a product is the product of the square roots of its factors.

In general, you can make the choice of what you want to call the inverse of f by looking at the graph of f , selecting a domain on which f is single-valued, and making that the range of f − 1 .

Exercises 1.8 For what values can you define the inverse of the function cos ⁡ ( sin ⁡ x ) . (Hint: set f = cos ⁡ ( sin ⁡ x ) look at its inverse and figure out the answer.) Solution

## 1.4 Inverse Functions

An inverse function reverses the operation done by a particular function. In other words, whatever a function does, the inverse function undoes it. In this section, we define an inverse function formally and state the necessary conditions for an inverse function to exist. We examine how to find an inverse function and study the relationship between the graph of a function and the graph of its inverse. Then we apply these ideas to define and discuss properties of the inverse trigonometric functions.

### Definition

Note that f −1 f −1 is read as “f inverse.” Here, the −1 −1 is not used as an exponent and f −1 ( x ) ≠ 1 / f ( x ) . f −1 ( x ) ≠ 1 / f ( x ) . Figure 1.37 shows the relationship between the domain and range of f and the domain and range of f −1 . f −1 .

### Example 1.28

#### Determining Whether a Function Is One-to-One

For each of the following functions, use the horizontal line test to determine whether it is one-to-one.

### Example 1.29

#### Solution

Follow the steps outlined in the strategy.

Step 2. Rewrite as y = 1 3 x + 4 3 y = 1 3 x + 4 3 and let y = f −1 ( x ) . y = f −1 ( x ) .

Therefore, f −1 ( x ) = 1 3 x + 4 3 . f −1 ( x ) = 1 3 x + 4 3 .

You can verify that f −1 ( f ( x ) ) = x f −1 ( f ( x ) ) = x by writing

### Example 1.31

#### Restricting the Domain

Consider the function f ( x ) = ( x + 1 ) 2 . f ( x ) = ( x + 1 ) 2 .

### Inverse Trigonometric Functions

The six basic trigonometric functions are periodic, and therefore they are not one-to-one. However, if we restrict the domain of a trigonometric function to an interval where it is one-to-one, we can define its inverse. Consider the sine function (Figure 1.34). The sine function is one-to-one on an infinite number of intervals, but the standard convention is to restrict the domain to the interval [ − π 2 , π 2 ] . [ − π 2 , π 2 ] . By doing so, we define the inverse sine function on the domain [ −1 , 1 ] [ −1 , 1 ] such that for any x x in the interval [ −1 , 1 ] , [ −1 , 1 ] , the inverse sine function tells us which angle θ θ in the interval [ − π 2 , π 2 ] [ − π 2 , π 2 ] satisfies sin θ = x . sin θ = x . Similarly, we can restrict the domains of the other trigonometric functions to define inverse trigonometric functions , which are functions that tell us which angle in a certain interval has a specified trigonometric value.

### Definition

To graph the inverse trigonometric functions, we use the graphs of the trigonometric functions restricted to the domains defined earlier and reflect the graphs about the line y = x y = x (Figure 1.41).

### Media

Go to the following site for more comparisons of functions and their inverses.

We now consider a composition of a trigonometric function and its inverse. For example, consider the two expressions sin ( sin −1 ( 2 2 ) ) sin ( sin −1 ( 2 2 ) ) and sin −1 ( sin ( π ) ) . sin −1 ( sin ( π ) ) . For the first one, we simplify as follows:

For the second one, we have

Similarly, for the cosine function,

Similar properties hold for the other trigonometric functions and their inverses.

### Example 1.32

#### Evaluating Expressions Involving Inverse Trigonometric Functions

Evaluate each of the following expressions.

### Student Project

#### The Maximum Value of a Function

In many areas of science, engineering, and mathematics, it is useful to know the maximum value a function can obtain, even if we don’t know its exact value at a given instant. For instance, if we have a function describing the strength of a roof beam, we would want to know the maximum weight the beam can support without breaking. If we have a function that describes the speed of a train, we would want to know its maximum speed before it jumps off the rails. Safe design often depends on knowing maximum values.

This project describes a simple example of a function with a maximum value that depends on two equation coefficients. We will see that maximum values can depend on several factors other than the independent variable x.

### Section 1.4 Exercises

For the following exercises, use the horizontal line test to determine whether each of the given graphs is one-to-one.

## No Inverse?

Let us see graphically what is going on here:

To be able to have an inverse we need unique values.

Just think . if there are two or more x-values for one y-value, how do we know which one to choose when going back?

 General Function No Inverse

Imagine we came from x1 to a particular y value, where do we go back to? x1 or x2?

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 "one-to-one correspondence" or Bijective, like this

 Bijective Function Has an Inverse

A function has to be "Bijective" to have an inverse.

So a bijective function follows stricter rules than a general function, which allows us to have an inverse.

## Examples

For more practice with the concepts covered in this tutorial, visit the Inverse Functions Problems page at the link below. The solutions to the problems will be posted after these chapters are covered in your calculus course.

To test your knowledge of inverse function problems, try taking the general inverse functions test on the iLrn website or the advanced inverse functions test at the link below.

Please forward any questions, comments, or problems you have experienced with this website to Alex Karassev.

I know that the mod function is defined by floor function (a - n floor(na). But I'm still unable to create an inverse function with one inverse function for all values. I made a piecewise function that works for those values, but I feel like there should be a better method. Does anybody know an inverse rule for either floor or modulus?

The 'mod' function? There is no such thing.

Just do it as for any other function:

y=4x+3, hence x= (y-3)/4, so what is division by 4 in mod 7 arithmetic?

The first thing you should do is calculate the actual values of f:
f(0)= 4(0)+ 3= 3. f(1)= 4(1)+ 3= 7= 0 mod 7. f(2)= 4(2)+ 3= 11= 4 mod 7.
f(3)= 4(3)+ 3= 15= 1 mod y. f(4)= 4(4)+ 3= 19= 5 mod 7. f(5)= 4(5)+ 3= 23= 2 mod 7. f(6)= 4(6)+ 3= 27= 6 mod 7

So you want f -1 (0)= 1, f -1 (1)= 3, f -1 (2)= 5, f -1 (3)= 0, f -1 (4)= 2, f -1 (5)= 4, f -1 (6)= 6.

Yes, that's exactly what your function gives. Good work! That is, by the way, a single "function" and I have no problem with it. However, you might remember that -6 is 1 in mod 7. Since the first formula was given "mod 7" why not do the same with this:
f -1 (x)= 2x+ 1 (mod 7) or f -1 (x)= 2x- 6 (mod 7)
They are exactly the same.

The 'mod' function? There is no such thing.

My apologies. mod "operator" (or modulo operator) I was just asking if there was an inverse operator that I could use so I could switch the x and y variables and solve for x instead of switching the values around and making an entirely new function.

## 1.6.6E: Inverse Functions

In the last example from the previous section we looked at the two functions (fleft( x ight) = 3x - 2) and (gleft( x ight) = frac <3>+ frac<2><3>) and saw that

[left( ight)left( x ight) = left( ight)left( x ight) = x]

and as noted in that section this means that there is a nice relationship between these two functions. Let’s see just what that relationship is. Consider the following evaluations.

In the first case we plugged (x = - 1) into (fleft( x ight)) and got a value of (-5). We then turned around and plugged (x = - 5) into (gleft( x ight)) and got a value of -1, the number that we started off with.

In the second case we did something similar. Here we plugged (x = 2) into (gleft( x ight)) and got a value of(frac<4><3>), we turned around and plugged this into (fleft( x ight)) and got a value of 2, which is again the number that we started with.

Note that we really are doing some function composition here. The first case is really,

[left( ight)left( < - 1> ight) = gleft[ ight)> ight] = gleft[ < - 5> ight] = - 1]

and the second case is really,

Note as well that these both agree with the formula for the compositions that we found in the previous section. We get back out of the function evaluation the number that we originally plugged into the composition.

So, just what is going on here? In some way we can think of these two functions as undoing what the other did to a number. In the first case we plugged (x = - 1) into (fleft( x ight)) and then plugged the result from this function evaluation back into (gleft( x ight)) and in some way (gleft( x ight)) undid what (fleft( x ight)) had done to (x = - 1) and gave us back the original (x) that we started with.

Function pairs that exhibit this behavior are called inverse functions. Before formally defining inverse functions and the notation that we’re going to use for them we need to get a definition out of the way.

A function is called one-to-one if no two values of (x) produce the same (y). Mathematically this is the same as saying,

So, a function is one-to-one if whenever we plug different values into the function we get different function values.

Sometimes it is easier to understand this definition if we see a function that isn’t one-to-one. Let’s take a look at a function that isn’t one-to-one. The function (fleft( x ight) = ) is not one-to-one because both (fleft( < - 2> ight) = 4) and (fleft( 2 ight) = 4). In other words, there are two different values of (x) that produce the same value of (y). Note that we can turn (fleft( x ight) = ) into a one-to-one function if we restrict ourselves to (0 le x < infty ). This can sometimes be done with functions.

Showing that a function is one-to-one is often tedious and/or difficult. For the most part we are going to assume that the functions that we’re going to be dealing with in this course are either one-to-one or we have restricted the domain of the function to get it to be a one-to-one function.

Now, let’s formally define just what inverse functions are. Given two one-to-one functions (fleft( x ight)) and (gleft( x ight)) if

then we say that (fleft( x ight)) and (gleft( x ight)) are inverses of each other. More specifically we will say that (gleft( x ight)) is the inverse of (fleft( x ight)) and denote it by

[gleft( x ight) = >left( x ight)]

Likewise, we could also say that (fleft( x ight)) is the inverse of (gleft( x ight)) and denote it by

[fleft( x ight) = >left( x ight)]

The notation that we use really depends upon the problem. In most cases either is acceptable.

For the two functions that we started off this section with we could write either of the following two sets of notation.

[eginfleft( x ight) & = 3x - 2hspace<0.25in>hspace <0.25in>& >left( x ight) & = frac <3>+ frac<2><3> & & & gleft( x ight) & = frac <3>+ frac<2><3>hspace<0.25in>hspace<0.25in>& >left( x ight) & = 3x - 2end]

Now, be careful with the notation for inverses. The “-1” is NOT an exponent despite the fact that it sure does look like one! When dealing with inverse functions we’ve got to remember that

This is one of the more common mistakes that students make when first studying inverse functions.

The process for finding the inverse of a function is a fairly simple one although there are a couple of steps that can on occasion be somewhat messy. Here is the process

#### Finding the Inverse of a Function

Given the function (fleft( x ight)) we want to find the inverse function, (>left( x ight)).

1. First, replace (fleft( x ight)) with (y). This is done to make the rest of the process easier.
2. Replace every (x) with a (y) and replace every (y) with an (x).
3. Solve the equation from Step 2 for (y). This is the step where mistakes are most often made so be careful with this step.
4. Replace (y) with (>left( x ight)). In other words, we’ve managed to find the inverse at this point!
5. Verify your work by checking that [left( >> ight)left( x ight) = x] and [left( <> circ f> ight)left( x ight) = x] are both true. This work can sometimes be messy making it easy to make mistakes so again be careful.

That’s the process. Most of the steps are not all that bad but as mentioned in the process there are a couple of steps that we really need to be careful with since it is easy to make mistakes in those steps.

In the verification step we technically really do need to check that both (left( >> ight)left( x ight) = x) and (left( <> circ f> ight)left( x ight) = x) are true. For all the functions that we are going to be looking at in this course if one is true then the other will also be true. However, there are functions (they are beyond the scope of this course however) for which it is possible for only one of these to be true. This is brought up because in all the problems here we will be just checking one of them. We just need to always remember that technically we should check both.

Now, we already know what the inverse to this function is as we’ve already done some work with it. However, it would be nice to actually start with this since we know what we should get. This will work as a nice verification of the process.

So, let’s get started. We’ll first replace (fleft( x ight)) with (y).

Next, replace all (x)’s with (y)and all (y)’s with (x).

Finally replace (y) with (>left( x ight)).

Now, we need to verify the results. We already took care of this in the previous section, however, we really should follow the process so we’ll do that here. It doesn’t matter which of the two that we check we just need to check one of them. This time we’ll check that (left( >> ight)left( x ight) = x) is true.

[eginleft( >> ight)left( x ight) & = fleft[ <>left( x ight)> ight] & = fleft[ <3>+ frac<2><3>> ight] & = 3left( <3>+ frac<2><3>> ight) - 2 & = x + 2 - 2 & = xend]

The fact that we’re using (gleft( x ight)) instead of (fleft( x ight)) doesn’t change how the process works. Here are the first few steps.

Now, to solve for (y)we will need to first square both sides and then proceed as normal.

Finally let’s verify and this time we’ll use the other one just so we can say that we’ve gotten both down somewhere in an example.

So, we did the work correctly and we do indeed have the inverse.

The next example can be a little messy so be careful with the work here.

The first couple of steps are pretty much the same as the previous examples so here they are,

Now, be careful with the solution step. With this kind of problem it is very easy to make a mistake here.

[eginxleft( <2y - 5> ight) & = y + 4 2xy - 5x & = y + 4 2xy - y & = 4 + 5x left( <2x - 1> ight)y & = 4 + 5x y & = frac<<4 + 5x>><<2x - 1>>end]

So, if we’ve done all of our work correctly the inverse should be,

Finally, we’ll need to do the verification. This is also a fairly messy process and it doesn’t really matter which one we work with.

Okay, this is a mess. Let’s simplify things up a little bit by multiplying the numerator and denominator by (2x - 1).

Wow. That was a lot of work, but it all worked out in the end. We did all of our work correctly and we do in fact have the inverse.

There is one final topic that we need to address quickly before we leave this section. There is an interesting relationship between the graph of a function and the graph of its inverse.

Here is the graph of the function and inverse from the first two examples.

In both cases we can see that the graph of the inverse is a reflection of the actual function about the line (y = x). This will always be the case with the graphs of a function and its inverse.

1. Replace f(x) by y .
2. Switch the roles of colorx and colory . In other words, interchange x and y in the equation.
3. Solve for y in terms of x .
4. Replace y by >left( x ight) to get the inverse function
5. Sometimes, it is helpful to use the domain and range of the original function to identify the correct inverse function out of two possibilities. This happens when you get a “plus or minus” case in the end.

#### Examples of How to Find the Inverse Function of a Quadratic Function

Example 1: Find the inverse function of fleft( x ight) = + 2 , if it exists. State its domain and range.

The first thing I realize is that this quadratic function doesn’t have a restriction on its domain. I am sure that when I graph this, I can draw a horizontal line that will intersect it more than once. Therefore the inverse is not a function. I will not even bother applying the key steps above to find its inverse.

The diagram shows that it fails the Horizontal Line Test, thus the inverse is not a function. I will stop here.

Example 2: Find the inverse function of fleft( x ight) = + 2,,,x ge 0 , if it exists. State its domain and range.

This same quadratic function, as seen in Example 1, has a restriction on its domain which is x ge 0 . After plotting the function in xy- axis, I can see that the graph is a parabola cut in half for all x values equal to or greater than zero. This should pass the Horizontal Line Test which tells me that I can actually find its inverse function by following the suggested steps.

In its graph below, I clearly defined the domain and range because I will need this information to help me identify the correct inverse function in the end.

Remember that the domain and range of the inverse function come from the range, and domain of the original function, respectively. It’s called the swapping of domain and range.

Even without solving for the inverse function just yet, I can easily identify its domain and range using the information from the graph of the original function: domain is x ≥ 2 and range is y ≥ 0.

Do you see how I interchange the domain and range of the original function to get the domain and range of its inverse?

Now, let’s go ahead and algebraically solve for its inverse.

Graphing the original function with its inverse in the same coordinate axis…

Example 3: Find the inverse function of fleft( x ight) = - - 1,,,x le 0 , if it exists. State its domain and range.

This problem is very similar to Example 2. The range starts at colory=-1 , and it can go down as low as possible.

Now, these are the steps on how to solve for the inverse.

Applying square root operation results in getting two equations because of the positive and negative cases. To pick the correct inverse function out of the two, I suggest that you find the domain and range of each possible answer. Now, the correct inverse function should have a domain coming from the range of the original function and a range coming from the domain of the same function.

The following are the graphs of the original function and its inverse on the same coordinate axis.

Example 4: Find the inverse of the function below, if it exists. State its domain and range.

I would graph this function first and clearly identify the domain and range. Notice that the restriction in the domain cuts the parabola into two equal halves. I will deal with the left half of this parabola. Clearly, this has an inverse function because it passes the Horizontal Line Test.

Proceed with the steps in solving for the inverse function. In fact, there are two ways how to work this out.

where a , b , and c can contain variables.

This is expected since we are solving for a function, not exact values.

The key step here is to pick the appropriate inverse function in the end because we will have the plus (+) and minus (−) cases. We can do that by finding the domain and range of each and compare that to the domain and range of the original function. Remember that we swap the domain and range of the original function to get the domain and range of its inverse.

• The method of completing the squares allows us to isolate the variable in a quadratic trinomial. As you will see in the steps, the quadratic trinomial is converted into linear binomial raised to the power of 2 . Obviously, we can apply the square root operation to get rid of the exponent 2 , therefore leaving us with an easy equation to solve.

If you observe, the graphs of the function and its inverse are actually symmetrical along the line y = x (see dashed line). They are like mirror images of each other.

I hope that you gain some level of appreciation on how to find the inverse of a quadratic function. Although it can be a bit tedious, as you can see, overall it is not that bad. I recommend that you check out the related lessons on how to find inverses of other kinds of functions.