Theory of Inverse Functions and Graph of Inverse Function

Inverse Functions:


The inverse of function f is symbolized by f-1 (if your browser does not support superscripts, that is looks similar to f with an exponent of -1) and is pronounced as ‘f inverse’. Though the inverse of a function looks similar to you are raising the function to the -1 power, it is not. The inverse of a function doesn’t signify the reciprocal of a function.


The function usually tells you what y is if you know what x is. The inverse of a function will state you what x had to be to obtain that value of y.

The function f-1 is the inverse of f if:

a) For each and every x in the domain of f, f-1[f(x)] = x, and
b) For each and every x in the domain of f-1, f[f-1(x)] = x

The domain of f is a range of f-1 and the range of f is the domain of f-1.

Graph of Inverse Function:

The inverse of a function varies from the function in that all x-coordinates and y-coordinates have been switched. That is, when (4,6) is a point on the graph of function, then (6,4) is a point on the graph of inverse function.

Points on an identity function (y = x) will remain on the identity function whenever switched. All the other points will have their coordinates switched and move positions.

The graph of a function and its inverse are the mirror images of one other. They are reflected regarding the identity function y = x.
Existence of an Inverse Function:

The function states that for each and every x, there is precisely one y. That is, y values can be duplicated however x values cannot be repeated.

When the function consists an inverse that is as well a function, then there can be only one y for every x.

The one-to-one function is a function in which for each and every x there is precisely one y and for each and every y, there is precisely one x. The one-to-one function consists of an inverse which is also a function.

There are functions that contain inverses which are not functions. There are as well inverses for relations. For most of the part, we disregard such, and deal only with the functions whose inverses are as well functions.

When the inverse of a function is as well a function, then the inverse relation should pass a vertical line test. As all the x-coordinates and y-coordinates are switched whenever finding out the inverse, stating that the inverse should pass a vertical line test is similar as stating the original function should pass a horizontal line test.

When a function passes both the vertical line test (and hence it is a function in first place) and the horizontal line test (and hence its inverse is a function), then the function is one-to-one and consists of an inverse function.

Finding Inverses Informally:

The inverse of certain functions, particularly those where there is just one occurrence of the independent variable, can be resolved by undoing the operations. To undo operations, you should not only reverse the order, however utilize the inverse operation.


The function f(x) = 5x-2

a) Begin with x: x
b) Multiply by 5: 5x
c) Subtract 2: 5x-2

The inverse f-1(x) = (x+2)/5

a) Begin with x: x
b) Add 2: x+2
c) Divide by 5: (x+2)/5


The function f(x) = 2(x-3)2 - 5, x≥3

Note that there is a restriction on the x.

a) Begin with x: x
b) Subtract 3: x-3
c) Square: (x-3)2
d) Multiply by 2: 2(x-3)2
e) Subtract 5: 2(x-3)2-5

The inverse f-1(x) = 3 + sqrt[(x+5)/2]

a) Begin with x: x
b) Add 5: x+5
c) Divide by 2: (x+5)/2
d) Take the square root: ± sqrt[(x+5)/2]
e) Add 3: 3 ± sqrt[(x+5)/2]

That inverse is not a function as there are two values of y for each and every x. That is because of the ± which appeared whenever we took the square root of both sides. Now we go back to the original domain of x≥3. That signifies that for the inverse, the range is y≥3. As y should be at least 3, we require the positive square root and not the negative. Devoid of the restriction on x in an original function, it would not have had an inverse function: 3 + sqrt[(x+5)/2]


The function f(x) = x2 - 4x + 6, x≤2

What happens if there is more than one occurrence of independent variable in the function? You do not know what you did to x as you did it to two different x's and you did not do similar thing to both of them.

Finding Inverses Formally:

Now, that last illustration is not to be said it cannot be completed, but it includes completing the square to get f(x) = (x-2)2+2, and then inversing it and hence you get f-1(x) = 2-sqrt(x-2).

Though, there is another way which does not rely so much on informality and will work whether or not you can figure out precisely what you did with precisely one x.

a) Begin with the function.

b) Substitute f(x) by y if essential.

c) Switch x's and y's. At this point you are dealing with inverse.

d) Solve for y.

e) Substitute y with f -1(x) if the inverse is as well a function, or else leave it as y.


The function f(x) = x2/(x2+1), x≥0

The restriction is significant to make it 1-1.

a) Begin with the function: f(x) = x2/(x2+1), x≥0

b) Substitute f(x) by y: y = x2/(x2+1), x≥0

c) Switch the x's and y's: x = y2/(y2+1), y ≥0

d) Solve for y:

•    Multiply by the denominator: x(y2+1) = y2
•    Distribute: xy2 + x = y2
•    Move y's to one side and all else to other: xy2-y2 = -x
•    Factor: y2(x-1) = -x
•    Divide by the coefficient on y2: y2 = -x/(x-1)
•    Simplify the right hand side: y2 = x/(1-x)
•    Take the square root: y = ± sqrt[x/(1-x)]
•    As y≥0, we require the positive square root: y = sqrt[x/(1-x)]

e) Call this f-1(x): f-1(x) = sqrt[x/(1-x)]

On this very last function, the implied domain of inverse is [0,1). That signifies that the range of original function should have been [0,1), as well. Ensure it on your calculator, and you will see it is.

Sometimes the instructions state if the function is not one-to-one, then do not find out the inverse function (as there is not one). Therefore, always check before wasting time trying to find out the inverse function. Now, if you are assumed to find out the inverse, regardless of whether it is the function or not, then go ahead.

Good Stuff:

One-to-One Functions are amazing things.

Whenever solving equations, you can add up the same thing to both sides, subtract similar thing from both sides, multiply both sides by similar non-zero thing, and divide both sides by similar non-zero thing and still get similar solution devoid of worrying about having to check your outcome or answer.

You can as well apply a one-to-one function to both the sides of an equation devoid of worrying about introducing the extraneous solutions (solutions which work after doing something which did not work before). This is not essentially true with functions that are not one-to-one similar to the squaring function where you must always check answers after you square both the sides of an equation. For illustration, the equation sqrt(x) = -2 has no solution, however if you square both the sides, you obtain x = 4, however it does not check in the original problem. With one-to-one functions, you won't be introducing any of extraneous solutions.

Assume that exp(x) is a one-to-one function and is the inverse of ln(x).

ln(x) = 3

Resolve for x.

exp[ ln(x)] = exp [3]

‘Wait a minute, Mr. Jones’ is your reply. You have never seen such a beast. That is okay. Take the inverse of function, and apply it to both the sides.

x = exp (3)

Go back to the definition of an inverse at top of this document. Whenever inverses are applied to one other, they inverse one other out and you are just left with the argument (that is, input) to the function.

x = e3

On calculator, the exp(x) function is written as e^x, and is found on [2nd] [ln] key.

More cohesiveness. The inverse of a function is found out by taking the [2nd] function. Look at it for other things on calculator.

The square root is an inverse of square. When you look at three trigonometric keys [sin], [cos], and [tan], then their inverses are all found out by utilizing the [2nd] key.

Latest technology based Algebra Online Tutoring Assistance

Tutors, at the, take pledge to provide full satisfaction and assurance in Algebra help via online tutoring. Students are getting 100% satisfaction by online tutors across the globe. Here you can get homework help for Algebra, project ideas and tutorials. We provide email based Algebra help. You can join us to ask queries 24x7 with live, experienced and qualified online tutors specialized in Algebra. Through Online Tutoring, you would be able to complete your homework or assignments at your home. Tutors at the TutorsGlobe are committed to provide the best quality online tutoring assistance for Algebra Homework help and assignment help services. They use their experience, as they have solved thousands of the Algebra assignments, which may help you to solve your complex issues of Algebra. TutorsGlobe assure for the best quality compliance to your homework. Compromise with quality is not in our dictionary. If we feel that we are not able to provide the homework help as per the deadline or given instruction by the student, we refund the money of the student without any delay.

2015 ©TutorsGlobe All rights reserved. TutorsGlobe Rated 4.8/5 based on 34139 reviews.