# Inverse Function

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Many times, in maths we are required to find the inverse of some function. For example, we know that distance travelled by a car can be represented as some function of time. In order to find the distance travelled by the car at some particular instant, we input that particular value of time to the function. But we may also ask the inverse question e.g., what is the time at which the distance travelled by the car is, say, 120 meters. See how valuable such information can be. Therefore, inverse of a function allows us to ask the reverse question.

Let us first revisit the concept of a function briefly.

## Function

A function is a mathematical relation that takes some value as input and based on a rule outputs another value. The rule is such that for a given input, there is only a single output

A function is generally represented by a symbol with round brackets enclosing the input variable such as f(x), g(x), h(x) etc

An example of a function is f(x) = x + 1. The rule for this function is simple. Take an input and output a number one greater than the input. For an input 3, the output is 4. Similarly, for input of -1.423, the output is -0.423.

Dependent and Independent variables: These terms are usually mentioned with the function. We input the values of the independent variable in the function. Independent variable is generally represented by the symbol x. The reason why it is named so is that we can input whatever we want (from the set of Domain)! Whereas the output of the function is represented in the form of dependent variable (because its value depends on the input/independent variable). Dependent variable is usually represented with the symbol y.

## Domain and Range

The set of values of the input for which the function is defined is called the Domain of the function. The set of corresponding output values is known as the Range of the function. The set of domain and range are denoted by X and Y respectively.

The domain of the aforementioned function $f(x)=x+1$ is the set of real numbers (we can input any number from R), whereas the range is also the set of real number.

For the function $f(x) = \sqrt{x}$, the domain is set of positive real numbers only. While the range is also set of positive real numbers.

Exercise: Find the domain and range of $f(x) = 2x^{2}$.

## The idea of inverse function

A function can be thought of as a transformation that maps the set X (domain) to set Y (range). Symbolically, it is asserted as $f: X\rightarrow Y$. The set Y is also called the image of X under the transformation f. A natural question arises here. What if there is a transformation or a rule that maps Y to X? An inverse transformation that does “everything in reverse”. For example, the function that we have been discussing (i.e., $f(x) = x+1$) takes a value and adds 1 to it. So, what should the inverse do? That’s correct. Subtract a 1 from the input.

Seems simple, right? But there is much more to it!

Notation: Inverse of a function is represented as $f^{-1}(x)$.

### Definition

The inverse of a function f(x) is a transformation that maps the range of f(x) to its domain. In other words, it reverses the action of f(x)

Notation: Inverse function is generally denoted as: $f^{-1}(x)$.

For example, we have a function $f(x) = 3x+1$. Its inverse function is $f^{-1}(x) = \frac{x-1}{3}$. You do not have to worry about how the inverse function is figured out. We will discuss it in detail.

## Is inverse of a function always another function?

The interesting fact that adds complexity is that inverse of a function is not always a function. You may recall the special condition for being a function i.e., a specific input should always yield a single output.

Now we see how the inverse of a function is not always a function itself. Take the example of the function $f(x) = 2x^{2}$. We see that it maps an input of 2 to an output of 8. An input of -2 is mapped to 8 also (Note that this does not violate the condition for being a function. Either of the input has only a single output i.e., 8). Therefore, $2x^{2}$ represents a function.

However, consider what should the inverse $f^{-1}(x)$ do? Undo the action of the original function, right? So, if we input 8 to $f^{-1}(x)$, it should output 2, correct? But an output of -2 is also correct! In other words, there are two outputs for one specific input. Therefore, $f^{-1}(x)$ does not qualify as a function!

You may be asking that how do we know if inverse of a function is also a function. In other words, how do we know that a given function is invertible? Well, there are a couple of methods to find that out. We will discuss those methods as we move along.

Domain and Range of Inverse Function: Here is a question. Is the domain of the inverse function same as the domain of original function? The answer is, “No”. On the contrary, the domain of the inverse function is basically the range of the original function, while the range of the inverse function is the domain of the original function.

Inverse of the Inverse Function: The inverse of the inverse function is the original function itself i.e., if $f^{-1}(x) = g(x)$, then $g^{-1}(x) = f(x)$. Yeah, it looks too simple to be stated. But later in this article, we will see the significance of this statement!

Well-known inverse functions: There are several well-known inverse functions in mathematics. They include x which is inverse function of exponential function $a^{x}$, the principal inverse trigonometric functions $Sin^{-1}(x)$, $Cos^{-1}(x)$, $Tan^{-1}(x)$ etc.

Multiplicative inverse of a function vs Inverse function: The notation $f^{-1}(x)$ can be misleading for some. It should be kept in mind that $f^{-1}(x)$ is used to represent the Inverse function whereas, the notation$(f(x))^{-1}$ is used to represent the multiplicative inverse of f(x) i.e., $(f(x))^{-1} = \frac{1}{f(x)}$

## Finding the inverse function

It is pretty straightforward. Given the function $f(x) = \sqrt{x+1} -2$, the first step is to replace f(x) by the symbol y. Thus, we have

$y= \sqrt{x+1} -2$

Next, the goal is to isolate the variable x on the left-hand side. For the equation above, adding 2 on both sides gives,

$y + 2 = \sqrt{x+1}$

Square both sides,

$(y + 2) ^{2}= x+1$

Subtract 1 from both sides,

$(y + 2) ^{2} - 1= x$

Or,

$x = (y + 2) ^{2} - 1$

Now is the awkward part. Swap the symbols x and y with each other.

$y = (x + 2) ^{2} - 1$

Replace the symbol y with $f^{-1}(x)$, we get the inverse.

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

You should revisit all these steps. Try to digest the part where we swap the symbols. It will gradually make sense as to why it is done.

Now that we have the expression for $f^{-1}(x)$, we can identify whether it is a function or not (This is one of the possible methods of finding whether the inverse would be a function or not!).

Exercise: Find the inverse of the following functions. Also find whether the respective inverses are eligible to be called as functions (i.e., Inverse Functions).

• $f(x) = x^{4} + 2$
• $f(x) = (x-1)^{2}$

Bonus exercise: Given the inverse function $f^{-1}(x) = (x - 1) ^{3} + 3$, find the original function f(x). (Hint: Inverse of Inverse function is the original function)

It should be noted here that this method cannot always be used to find inverse of a function. Success of this method relies on whether x can be isolated or not. For example, for the function $f(x) = \sqrt{x} + x^{3}$, the symbol x cannot be isolated.

Self-inverse: Is there any function that is inverse of itself? Can you think of any such function?

As a matter of fact, there is one such function. It is $f(x) = x = f^{-1}(x)$

## Uniqueness of Inverse function

The inverse of a function is always unique. It can be understood mathematically as

$f(f^{-1}(x)) = x = f^{-1}(f(x))$ (1)

Therefore, given a function f(x), its inverse $f^{-1}(x)$ will be unique!

## Inverse Composition

You may recall that two functions can be represented in composite form as

$(f\,o\,g)(x) = f(g(x))$ (2)

Or as,

$(g\,o\,f)(x) = g(f(x))$ (3)

Note that in general,

$f(g(x))\neq g(f(x))$

What about the inverse of the composition i.e., $(f\,o\,g)^{-1}(x)$ and $(g\,o\,f)^{-1}(x)$? How do they relate to $f^{-1}(x)$ and $g^{-1}(x)$?

The answers can be found below.

$(f\,o\,g)^{-1}(x) = (g^{-1}\,o\,f^{-1})(x) = g^{-1}(f^{-1}(x))$

And,

$(g\,o\,f)^{-1}(x) = (f^{-1}\,o\,g^{-1})(x) = f^{-1}(g^{-1}(x))$

Yeah, it seems troublesome at first. Let us solve an example to get acquainted to these relations. Given the functions $f(x) = 2\sqrt{x}$ and g(x) = 2x + 3, we have,

$(f\,o\,g)(x) = h(x) = 2\sqrt{2x+3}$

$(g\,o\,f)(x) = k(x) =4\sqrt{x} + 3$

$f^{-1}(x) = \frac{x^{2}}{4}$

$g^{-1}(x) = \frac{x-3}{2}$

$(f\,o\,g)^{-1}(x) = h^{-1}(x) = \frac{x^{2}-12}{8}$

$(g\,o\,f)^{-1}(x) = k^{-1}(x) =\frac{(x-3)^{2}}{16}$

Note that in order to prevent confusion, we have represented (f o g)(x) as another function h(x) (Similarly we have represented (g o f)(x) as k(x)). This technique eases the computations.

Now we may verify the relations (4) and (5). Using the relations for $f^{-1}(x)$ and $g^{-1}(x)$, we have

$f^{-1}(g^{-1}(x) ) = f^{-1}(\frac{x-3}{2})= \frac{(\frac{x-3}{2})^{2}}{4}$

$= \frac{(x-3)^{2}}{16} = (g\,o\,f)^{-1}(x)$

Also,

$g^{-1}(f^{-1}(x) ) = g^{-1}(\frac{x^{2}}{4}) = \frac{(\frac{x^{2}}{4})-3}{2}$

$= \frac{x^{2}-12}{8}=(f\,o\,g)^{-1}(x)$

It can also be observed that $(f\,o\,g)^{-1}(x) \neq (g\,o\,f)^{-1}(x)$.

Exercise: Show that the relations (4) and (5) also hold for the following functions.

• $f(x) = \frac{3}{x} - 5$
• $g(x)=2x$

## Graphing Inverse Function

It should be noted that an inverse function $f^{-1}(x)$ of any function f(x) can be regarded as an individual function i.e., $f^{-1}(x) = g(x)$. It can also be graphed, just like you would plot any other function.

To graph an inverse function, there are a number of methods. You can do it by using a calculator and evaluating the expression of the inverse function at different values of input. In this way, you get several coordinate points (x, f(x)). The points can be plotted on a graph after properly defining the scale on the x and y axes. You can define the scale as one small box equal to 1 unit or 10 units or 50 units, whatever suits you. To mark the coordinate point, say (2,-10), we move 2 units right on the x-axis from the origin and then 10 units vertically down. Similarly, other points can be marked. Then the last step is to join all these points with a smooth curve.

Software plotting tools can also be utilized to have a plot for any inverse function (or any expression in general). You would only need to type the expression and the graph will be plotted automatically. There are several customization options that these tools offer, such as zooming in/out, changing scale etc.

Let us visit an example. Given a function $f(x) = \sqrt{x}$, the inverse function is $f^{-1}(x) = x^{2}$. As you can see, the domain of f(x) is the set of positive real numbers (including 0), whereas the range of f(x) is also the set of positive real numbers. What about the domain of $f^{-1}(x) = x^{2}$? You might be tempted to say that its domain is the set of real numbers (both positive and negative) because $x^{2}$ is defined for all real numbers. But this is not true.

We had stated earlier that the inverse of inverse function is the original function. Whereas the inverse of $x^{2}$ is not $\sqrt{x}$ UNLESS we restrict the domain of $\sqrt{x}$ to only positive real numbers. Thus, the domain of $f^{-1}(x) = x^{2}$ is the set of positive real numbers only (otherwise its inverse will not be our original function $\sqrt{x}$). Reread the paragraph again to understand this point.

This can also be shown from another point of view. As we know that the domain of the inverse function is the range of the original function. Therefore, the domain of $f^{-1}(x) = x^{2}$ is the range of $f(x) = \sqrt{x}$, which is the set of positive real numbers.

The domain should always be considered while plotting a graph for any expression!

The graph of the inverse function $f^{-1}(x)$ and its original function f(x) is plotted as

Note that how the domain for $f^{-1}(x)$ has been taken care of i.e., negative values of x have not been considered.

It should also be noted that we can extract information about the domain and range of an inverse function (or function also) from its graph. Simply observe the part of x-axis and y-axis, where the graph is defined, and you know the domain and range.

You can also observe that both $f^{-1}(x)$ and f(x) pass the vertical line test (i.e., both qualify the criteria for being a function).

Vertical line test: It is used to test whether a curve represents a function or not. In vertical line test, we seek whether a curve intersects any vertical line along the x-axis at more than a single point. If the curve does intersect a vertical line at more than a single point then it does not represent a function.

Horizontal line test: Given the graph of a function, we can also determine whether its inverse would also be a function or not. If a horizontal line intersects the curve of f(x) at more than a single point then we know for sure that its inverse is not a function.

Now, take a look at the graph again. The purple curve represents $f^{-1}(x)$ whereas, the red one represents f(x). Do you observe any kind of connection between the two? You should be able to tell that both are reflections of each other along the $45^{\circ}$ line i.e., y = x. To help you clearly observe this phenomenon. The graph is replotted with the $45^{\circ}$ line.

Exercise: Graph the inverse function$f^{-1}(x) = x^{\frac{1}{3}}$. Find its original function and graph the function also.

## Real World Examples

Congratulations! You now possess sound knowledge about inverse functions. Let us now take a look at some real-world examples of inverse function.

Volume of Sphere: You may recall that volume of a sphere $V_{s}$ is given by:

$V_{s} = \frac{4}{3}\pi r^{3}$

Since, the volume is dependent on the radius of the sphere, it can be said to be a function of radius r.

$V_{s} = f(r) = \frac{4}{3}\pi r^{3}$

If you are asked to find the radius for a particular value of Volume of sphere, what will you do? Yes, that is correct! You will find the inverse function.

$r = f^{-1}(V_{s}) = (\frac{3V_{s}}{4\pi})^{\frac{1}{3}}$

Radioactive decay: The number of particles (atoms) of a radioactive element keep on decreasing as some unstable atoms decay continuously and convert into atoms of other elements. This phenomenon is called the radioactive decay.

The number of atoms left after a particular amount of time t can be expressed in the form of an exponential relation.

$N = f(t) = N_{0}e^{-\lambda t}$

where $N_{0}$ is initial number of atoms and λ is a constant.

Note how the negative exponent i.e., -λt depicts the decay of the original number of particles (atoms) with respect to time. If we want to find the time at which, say, a certain number of atoms remain, then we are basically asking about the inverse function,

$t = f^{-1}(N) = -\frac{1}{\lambda }\ln \ln (\frac{N}{N_{0}})$

Note that the natural log (ln) comes out to be negative for $\frac{N}{N_{0}}<1$. This negative sign and the one in the expression cancel out. Thus, the value of time evaluates to be a positive number.

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