Suppose a set of ordered pairs defines a relation. Then the inverse relation is obtained by reversing the coordinates of each ordered pair. This means if (a, b) is an element of a relation then (b, a) is the corresponding element in the inverse relation. In short inverse relation switches input and output. If both the relation and its inverse are functions then two functions serve as inverse functions of each other.

The function g which is the inverse of f can be denoted as f-1 and read as "f inverse".

The Domain and Range of function switch the roles as Range and Domain of the inverse function. This means all functions do not have inverses. For an inverse function to be defined the original function needs to be one-to-one, that is no image in the range can have more than one pre image in the domain.
Definition: Suppose f(x) is an one-to-one function with domain D and range R. Then its inverse function f-1(x) has domain R and range D and defined by
f-1(y) = x ⇔ f(x) = y for any y ∈ R.

  1. Suppose f and f-1 are inverse functions. Then f(a) = b if and only if f-1(b) = a.
  2. The domain and range of a function switch the roles as range and domain for the inverse functions.
  3. Two functions f and g are inverse functions of each other if and only if both the compositions are identity functions.
    [f o g](x) = x and [g o f](x) = x.
  4. The graph of f-1 is the reflection of the graph of f over the line y = x.
  5. If a function is one-to-one, then any horizontal line drawn will intersect the graph of the function at not more than one point. This is property is used in horizontal line test to determine whether the inverse of a function exists.
The following steps are followed in finding inverse of a given function.
  1. Replace f(x) with y in the equation given.
  2. Swap x and y in the equation.
  3. Solve for y.
  4. Replace y with f-1.

Solved Example

Question: Find the inverse of the function f(x) = $\frac{x}{3}$$-5$
y = $\frac{x}{3}$$-5$                     f(x) replaced with y.

                                                    x = $\frac{y}{3}$$-5$                     x and y swapped.

                                                  3x = y - 15

                                                    y = 3x + 15                                Equation solved for y

                                              f-1(x) = 3x + 15                                 Inverse function.

The condition for two functions to be inverse of each other is that their compositions either way outputs the input itself.
That is f(g(x)) = x and g(f(x)) = x.
We check this condition in verifying whether the two given functions are inverses of each other.

Solved Example

Question: Verify f(x) = x3 - 5   and g(x) = $\sqrt[3]{x+5}$ are inverses of each other.
[f o g](x) = f(g(x)) = $(\sqrt[3]{x+5})^{3}-5$ = x + 5 - 5 = x

[g o f](x) = g(f(x)) = $\sqrt[3]{x^{3}-5+5}=\sqrt[3]{x^{3}}$ = x

As both the compositions yield x, f(x) and g(x) are inverses of each other.

How to find the inverse function from the graph of a function?
We use the rule that if (a, b) is a point on the graph of a function then (b, a) is a point on the graph of its inverse.
Let f(x) = x2 where x = 0. Graph f(x) and its inverse.
Let us make a table for graphing f(x) and f-1(x). To find the points on f(x) we reverse the coordinates of each ordered pair of f(x).

x f(x) [x, f(x)] [x,f-1(x)]
0 -4 (0, -4) (-4, 0)
1 -3 (1, -3) (-3, 1)
2 0 (2,0) (0,2)
3 5 (3, 5) (5, 3)

Inverse Function
The graph of f(x) is shown in green and f-1(x) in purple. You may also find the corresponding points on the two graphs are mirror images on the line y = x.
f-1(x) can be algebraically found as f-1(x) = $\sqrt{x+4}$
In calculus the derivative of inverse function is derived in terms of the derivative of the original function applying chain rule.
If the function f(x) is differentiable in its domain and g(x) is the inverse of f(x), then
g'(x) = $\frac{1}{f'(g(x))}$.
Let us consider the two linear functions f(x) = $\frac{x}{3}$$-5$ and g(x) = 3x +15 which are inverses of each other.

According to the formula given,

g'(x) = $\frac{1}{f'(g(x))}$

Differentiating f we get f'(x) = $\frac{1}{3}$. This being a constant f'(g(x)) = $\frac{1}{3}$.

Thus g'(x) = $\frac{1}{\frac{1}{3}}$ = 3.

We get the same differentiating g(x) direct.

Solved Examples

Question 1: Find the inverse of f(x) = $\frac{4x-1}{2x+3}$ and verify your answer.
y = $\frac{4x-1}{2x+3}$

x = $\frac{4y-1}{2y+3}$ x and y swapped.

(2y+3)x = 4y -1 Cross multiplication.

2xy + 3x = 4y -1

3x + 1 = 4y -2xy

y(4-2x) = 3x +1

y = $\frac{3x+1}{4-2x}$

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


[f o f-1](x) = f(f-1(x)) = $\frac{\frac{4(3x+1)}{4-2x}-1}{\frac{2(3x+1)}{4-2x}+3}$
= $\frac{\frac{12x+4-4+2x}{4-2x}}{\frac{6x+2+12-6x}{4-2x}}$ = $\frac{14x}{14}$ = x

[f-1 o f](x) = f-1(f(x)) = $\frac{\frac{3(4x-1)}{2x+3}+1}{4-\frac{2(4x-1)}{2x+3}}$
= $\frac{\frac{12x-3+2x+3}{2x+3}}{\frac{8x+12-8x+2}{2x+3}}$ = $\frac{14x}{14}$ = x

Hence verified.


Question 2: Find the inverse of f(x) = 4x2 + 7 for x = 0.
As the domain is limited to non-negative real numbers, f(x) is one-to-one and the inverse exists.

The range of f(x) is f(x) = 7.

f(x) = 4x2 + 7

y = 4x2 + 7

x = 4y2 + 7 x and y swapped.

4y2 = x - 7

y = $\frac{1}{2}$$\sqrt{x-7}$ Solved for y.

f-1(x) = $\frac{1}{2}$$\sqrt{x-7}$ for x = 7.