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# Inverse of a Function – Explanation & Examples

## What is an inverse function?

**In mathematics, an inverse function is a function that undoes the action of another function.**

**For example**, addition and multiplication are the inverse of subtraction and division, respectively.

The inverse of a function can be viewed as reflecting the original function over the line y = x. In simple words, the inverse function is obtained by swapping the (x, y) of the original function to (y, x).

We use the symbol f^{ − 1} to denote an inverse function. For example, if f (x) and g (x) are inverses of each other, then we can symbolically represent this statement as:

g(x) = f^{ − 1}(x) or f(x) = g^{−1}(x)

One thing to note about the inverse function is that the inverse of a function is not the same as its reciprocal, i.e., f^{ – 1} (x) ≠ 1/ f(x). This article will discuss how to find the inverse of a function.

Since not all functions have an inverse, it is therefore important to check whether a function has an inverse before embarking on determining its inverse.

We check whether or not a function has an inverse in order to avoid wasting time trying to find something that does not exist.

### One-to-one functions

**So how do we prove that a given function has an inverse? Functions that have inverse are called one-to-one functions. **

A function is said to be one-to-one if, for each number y in the range of f, there is exactly one number x in the domain of f such that f (x) = y.

*In other words, the domain and range of one-to-one function have the following relations:*

- Domain of f
^{−1 }= Range of f.

- Range of f
^{−1}= Domain of f.

For example, to check if f(x) = 3x + 5 is one to one function given, f(a) = 3a + 5 and f(b) = 3b + 5.

⟹ 3a + 5 = 3b + 5

⟹ 3a = 3b

⟹ a = b.

Therefore, f (x) is one-to-one function because, a = b.

Consider another case where a function f is given by f = {(7, 3), (8, –5), (–2, 11), (–6, 4)}. This function is one-to-one because none of its y - values appear more than once.

What about this other function h = {(–3, 8), (–11, –9), (5, 4), (6, –9)}? Function h is not one-to-one because the y- value of –9 appears more than once.

You can also graphically check one-to-one function by drawing a vertical line and horizontal line through a function graph. A function is one-to-one if both the horizontal and vertical line passes through the graph once.

## How to Find the Inverse of a Function?

Finding the inverse of a function is a straightforward process, though we really need to be careful with a couple of steps. In this article, we are going to assume that all functions we are going to deal with are one to one.

*Here is the procedure of finding of the inverse of a function f(x):*

- Replace the function notation f(x) with y.
- Swap x with y and vice versa.
- From step 2, solve the equation for y. Be careful with this step.
- Finally, change y to f
^{−1}(x). This is the inverse of the function. - You can verify your answer by checking if the following two statements are true:

⟹ (f ∘ f^{−1}) (x) = x

⟹ (f^{−1 }∘ f) (x) = x

Let’s work a couple of examples.

*Example 1*

Given the function f (x) = 3x − 2, find its inverse.

__Solution__

f(x) = 3x − 2

Replace f(x) with y.

⟹ y = 3x − 2

Swap x with y

⟹ x = 3y − 2

Solve for y

x + 2 = 3y

Divide through by 3 to get;

1/3(x + 2) = y

x/3 + 2/3 = y

Finally, replace y with f^{−1}(x).

f^{−1}(x) = x/3 + 2/3

Verify (f ∘ f^{−1}) (x) = x

(f ∘ f^{−1}) (x) = f [f ^{−1 }(x)]

= f (x/3 + 2/3)

⟹ 3(x/3 + 2/3) – 2

⟹ x + 2 – 2

= x

Hence, f ^{−1 }(x) = x/3 + 2/3 is the correct answer.

*Example 2*

Given f(x) = 2x + 3, find f^{−1}(x).

__Solution__

f(x) = y = 2x + 3

2x + 3 = y

Swap x and y

⟹2y + 3 = x

Now solve for y

⟹2y = x – 3

⟹ y = x/2 – 3/2

Finally substitute y with f ^{−1}(x)

⟹ f ^{−1 }(x) = (x– 3)/2

*Example 3*

Give the function f (x) = log_{10 }(x), find f ^{−1 }(x).

__Solution__

f (x) = log₁₀ (x)

Replaced f (x) with y

⟹ y = log_{10 }(x) ⟹ 10 ^{y} = x

Now swap x with y to get;

⟹ y = 10 ^{x}

Finally, substitute y with f^{−1}(x).

f ^{-1 }(x) = 10 ^{x}

Therefore, the inverse of f(x) = log_{10}(x) is f^{-1}(x) = 10^{x}

*Example 4*

Find the inverse of the following function g(x) = (x + 4)/ (2x -5)

__Solution__

g(x) = (x + 4)/ (2x -5) ⟹ y = (x + 4)/ (2x -5)

Interchange y with x and vice versa

y = (x + 4)/ (2x -5) ⟹ x = (y + 4)/ (2y -5)

⟹ x(2y−5) = y + 4

⟹ 2xy − 5x = y + 4

⟹ 2xy – y = 4 + 5x

⟹ (2x − 1) y = 4 + 5x

Divide both side of the equation by (2x − 1).

⟹ y = (4 + 5x)/ (2x − 1)

Replace y with g ^{– 1}(x)

= g ^{– 1}(x) = (4 + 5x)/ (2x − 1)

Proof:

(g ∘ g^{−1}) (x) = g [g ^{−1}(x)]

= g [(4 + 5x)/ (2x − 1)]

= [(4 + 5x)/ (2x − 1) + 4]/ [2(4 + 5x)/ (2x − 1) − 5]

Multiply the both the numerator and denominator by (2x − 1).

⟹ (2x − 1) [(4 + 5x)/ (2x − 1) + 4]/ [2(4 + 5x)/ (2x − 1) − 5] (2x − 1).

⟹ [4 + 5x + 4(2x − 1)]/ [ 2(4 + 5x) − 5(2x − 1)]

⟹ [4 + 5x + 8x−4]/ [8 + 10x − 10x + 5]

⟹13x/13 = x

Therefore, g ^{– 1 }(x) = (4 + 5x)/ (2x − 1)

*Example 5*

Determine the inverse of the following function f(x) = 2x – 5

__Solution__

Replace f(x) with y.

f(x) = 2x – 5⟹ y = 2x – 5

Switch x and y to get;

⟹ x = 2y – 5

Isolate the variable y.

2y = x + 5

⟹ y = x/2 + 5/2

Change y back to f ^{–1}(x).

⟹ f ^{–1}(x) = (x + 5)/2

*Example 6*

Find the inverse of the function h(x) = (x – 2)^{3}.

__Solution__

Change h (x) to y to get;

h(x) = (x – 2)^{3}⟹ y = (x – 2)^{3}

Swap x and y

⟹ x = (y – 2)^{3}

Isolate y.

y^{3 }= x + 2^{3}

Find the cube root of both sides of the equation.

^{3}√y^{3} = ^{3}√x^{3} + ^{3}√2^{3}

y = ^{3}√ (2^{3}) + 2

Replace y with h^{ – 1}(x)

h^{ – 1}(x) = ^{3}√ (2^{3}) + 2

*Example 7*

Find the inverse of h (x) = (4x + 3)/(2x + 5)

__Solution__

Replace h (x) with y.

h (x) = (4x+3)/(2x+5) ⟹ y = (4x + 3)/(2x + 5)

Swap x and y.

⟹ x = (4y + 3)/ (2y + 5).

Solve for y in the above equation as follows:

⟹ x = (4y + 3)/ (2y + 5)

Multiply both sides by (2y + 5)

⟹ x (2y + 5) = 4y + 3

Distribute the x

⟹ 2xy + 5x = 4y + 3

Isolate y.

⟹ 2xy – 4y = 3 – 5x

⟹ y (2x – 4) = 3 – 5x

Divide through by 2x – 4 to get;

⟹ y = (3 – 5x)/ (2x – 4)

Finally replace y with h^{ – 1}(x).

⟹ h^{ – 1 }(x) = (3 – 5x)/ (2x – 4)