Inverse Function

Let f be defined a function from A to B such that for every element of B their exist a image. Let y be an arbitrary element of B. Then, f being onto, there exists an element x $\in$ A such that f(x) = y. Also, f being one-one, this x must be unique. Thus, for each y $\in$ B, there exists a unique element x $\in$ A such that f(x) = y. So, we may define a function, 

$f^{-1}$ B → A

So, $f^{-1}$(y) = x ⇔ f(x) = y

The above function $f^{-1}$ is called the inverse of f . 

mapping representation: An inverse function reverses the inputs and outputs.

Note:

• Inverse of bijective function is also bijective function. 
• If the inverse of f exist, then f is called an invertible function. i.e., A function f is invertible if and only if f is one-one onto.

Sample Problem 1

Let f : R → R be a function defined by f(x) = 2x - 3. Then, $f^{-1}$ is equal to

(a) x + 3

(b) $\frac{x+3}{2}$

(c) $\frac{x-3}{2}$

(d) None of these

Answer: (b)

Let, y = 2x - 3 

⇒ 2x = y + 3

⇒ x = $\frac{y+3}{2}$

So, $f^{-1}$(x) = $\frac{x+3}{2}$

Sample Problem 2

Let f : (4, 6) → (6, 8)  be a function defined by f(x) = x + $\left[\frac{x}{2}\right]$, where [$\cdot$] denotes the greatest integer function, then $f^{-1}$(x) is equal to

(a) x + $\left[\frac{x}{2}\right]$

(b) - x - 2

(c) x - 2

(d) $\frac{1}{x+\left[\frac{\pi}{2}\right]}$

Answer: (c)

Since, x $\in$ (4, 6)

So, f(x) = x + $\left[\frac{x}{2}\right]$ = x + 2

Let, f(x) = y

so, y = x + 2

⇒ x = y - 2

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

Sample Problem 3

Let f(x) = $x^2$ - x + 1, x $\geqslant \frac{1}{2}$ then the solution of the equation f(x) = $f^{-1}$(x) is
(a) x = 1
(b) x = 2
(c) x = $\frac{1}{2}$
(d) None of these

Answer:

Let, y = $x^2$ - x + 1

⇒ x = $\frac{1 \pm \sqrt{1-4(1-y)}}{2}$, x > $\frac{1}{2}$

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

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

Now, $x^2$ - x + 1 = $\frac{1}{2}$ + $\sqrt{x-\frac{3}{4}}$

⇒ x = 1

Sample Problem 4

Find the inverse for the function f(x) = $\frac{3x+2}{x-1}$ 

Answer:

First, replace f(x) with y and the function becomes,

y = $\frac{3x+2}{x-1}$ 

By replacing x with y we get,

x = $\frac{3y+2}{y-1}$ 

Now, solve y in terms of x :

x (y – 1) = 3y + 2

⇒ xy – x = 3y +2

⇒ xy – 3y = 2 + x

⇒ y (x – 3) = 2 + x

⇒ y = $\frac{x+2}{x-2}$

So, y = f^-1(x) = $\frac{x+2}{x-2}$

Contents:

• Ordered Pair
• Cartesian Product
• Relations
• Different Types of Relations
• Composition of Relations
• Functions or Mappings
• Classification of Functions 
• Modulus Function
• Greatest Integer Function
• Inverse Function
• Algebra of Real Functions
• Composition of Functions