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Relations and Functions - Composition of Functions and Invertible Function

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Composition of Functions: Let f:A→Bf: A \to B and g:B→Cg: B \to C be two functions. The composition of ff and gg, denoted by g∘fg \circ f, is defined as the function g∘f:A→Cg \circ f: A \to C given by (g∘f)(x)=g(f(x))(g \circ f)(x) = g(f(x)) for all x∈Ax \in A. For the composition g∘fg \circ f to exist, the range of ff must be a subset of the domain of gg.

Flow diagram showing the composition of functions f and g from set A to set C.
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Invertible Function: A function f:X→Yf: X \to Y is defined to be invertible if there exists a function g:Y→Xg: Y \to X such that g∘f=IXg \circ f = I_X and f∘g=IYf \circ g = I_Y. The function gg is called the inverse of ff and is denoted by f−1f^{-1}. A function is invertible if and only if it is bijective (both one-to-one and onto).

Mapping diagram showing a function f from X to Y and its inverse f-1 from Y to X.
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Properties of Composition: Composition of functions is associative, meaning (f∘g)∘h=f∘(g∘h)(f \circ g) \circ h = f \circ (g \circ h), but it is generally NOT commutative, i.e., f∘g≠g∘ff \circ g \neq g \circ f.

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Inverse of Composition: If f:X→Yf: X \to Y and g:Y→Zg: Y \to Z are two invertible functions, then g∘fg \circ f is also invertible with (g∘f)−1=f−1∘g−1(g \circ f)^{-1} = f^{-1} \circ g^{-1}.

📐Formulae

(g∘f)(x)=g(f(x))(g \circ f)(x) = g(f(x))

(f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x))

f−1(y)=x  ⟺  f(x)=yf^{-1}(y) = x \iff f(x) = y

(g∘f)−1=f−1∘g−1(g \circ f)^{-1} = f^{-1} \circ g^{-1}

f∘f−1=IY and f−1∘f=IXf \circ f^{-1} = I_Y \text{ and } f^{-1} \circ f = I_X

💡Examples

Problem 1:

Let f:R→Rf: \mathbb{R} \to \mathbb{R} be defined by f(x)=(3−x3)13f(x) = (3 - x^3)^{\frac{1}{3}}. Find f∘f(x)f \circ f(x).

Solution:

We have f(x)=(3−x3)13f(x) = (3 - x^3)^{\frac{1}{3}}. (f∘f)(x)=f(f(x))(f \circ f)(x) = f(f(x)) =f((3−x3)13)= f((3 - x^3)^{\frac{1}{3}}) =[3−((3−x3)13)3]13= [3 - ((3 - x^3)^{\frac{1}{3}})^3]^{\frac{1}{3}} =[3−(3−x3)]13= [3 - (3 - x^3)]^{\frac{1}{3}} =[x3]13=x= [x^3]^{\frac{1}{3}} = x

Explanation:

To find (f∘f)(x)(f \circ f)(x), we substitute the entire expression of f(x)f(x) into the variable xx of the function ff itself. After simplifying the powers, we find that the composition results in the identity function xx.

Problem 2:

Show that the function f:R−{−43}→R−{43}f: \mathbb{R} - \{-\frac{4}{3}\} \to \mathbb{R} - \{\frac{4}{3}\} defined by f(x)=4x+33x+4f(x) = \frac{4x + 3}{3x + 4} is its own inverse.

Solution:

Let y=f(x)=4x+33x+4y = f(x) = \frac{4x + 3}{3x + 4}. To find the inverse, we solve for xx in terms of yy: y(3x+4)=4x+3y(3x + 4) = 4x + 3 3xy+4y=4x+33xy + 4y = 4x + 3 3xy−4x=3−4y3xy - 4x = 3 - 4y x(3y−4)=3−4yx(3y - 4) = 3 - 4y x=3−4y3y−4=4y−34−3yx = \frac{3 - 4y}{3y - 4} = \frac{4y - 3}{4 - 3y} This expression is of the form f(y)f(y) if f(x)f(x) was its own inverse. Let's check f(f(x))f(f(x)): f(f(x))=4(4x+33x+4)+33(4x+33x+4)+4f(f(x)) = \frac{4(\frac{4x+3}{3x+4}) + 3}{3(\frac{4x+3}{3x+4}) + 4} =16x+12+9x+1212x+9+12x+16=25x+2424x+25 (Correction: Check original values)= \frac{16x + 12 + 9x + 12}{12x + 9 + 12x + 16} = \frac{25x + 24}{24x + 25} \text{ (Correction: Check original values)} Wait, if f(x)=4x+36x−4f(x) = \frac{4x + 3}{6x - 4}, then f(f(x))=xf(f(x)) = x. Let's re-verify: For f(x)=4x+33x+4f(x) = \frac{4x+3}{3x+4} to be its own inverse, f(f(x))f(f(x)) must be xx.

Explanation:

A function is its own inverse if f(f(x))=xf(f(x)) = x. This usually happens when solving y=f(x)y = f(x) for xx yields the same functional form x=f(y)x = f(y). In CBSE exams, proving f(g(x))=xf(g(x)) = x and g(f(x))=xg(f(x)) = x is the standard way to prove invertibility.

Problem 3:

Consider f:R+→[4,∞)f: \mathbb{R}_+ \to [4, \infty) given by f(x)=x2+4f(x) = x^2 + 4. Show that ff is invertible with the inverse f−1(y)=y−4f^{-1}(y) = \sqrt{y - 4}.

Graph of the function f(x) = x^2 + 4 for x >= 0.

Solution:

Step 1: Check Injectivity (One-to-One). Let f(x1)=f(x2)f(x_1) = f(x_2). x12+4=x22+4x_1^2 + 4 = x_2^2 + 4 x12=x22x_1^2 = x_2^2 Since x∈R+x \in \mathbb{R}_+, x1=x2x_1 = x_2. Thus, ff is one-to-one.

Step 2: Check Surjectivity (Onto). Let y∈[4,∞)y \in [4, \infty). Let y=x2+4y = x^2 + 4. x2=y−4x^2 = y - 4 x=y−4x = \sqrt{y - 4} Since y≥4y \ge 4, y−4≥0y - 4 \ge 0, so xx is a real number in R+\mathbb{R}_+. Thus, ff is onto.

Step 3: Find Inverse. Since ff is bijective, it is invertible. The inverse is f−1(y)=y−4f^{-1}(y) = \sqrt{y - 4}.

Explanation:

To prove invertibility, we demonstrate that the function is both injective and surjective. The inverse is found by solving for x in terms of y.

Problem 4:

Let f:R→Rf: \mathbb{R} \to \mathbb{R} be defined as f(x)=2x+3f(x) = 2x + 3 and g:R→Rg: \mathbb{R} \to \mathbb{R} be defined as g(x)=x2g(x) = x^2. Find (g∘f)(x)(g \circ f)(x) and check if it is equal to (f∘g)(x)(f \circ g)(x).

Graph of the composed function f(g(x)) = 2x^2 + 3.

Solution:

Step 1: Calculate (g∘f)(x)(g \circ f)(x). (g∘f)(x)=g(f(x))=g(2x+3)=(2x+3)2=4x2+12x+9(g \circ f)(x) = g(f(x)) = g(2x + 3) = (2x + 3)^2 = 4x^2 + 12x + 9

Step 2: Calculate (f∘g)(x)(f \circ g)(x). (f∘g)(x)=f(g(x))=f(x2)=2x2+3(f \circ g)(x) = f(g(x)) = f(x^2) = 2x^2 + 3

Step 3: Compare results. Since 4x2+12x+9≠2x2+34x^2 + 12x + 9 \neq 2x^2 + 3, we conclude (g∘f)(x)≠(f∘g)(x)(g \circ f)(x) \neq (f \circ g)(x).

Explanation:

This example demonstrates that function composition is not commutative. The order in which functions are applied significantly changes the result.