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Relations and Functions - Inverse of a Function

Grade 12ICSE

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

🔑Concepts

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A function f:A→Bf: A \rightarrow B is invertible if and only if it is a bijection, meaning it is both one-to-one (injective) and onto (surjective). Visually, this ensures that every output has exactly one corresponding input, and every possible output in set BB is mapped to.

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The one-to-one property (Injectivity) requires that f(x1)=f(x2)f(x_1) = f(x_2) implies x1=x2x_1 = x_2. Geometrically, this is verified by the Horizontal Line Test: any horizontal line drawn through the graph of the function must intersect the curve at most once.

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The onto property (Surjectivity) requires that for every y∈By \in B, there exists an x∈Ax \in A such that f(x)=yf(x) = y. Visually, this means the range of the function covers the entire codomain; if you project the graph onto the y-axis, there should be no gaps within the specified codomain.

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The domain and range relationship is inverted: the domain of ff becomes the range of f−1f^{-1}, and the range of ff becomes the domain of f−1f^{-1}. If the point (a,b)(a, b) lies on the graph of ff, then the point (b,a)(b, a) must lie on the graph of f−1f^{-1}.

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Geometrically, the graph of an inverse function y=f−1(x)y = f^{-1}(x) is the reflection of the original function y=f(x)y = f(x) about the identity line y=xy = x. Imagine the line y=xy=x acting as a mirror; the shapes of the two functions will be symmetric across this diagonal axis.

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The composition of a function and its inverse yields the identity function. Specifically, (f∘f−1)(x)=x(f \circ f^{-1})(x) = x for all xx in the domain of f−1f^{-1}, and (f−1∘f)(x)=x(f^{-1} \circ f)(x) = x for all xx in the domain of ff. This is the mathematical 'undoing' of the original operation.

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Monotonicity and Invertibility: A continuous function is invertible if it is strictly monotonic (either strictly increasing or strictly decreasing) on its entire domain. Visually, a strictly increasing graph always moves upward from left to right, preventing any horizontal line from crossing it twice.

📐Formulae

f:A→B  ⟺  f−1:B→Af: A \rightarrow B \iff f^{-1}: B \rightarrow A

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

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

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

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

💡Examples

Problem 1:

Let f:R−{75}→R−{35}f: \mathbb{R} - \{\frac{7}{5}\} \rightarrow \mathbb{R} - \{\frac{3}{5}\} be defined by f(x)=3x+45x−7f(x) = \frac{3x + 4}{5x - 7}. Find f−1(x)f^{-1}(x).

Solution:

Step 1: Set y=f(x)y = f(x). Hence, y=3x+45x−7y = \frac{3x + 4}{5x - 7}. \nStep 2: Solve for xx in terms of yy. \nMultiply both sides by (5x−7)(5x - 7): \ny(5x−7)=3x+4y(5x - 7) = 3x + 4 \n5xy−7y=3x+45xy - 7y = 3x + 4 \nStep 3: Collect all terms involving xx on one side: \n5xy−3x=7y+45xy - 3x = 7y + 4 \nx(5y−3)=7y+4x(5y - 3) = 7y + 4 \nStep 4: Isolate xx: \nx=7y+45y−3x = \frac{7y + 4}{5y - 3} \nStep 5: Replace xx with f−1(y)f^{-1}(y): \nf−1(y)=7y+45y−3f^{-1}(y) = \frac{7y + 4}{5y - 3} \nReplacing yy with xx, we get: \nf−1(x)=7x+45x−3f^{-1}(x) = \frac{7x + 4}{5x - 3}.

Explanation:

To find the inverse, we express the independent variable xx as a function of the dependent variable yy. This algebraic manipulation swaps the roles of input and output.

Problem 2:

Show that the function f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} defined by f(x)=4x+3f(x) = 4x + 3 is invertible and find its inverse.

Solution:

Step 1: Prove ff is one-to-one (Injective). \nLet f(x1)=f(x2)f(x_1) = f(x_2). \n4x1+3=4x2+3  ⟹  4x1=4x2  ⟹  x1=x24x_1 + 3 = 4x_2 + 3 \implies 4x_1 = 4x_2 \implies x_1 = x_2. Thus, ff is injective. \nStep 2: Prove ff is onto (Surjective). \nLet y∈Ry \in \mathbb{R}. We seek xx such that y=4x+3y = 4x + 3. \nx=y−34x = \frac{y - 3}{4}. Since yy is any real number, xx is also a real number. Thus, ff is surjective. \nStep 3: Since ff is a bijection, f−1f^{-1} exists. \nFrom Step 2, x=f−1(y)=y−34x = f^{-1}(y) = \frac{y - 3}{4}. \nTherefore, f−1(x)=x−34f^{-1}(x) = \frac{x - 3}{4}.

Explanation:

A function is invertible only if it is a bijection. We first verify injectivity (one-to-one) and surjectivity (onto) before calculating the inverse expression by solving for xx.