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Functions - Composite and inverse functions

Grade 10IB

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

🔑Concepts

Composite Functions: A composite function, denoted as (fg)(x)(f \circ g)(x) or f(g(x))f(g(x)), is formed by applying one function to the result of another. Visually, this can be imagined as a 'function machine' chain where the output of g(x)g(x) becomes the raw input for ff. The inner function gg is evaluated first, and its result is substituted into the outer function ff.

Order of Composition: The order in which functions are composed is crucial because composition is generally not commutative. This means f(g(x))f(g(x)) is usually not equal to g(f(x))g(f(x)). On a mapping diagram, this is seen as following different paths between sets of values, leading to different final destinations.

Inverse Functions: An inverse function, written as f1(x)f^{-1}(x), performs the opposite operation of the original function f(x)f(x). If ff maps xx to yy, then f1f^{-1} maps yy back to xx. Visually, this 'undoes' the transformation, returning the value to its starting point in the domain.

Graphical Relationship of Inverses: The graph of an inverse function f1(x)f^{-1}(x) is a reflection of the original function f(x)f(x) across the identity line y=xy = x. If a point (a,b)(a, b) lies on the graph of ff, then the point (b,a)(b, a) must lie on the graph of f1f^{-1}. This creates a symmetrical mirror image relative to the diagonal line passing through the origin.

Existence of an Inverse (One-to-One): A function has an inverse only if it is a 'one-to-one' (injective) function, meaning every yy-value corresponds to exactly one xx-value. Visually, you can use the Horizontal Line Test: if any horizontal line crosses the graph of f(x)f(x) more than once, the function does not have an inverse over that domain.

Domain and Range Swap: When finding an inverse, the domain of the original function ff becomes the range of the inverse function f1f^{-1}, and the range of ff becomes the domain of f1f^{-1}. This reciprocal relationship is a direct result of swapping the xx and yy coordinates.

Identity Property: When a function is composed with its own inverse, they cancel each other out to yield the original input. This is expressed as f(f1(x))=xf(f^{-1}(x)) = x and f1(f(x))=xf^{-1}(f(x)) = x. Graphically, the result of this composition is always the straight line y=xy = x.

📐Formulae

(fg)(x)=f(g(x))(f \circ g)(x) = f(g(x)) (The composite function of ff and gg)

f(f1(x))=xf(f^{-1}(x)) = x (Identity property of inverses)

f1(f(x))=xf^{-1}(f(x)) = x (Identity property of inverses)

To find f1(x)f^{-1}(x): set y=f(x)y = f(x), swap xx and yy, then solve for yy

💡Examples

Problem 1:

Given f(x)=2x+3f(x) = 2x + 3 and g(x)=x21g(x) = x^2 - 1, find the expression for (fg)(x)(f \circ g)(x) and evaluate (fg)(3)(f \circ g)(3).

Solution:

  1. Substitute the expression for g(x)g(x) into ff: (fg)(x)=f(g(x))=f(x21)(f \circ g)(x) = f(g(x)) = f(x^2 - 1) \ 2. Apply the rule for ff: f(x21)=2(x21)+3f(x^2 - 1) = 2(x^2 - 1) + 3 \ 3. Expand and simplify: 2x22+3=2x2+12x^2 - 2 + 3 = 2x^2 + 1 \ 4. Evaluate for x=3x = 3: (fg)(3)=2(3)2+1=2(9)+1=18+1=19(f \circ g)(3) = 2(3)^2 + 1 = 2(9) + 1 = 18 + 1 = 19

Explanation:

We treat the entire function g(x)g(x) as the input for ff. After simplifying the algebraic expression for the composite function, we substitute the specific value of 3 to find the final numerical result.

Problem 2:

Find the inverse function f1(x)f^{-1}(x) for f(x)=4x53f(x) = \frac{4x - 5}{3}.

Solution:

  1. Replace f(x)f(x) with yy: y=4x53y = \frac{4x - 5}{3} \ 2. Swap xx and yy variables: x=4y53x = \frac{4y - 5}{3} \ 3. Multiply both sides by 3: 3x=4y53x = 4y - 5 \ 4. Add 5 to both sides: 3x+5=4y3x + 5 = 4y \ 5. Divide by 4 to isolate yy: y=3x+54y = \frac{3x + 5}{4} \ 6. State the final inverse function: f1(x)=3x+54f^{-1}(x) = \frac{3x + 5}{4}

Explanation:

To find the inverse, we switch the roles of the input and output (x and y) and then use algebraic manipulation to solve for the new output variable.