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Algebra - Functions and Composite Functions

Grade 10IGCSE

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

🔑Concepts

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Definition of a Function: A rule that maps an input (domain) to exactly one output (range). Usually written as f(x)f(x) or f:x↦yf: x \mapsto y.

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Domain and Range: The domain is the set of all possible input values (xx), and the range is the set of all possible output values (f(x)f(x)).

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Substitution: To find the value of a function for a specific number, replace the variable xx with that number.

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Composite Functions: Applying one function to the result of another. fg(x)fg(x) means applying gg first, then ff. Note that fg(x)≠gf(x)fg(x) \neq gf(x) in most cases.

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Inverse Functions: The function that 'undoes' the original function, denoted as f−1(x)f^{-1}(x). The domain of ff becomes the range of f−1f^{-1}.

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Solving Function Equations: Setting the function expression equal to a value (e.g., f(x)=10f(x) = 10) and solving for xx.

📐Formulae

f(x)=yf(x) = y (Basic function notation)

fg(x)=f(g(x))fg(x) = f(g(x)) (Composite function formula)

ff−1(x)=xf f^{-1}(x) = x (Identity property of inverse functions)

To find f−1(x)f^{-1}(x): 1. Let y=f(x)y = f(x), 2. Swap xx and yy, 3. Rearrange to make yy the subject.

💡Examples

Problem 1:

Given f(x)=3x−5f(x) = 3x - 5, find f(4)f(4).

Solution:

f(4)=3(4)−5=12−5=7f(4) = 3(4) - 5 = 12 - 5 = 7

Explanation:

Substitute the value 4 into the expression wherever xx appears and simplify.

Problem 2:

If f(x)=2x+1f(x) = 2x + 1 and g(x)=x2g(x) = x^2, find the expression for fg(x)fg(x).

Solution:

fg(x)=f(g(x))=f(x2)=2(x2)+1=2x2+1fg(x) = f(g(x)) = f(x^2) = 2(x^2) + 1 = 2x^2 + 1

Explanation:

To find fg(x)fg(x), substitute the entire expression for g(x)g(x) into the xx position of f(x)f(x).

Problem 3:

Find the inverse function f−1(x)f^{-1}(x) for f(x)=x+32f(x) = \frac{x + 3}{2}.

Solution:

  1. Let y=x+32y = \frac{x + 3}{2}. 2. Swap variables: x=y+32x = \frac{y + 3}{2}. 3. Solve for yy: 2x=y+3⇒y=2x−32x = y + 3 \Rightarrow y = 2x - 3. Therefore, f−1(x)=2x−3f^{-1}(x) = 2x - 3.

Explanation:

The inverse function is found by reversing the operations. We swap xx and yy and isolate yy to find the new rule.

Problem 4:

Given h(x)=5x−2h(x) = 5x - 2, find xx when h(x)=13h(x) = 13.

Solution:

5x−2=13⇒5x=15⇒x=35x - 2 = 13 \Rightarrow 5x = 15 \Rightarrow x = 3

Explanation:

Set the algebraic expression for the function equal to the given value and solve the resulting linear equation for xx.