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Algebra - Mappings

Grade 10IB

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

🔑Concepts

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A mapping is a rule that assigns elements from one set (the domain) to elements in another set (the codomain).

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A function is a specific type of mapping where every element in the domain is mapped to exactly one element in the codomain.

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Domain: The set of all possible input values (xx-values).

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Codomain: The set of all potential output values.

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Range: The set of actual output values (yy-values) that are mapped from the domain. The range is a subset of the codomain.

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One-to-one mapping: Each element in the domain maps to a unique element in the codomain. For example, f(x)=x+3f(x) = x + 3.

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Many-to-one mapping: Multiple elements in the domain map to the same element in the codomain. For example, f(x)=x2f(x) = x^2 where both 22 and −2-2 map to 44.

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One-to-many mapping: One element in the domain maps to multiple elements in the codomain. These are not functions.

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Inverse Function: If a function ff is one-to-one, the inverse function f−1f^{-1} reverses the mapping, such that f−1(y)=xf^{-1}(y) = x.

📐Formulae

f:x↦yf: x \mapsto y

f(x)=ax+bf(x) = ax + b

f−1(x) (Inverse notation)f^{-1}(x) \text{ (Inverse notation)}

fg(x)=f(g(x)) (Composite function)fg(x) = f(g(x)) \text{ (Composite function)}

💡Examples

Problem 1:

Given the function f(x)=3x−5f(x) = 3x - 5 with the domain {1,2,3,4}\{1, 2, 3, 4\}, find the range.

Solution:

f(1)=3(1)−5=−2f(1) = 3(1) - 5 = -2 f(2)=3(2)−5=1f(2) = 3(2) - 5 = 1 f(3)=3(3)−5=4f(3) = 3(3) - 5 = 4 f(4)=3(4)−5=7f(4) = 3(4) - 5 = 7 Range = {−2,1,4,7}\{-2, 1, 4, 7\}

Explanation:

To find the range, substitute each value from the domain into the function expression to find the corresponding outputs.

Problem 2:

Determine if the mapping y2=xy^2 = x is a function.

Solution:

If x=4x = 4, then y2=4y^2 = 4, which means y=2y = 2 or y=−2y = -2.

Explanation:

Since one input (x=4x = 4) results in more than one output (y=2y = 2 and y=−2y = -2), this is a one-to-many mapping and therefore is not a function.

Problem 3:

Find the inverse function f−1(x)f^{-1}(x) for f(x)=2x+13f(x) = \frac{2x + 1}{3}.

Solution:

Let y=2x+13y = \frac{2x + 1}{3}. Swap xx and yy: x=2y+13x = \frac{2y + 1}{3} Multiply by 3: 3x=2y+13x = 2y + 1 Subtract 1: 3x−1=2y3x - 1 = 2y Divide by 2: y=3x−12y = \frac{3x - 1}{2} Therefore, f−1(x)=3x−12f^{-1}(x) = \frac{3x - 1}{2}

Explanation:

To find the inverse, replace f(x)f(x) with yy, swap the variables xx and yy, and solve for the new yy.