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Algebra - Functions: domain, range, and notation

Grade 10IB

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

πŸ”‘Concepts

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A function is a relation where each input xx from the domain corresponds to exactly one output yy in the range. In a mapping diagram, this means every element in the first set has exactly one arrow pointing away from it.

Mapping diagram showing a one-to-one function where elements 1 and 2 map to A and B respectively.
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The Vertical Line Test is used to determine if a graph represents a function. If any vertical line x=cx = c intersects the graph at more than one point, the relation is not a function because one input would have multiple outputs.

A circle graph failing the vertical line test because a vertical line crosses it at two points.
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Domain refers to the set of all possible input values (xx-values). For a square root function f(x)=xf(x) = \sqrt{x}, the domain is xβ‰₯0x \geq 0 because we cannot take the square root of a negative number in real numbers.

Graph of y equals square root of x starting from the origin and extending to the right.
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Range refers to the set of all possible output values (yy-values). For the function f(x)=x2f(x) = x^2, the range is yβ‰₯0y \geq 0 because squaring any real number results in a non-negative value.

Graph of y equals x squared showing the range starting from y equals 0 upwards.

πŸ“Formulae

f(x)=yf(x) = y

f:x↦ax+bf: x \mapsto ax + b

D={x∣x∈R}D = \{x | x \in \mathbb{R}\}

R={y∣y∈R}R = \{y | y \in \mathbb{R}\}

x≠0 (for f(x)=1x)x \neq 0 \text{ (for } f(x) = \frac{1}{x})

πŸ’‘Examples

Problem 1:

Given the function f(x)=5x2βˆ’3x+1f(x) = 5x^2 - 3x + 1, evaluate f(βˆ’3)f(-3).

Solution:

  1. Substitute x=βˆ’3x = -3 into the function: f(βˆ’3)=5(βˆ’3)2βˆ’3(βˆ’3)+1f(-3) = 5(-3)^2 - 3(-3) + 1
  2. Calculate the square of βˆ’3-3: f(βˆ’3)=5(9)βˆ’3(βˆ’3)+1f(-3) = 5(9) - 3(-3) + 1
  3. Perform the multiplications: f(βˆ’3)=45+9+1f(-3) = 45 + 9 + 1
  4. Add the terms: f(βˆ’3)=55f(-3) = 55

Explanation:

To evaluate a function for a specific value, replace the variable xx with the given number and follow the order of operations (BIDMAS/BODMAS).

Problem 2:

Determine the domain and range for the relation R={(1,2),(3,4),(5,2),(7,8)}R = \{(1, 2), (3, 4), (5, 2), (7, 8)\}. Is this relation a function?

Solution:

  1. List the first elements for the Domain: D={1,3,5,7}D = \{1, 3, 5, 7\}
  2. List the unique second elements for the Range: R={2,4,8}R = \{2, 4, 8\}
  3. Check for uniqueness: Each input xx (1,3,5,71, 3, 5, 7) appears only once and is assigned to exactly one output.

Explanation:

The domain consists of all unique xx-coordinates, and the range consists of all unique yy-coordinates. Because no input value is repeated with a different output, the relation is a function.

Problem 3:

Identify the domain and range of the function f(x)=1xβˆ’2f(x) = \frac{1}{x-2} and sketch its behavior near the vertical asymptote.

Graph of 1 over x minus 2 showing a vertical asymptote at x equals 2.

Solution:

The denominator cannot be zero, so xβˆ’2β‰ 0x - 2 \neq 0, which means xβ‰ 2x \neq 2. Thus, Domain D={x∈R∣xβ‰ 2}D = \{x \in \mathbb{R} | x \neq 2\}. Since the numerator is non-zero, f(x)f(x) can never be 00. Thus, Range R={y∈R∣yβ‰ 0}R = \{y \in \mathbb{R} | y \neq 0\}.

Explanation:

The value x=2x=2 causes division by zero, creating a vertical asymptote. As xx approaches 22 from the right, f(x)f(x) goes to +∞+\infty; from the left, it goes to βˆ’βˆž-\infty.

Problem 4:

Determine the domain and range for the linear function f(x)=βˆ’2x+4f(x) = -2x + 4 on the restricted interval βˆ’1≀x≀3-1 \leq x \leq 3.

A line segment from (-1,6) to (3,-2) representing the function over a restricted domain.

Solution:

  1. Domain is given as [βˆ’1,3][-1, 3].
  2. Evaluate endpoints: f(βˆ’1)=βˆ’2(βˆ’1)+4=6f(-1) = -2(-1) + 4 = 6 and f(3)=βˆ’2(3)+4=βˆ’2f(3) = -2(3) + 4 = -2.
  3. Since it is a decreasing linear function, the range is [βˆ’2,6][-2, 6].

Explanation:

For a linear function, the range over a closed interval is simply the interval between the yy-values of the endpoints.