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Functions - Concept of functions

Grade 12IB_AI

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

πŸ”‘Concepts

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A function is a relation where every input xx in the domain corresponds to exactly one output yy in the range. Visually, this means a vertical line drawn anywhere on the graph of a function will cross the graph at most once (the Vertical Line Test).

Graph of a parabola showing the vertical line test with one intersection point.
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The domain is the set of all possible input values (xx-values) for which the function is defined. Restrictions typically arise from denominators (cannot be zero) and square roots (radicand must be non-negative).

Graph of y = sqrt(x-1) showing the domain starting at x=1.
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The range is the set of all resulting output values (yy-values). It represents the vertical extent of the graph.

Absolute value graph shifted up by 2 units, showing range starting at y=2.
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Function notation f(x)f(x) represents the value of the function at a specific xx. Mapping diagrams can visualize how elements from Set A (domain) relate to Set B (codomain).

Mapping diagram showing doubling function values.

πŸ“Formulae

y=f(x)y = f(x) strips the relation into input xx and output yy.

Domain: {x∈R∣conditions on x}\text{Domain: } \{x \in \mathbb{R} \mid \text{conditions on } x\}

Range: {y∈R∣conditions on y}\text{Range: } \{y \in \mathbb{R} \mid \text{conditions on } y\}

(f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x))

f(fβˆ’1(x))=xΒ andΒ fβˆ’1(f(x))=xf(f^{-1}(x)) = x \text{ and } f^{-1}(f(x)) = x

πŸ’‘Examples

Problem 1:

Given the function f(x)=2x2βˆ’3x+5f(x) = 2x^2 - 3x + 5, find the value of f(βˆ’2)f(-2).

Solution:

f(βˆ’2)=2(βˆ’2)2βˆ’3(βˆ’2)+5f(-2) = 2(-2)^2 - 3(-2) + 5 f(βˆ’2)=2(4)+6+5f(-2) = 2(4) + 6 + 5 f(βˆ’2)=8+6+5=19f(-2) = 8 + 6 + 5 = 19

Explanation:

To evaluate the function, substitute the value βˆ’2-2 into every instance of xx in the algebraic expression and simplify using the order of operations.

Problem 2:

Determine the domain of the function g(x)=5xβˆ’4g(x) = \frac{5}{\sqrt{x - 4}}.

Solution:

For the function to be defined:

  1. The denominator cannot be zero: xβˆ’4β‰ 0β€…β€ŠβŸΉβ€…β€Šxβ‰ 4\sqrt{x - 4} \neq 0 \implies x \neq 4
  2. The value inside the square root must be non-negative: xβˆ’4β‰₯0β€…β€ŠβŸΉβ€…β€Šxβ‰₯4x - 4 \geq 0 \implies x \geq 4 Combining these: x>4x > 4. Domain: {x∈R∣x>4}\{x \in \mathbb{R} \mid x > 4\}

Explanation:

In a rational function, the denominator cannot be zero. In a square root function, the radicand must be greater than or equal to zero. Since the root is in the denominator, it must be strictly greater than zero.

Problem 3:

If f(x)=3x+1f(x) = 3x + 1 and g(x)=x2g(x) = x^2, find (g∘f)(2)(g \circ f)(2).

Solution:

Step 1: Find f(2)f(2) f(2)=3(2)+1=7f(2) = 3(2) + 1 = 7 Step 2: Find g(f(2))g(f(2)), which is g(7)g(7) g(7)=(7)2=49g(7) = (7)^2 = 49 So, (g∘f)(2)=49(g \circ f)(2) = 49.

Explanation:

In a composite function, evaluate the inner function first. Use the output of f(2)f(2) as the input for the function gg.

Problem 4:

Determine the range of the function h(x)=βˆ’(xβˆ’3)2+4h(x) = -(x - 3)^2 + 4 by sketching its graph.

Graph of an inverted parabola with vertex at (3, 4).

Solution:

  1. Identify the vertex: The function is in vertex form a(xβˆ’h)2+ka(x-h)^2 + k, so the vertex is (3,4)(3, 4).
  2. Determine the direction: Since a=βˆ’1a = -1, the parabola opens downwards.
  3. Identify the maximum value: The highest point on the graph is y=4y = 4.
  4. State the range: The range is all real numbers less than or equal to 4, or {y∈R∣y≀4}\{y \in \mathbb{R} \mid y \leq 4\}.

Explanation:

The range of a downward-opening parabola is limited by its vertex's y-coordinate. All values below this peak are reachable.

Problem 5:

Find the domain of the rational function f(x)=1x+2f(x) = \frac{1}{x + 2} and identify its vertical asymptote.

Graph of 1 over x+2 showing a vertical asymptote at x = -2.

Solution:

  1. Set the denominator to zero: x+2=0x + 2 = 0.
  2. Solve for xx: x=βˆ’2x = -2.
  3. State the domain: The function is undefined at x=βˆ’2x = -2, so the domain is {x∈R∣xβ‰ βˆ’2}\{x \in \mathbb{R} \mid x \neq -2\}.
  4. Identify the asymptote: The vertical line x=βˆ’2x = -2 is where the function approaches infinity, representing a vertical asymptote.

Explanation:

In rational functions, the values that make the denominator zero are excluded from the domain and typically form vertical asymptotes.