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Functions - Function Concepts (Domain, Range, Inverse, Composite)

Grade 11IB

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

🔑Concepts

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Definition of a Function: A function is a specific type of relation where every input xx from the domain corresponds to exactly one output yy in the range. Visually, this is verified using the Vertical Line Test: if any vertical line intersects a graph more than once, the graph does not represent a function.

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Domain and Range: The domain is the set of all possible input values (independent variables, usually xx) for which the function is defined. The range is the set of all possible output values (dependent variables, usually yy). On a graph, the domain represents the horizontal span of the curve, while the range represents the vertical span.

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Composite Functions: A composite function (f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x)) is formed by substituting one function into another. The output of the inner function g(x)g(x) becomes the input for the outer function f(x)f(x). Visually, this can be seen as a multi-step transformation where the domain of the composite function is restricted to values of xx in the domain of gg such that g(x)g(x) is in the domain of ff.

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Inverse Functions: The inverse function f−1(x)f^{-1}(x) reverses the mapping of the original function f(x)f(x). A function has an inverse if and only if it is a one-to-one (bijective) mapping. Visually, the graph of f−1(x)f^{-1}(x) is a reflection of the graph of f(x)f(x) across the line y=xy = x.

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The Horizontal Line Test: This test is used to determine if a function is one-to-one and thus has an inverse. If any horizontal line intersects the graph of a function more than once, the function is many-to-one and does not have an inverse unless its domain is restricted.

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Domain-Range Relationship in Inverses: There is an inherent swap between a function and its inverse. The domain of f(x)f(x) becomes the range of f−1(x)f^{-1}(x), and the range of f(x)f(x) becomes the domain of f−1(x)f^{-1}(x). For example, if f(x)f(x) has a vertical asymptote at x=ax = a, then f−1(x)f^{-1}(x) will have a horizontal asymptote at y=ay = a.

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Self-Inverse Functions: A function is called a self-inverse if f(x)=f−1(x)f(x) = f^{-1}(x). Applying the function twice returns the original input, such that f(f(x))=xf(f(x)) = x. Geometrically, these functions are symmetric about the line y=xy = x (e.g., f(x)=1xf(x) = \frac{1}{x}).

📐Formulae

Function Notation: f:x↦yf: x \mapsto y or y=f(x)y = f(x)

Composite Function: (f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x))

Inverse Identity: f(f−1(x))=xf(f^{-1}(x)) = x and f−1(f(x))=xf^{-1}(f(x)) = x

Domain/Range Swap: Df=Rf−1D_{f} = R_{f^{-1}} and Rf=Df−1R_{f} = D_{f^{-1}}

Vertical Asymptote of Rational Function f(x)=ax+bcx+df(x) = \frac{ax+b}{cx+d}: x=−dcx = -\frac{d}{c}

Horizontal Asymptote of Rational Function f(x)=ax+bcx+df(x) = \frac{ax+b}{cx+d}: y=acy = \frac{a}{c}

💡Examples

Problem 1:

Given f(x)=2x+5x−3f(x) = \frac{2x + 5}{x - 3} for x≠3x \neq 3, find the inverse function f−1(x)f^{-1}(x) and state its domain.

Solution:

Step 1: Replace f(x)f(x) with yy: y=2x+5x−3y = \frac{2x + 5}{x - 3} Step 2: Swap xx and yy to begin finding the inverse: x=2y+5y−3x = \frac{2y + 5}{y - 3} Step 3: Solve for yy. Multiply both sides by (y−3)(y - 3): x(y−3)=2y+5x(y - 3) = 2y + 5 xy−3x=2y+5xy - 3x = 2y + 5 Step 4: Move all yy terms to one side: xy−2y=3x+5xy - 2y = 3x + 5 Step 5: Factor out yy: y(x−2)=3x+5y(x - 2) = 3x + 5 Step 6: Isolate yy: y=3x+5x−2y = \frac{3x + 5}{x - 2} Thus, f−1(x)=3x+5x−2f^{-1}(x) = \frac{3x + 5}{x - 2}. The domain of f−1(x)f^{-1}(x) is x∈R,x≠2x \in \mathbb{R}, x \neq 2.

Explanation:

To find an inverse, we swap the roles of xx and yy and solve for the new yy. The domain restriction of the inverse is determined by the value that makes the denominator zero (x=2x = 2), which also happens to be the horizontal asymptote of the original function.

Problem 2:

Let f(x)=3x−1f(x) = 3x - 1 and g(x)=x2+2g(x) = x^2 + 2. Find the value of (f∘g)(−2)(f \circ g)(-2).

Solution:

Step 1: Identify that (f∘g)(−2)(f \circ g)(-2) means f(g(−2))f(g(-2)). Step 2: Evaluate the inner function g(−2)g(-2) first: g(−2)=(−2)2+2=4+2=6g(-2) = (-2)^2 + 2 = 4 + 2 = 6 Step 3: Use the result from g(−2)g(-2) as the input for f(x)f(x): f(6)=3(6)−1=18−1=17f(6) = 3(6) - 1 = 18 - 1 = 17 Final Answer: (f∘g)(−2)=17(f \circ g)(-2) = 17.

Explanation:

Composite functions are solved from the inside out. First, calculate the output of the function closest to the variable, then plug that numerical result into the outer function.