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

Grade 12IB

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

🔑Concepts

Definition of a Function: A relation where every input xx in the domain maps to exactly one output yy in the range. Visually, a function must pass the Vertical Line Test, meaning any vertical line drawn through the graph intersects the curve at most once.

Domain and Range: The domain is the set of all possible input values (xx-axis coverage), and the range is the set of all possible output values (yy-axis coverage). On a graph, you can identify the domain by looking at the left-to-right span and the range by looking at the bottom-to-top span of the curve.

Composite Functions: Formed by substituting one function into another, denoted as (fg)(x)=f(g(x))(f \circ g)(x) = f(g(x)). This represents a sequence of transformations where the output of the inner function gg becomes the input for the outer function ff. For the composite to exist, the range of gg must be a subset of the domain of ff.

Inverse Functions: An inverse function f1(x)f^{-1}(x) reverses the operation of f(x)f(x). A function has an inverse only if it is one-to-one (bijective), which is verified visually using the Horizontal Line Test: a horizontal line should only touch the graph once.

Geometric Properties of Inverses: The graph of an inverse function f1(x)f^{-1}(x) is a reflection of the original function f(x)f(x) across the line y=xy = x. If a point (a,b)(a, b) lies on the graph of ff, then the point (b,a)(b, a) must lie on the graph of f1f^{-1}.

Domain Restrictions: Functions that are not naturally one-to-one, such as f(x)=x2f(x) = x^{2}, require a domain restriction (e.g., x0x \geq 0) to have an inverse. Visually, this involves selecting only one 'branch' of a curve (like the right side of a parabola) so it passes the Horizontal Line Test.

Self-Inverse Functions: Some functions are their own inverse, meaning f(x)=f1(x)f(x) = f^{-1}(x). A common visual example is the reciprocal function f(x)=1xf(x) = \frac{1}{x}, which is symmetrical about the line y=xy = x and y=xy = -x.

📐Formulae

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

f(f1(x))=x and f1(f(x))=xf(f^{-1}(x)) = x \text{ and } f^{-1}(f(x)) = x

Domain of f=Range of f1\text{Domain of } f = \text{Range of } f^{-1}

Range of f=Domain of f1\text{Range of } f = \text{Domain of } f^{-1}

For f(x)=ax+bcx+d, the inverse is f1(x)=dx+bcxa\text{For } f(x) = \frac{ax + b}{cx + d}, \text{ the inverse is } f^{-1}(x) = \frac{-dx + b}{cx - a}

Vertical Asymptote: x=dc\text{Vertical Asymptote: } x = -\frac{d}{c}

Horizontal Asymptote: y=ac\text{Horizontal Asymptote: } y = \frac{a}{c}

💡Examples

Problem 1:

Given f(x)=3x1x+2f(x) = \frac{3x - 1}{x + 2}, where x2x \neq -2, find the expression for f1(x)f^{-1}(x) and state its domain.

Solution:

  1. Let y=3x1x+2y = \frac{3x - 1}{x + 2}.\n2. Swap xx and yy to find the inverse: x=3y1y+2x = \frac{3y - 1}{y + 2}.\n3. Multiply both sides by (y+2)(y + 2): x(y+2)=3y1x(y + 2) = 3y - 1.\n4. Expand the bracket: xy+2x=3y1xy + 2x = 3y - 1.\n5. Move terms with yy to one side: xy3y=2x1xy - 3y = -2x - 1.\n6. Factor out yy: y(x3)=(2x+1)y(x - 3) = -(2x + 1).\n7. Solve for yy: y=(2x+1)x3y = \frac{-(2x + 1)}{x - 3} or y=2x+13xy = \frac{2x + 1}{3 - x}.\n8. Therefore, f1(x)=2x+13xf^{-1}(x) = \frac{2x + 1}{3 - x}.\n9. The domain of f1(x)f^{-1}(x) is all real numbers except where the denominator is zero: xR,x3x \in \mathbb{R}, x \neq 3.

Explanation:

To find an inverse, we swap the roles of xx and yy and solve for the new yy. The domain of the inverse is restricted by the vertical asymptote of the resulting expression, which corresponds to the horizontal asymptote of the original function.

Problem 2:

Let f(x)=2x+3f(x) = 2x + 3 and g(x)=x21g(x) = x^{2} - 1. Find (gf)(x)(g \circ f)(x) and solve for xx when (gf)(x)=0(g \circ f)(x) = 0.

Solution:

  1. Substitute f(x)f(x) into g(x)g(x): (gf)(x)=g(f(x))=(2x+3)21(g \circ f)(x) = g(f(x)) = (2x + 3)^{2} - 1.\n2. Expand the squared binomial: (gf)(x)=(4x2+12x+9)1(g \circ f)(x) = (4x^{2} + 12x + 9) - 1.\n3. Simplify: (gf)(x)=4x2+12x+8(g \circ f)(x) = 4x^{2} + 12x + 8.\n4. Set the equation to zero: 4x2+12x+8=04x^{2} + 12x + 8 = 0.\n5. Divide by 4: x2+3x+2=0x^{2} + 3x + 2 = 0.\n6. Factor the quadratic: (x+1)(x+2)=0(x + 1)(x + 2) = 0.\n7. Find the roots: x=1x = -1 and x=2x = -2.

Explanation:

Composite functions involve nesting the 'inner' function inside the 'outer' function. Once the expression is simplified, standard algebraic techniques like factoring are used to find the roots.