Flashcards · IB Diploma Programme Mathematics: Analysis and Approaches · Topic 2

Functions flashcards

79 cards on Topic 2, subtopics 2.1 to 2.16. Pick a subtopic or study them all.

Study mode needs JavaScript. Every card is listed below, so you can still cover the answers and test yourself.

All 79 cards

For a quick read-through or printing. Cover the right-hand side to test yourself. Cards marked HL are HL content only.

2.1 Equations of a straight line

Gradient of the line through (x₁, y₁) and (x₂, y₂): m = (y₂ − y₁)/(x₂ − x₁) Example: (3, −1) and (7, 11) give m = 12/4 = 3.
Three forms of the equation of a straight line: Gradient-intercept: y = mx + c General: ax + by + d = 0 (gradient −a/b) Point-gradient: y − y₁ = m(x − x₁)
Gradients of parallel and perpendicular lines: Parallel: m₁ = m₂. Perpendicular: m₁ × m₂ = −1, so the gradient is the negative reciprocal (−4/5 → 5/4).
Find the line through (2, −1) perpendicular to y = 4x + 3, in the form ax + by + d = 0: m = −1/4, so y + 1 = −(1/4)(x − 2). Multiply by 4: 4y + 4 = −x + 2, giving x + 4y + 2 = 0.
State the gradient and y-intercept of 4x − 2y + 7 = 0: Rearrange: y = 2x + 7/2. Gradient 2, y-intercept (0, 7/2). (Not 4 or −2.)

2.2 Concept of a function, domain, range and inverse

Function: A rule that gives exactly one output for each input in its domain.
Domain and range of a function: Domain: the allowed inputs (by default, the largest set of real numbers for which the rule works). Range: the set of outputs produced from the domain. Write it with f(x) or y, not x.
How do you find the domain of a function given by a formula?: Look for what is not allowed: the expression under a square root must be ≥ 0, and a denominator must be ≠ 0. (For a logarithm, the argument must be > 0.)
Inverse function f⁻¹: graph, and link between domain and range: f⁻¹ reverses f. Its graph is the reflection of y = f(x) in the line y = x. Domain of f⁻¹ = range of f, and range of f⁻¹ = domain of f.
f(x) = 5x − 3. Find f⁻¹(12) without finding f⁻¹(x).: Solve f(x) = 12: 5x − 3 = 12, so x = 3. Hence f⁻¹(12) = 3.
State the range of f(x) = 7 − x², x ∈ ℝ: The maximum value is 7 (at x = 0), so the range is f(x) ≤ 7.

2.3 The graph of a function

"Draw" vs "sketch" a graph: Draw: accurate, to scale, with plotted points joined smoothly (ruler for straight lines). Sketch: the correct general shape with the relevant features labelled.
What must you label on a sketch of a graph?: The axes and every key feature: intercepts, maxima and minima, the vertex and any asymptotes (as equations). Mark end points if the domain is restricted.
How do you copy a graph from a GDC screen correctly?: Match the viewing window: note where the curve enters and leaves the window, and mark end points if the domain is restricted. Label the key features you found.
Which graphs does the course expect you to produce with technology?: Graphs of functions, including sums and differences of functions such as y = f(x) + g(x), and graphs of two functions together to find where they meet.

2.4 Key features of graphs and points of intersection

Key features of a graph you should be able to identify: Maximum and minimum values, intercepts, symmetry, vertex, zeros of functions (roots of equations), and vertical and horizontal asymptotes.
Zeros, roots and x-intercepts: how are they linked?: The zeros of f are the roots of the equation f(x) = 0, which are the x-intercepts of the graph y = f(x).
How do you find key features and intersections with a GDC?: 1. Graph in a window showing all turning points and intercepts. 2. Use the zero, maximum, minimum and intersect tools; f(0) gives the y-intercept. 3. Give values to 3 s.f. and points as coordinates.
A GDC graph of y = 3 + 2/(x − 1). How do you state its asymptotes?: As equations, not numbers: vertical asymptote x = 1 and horizontal asymptote y = 3.

2.5 Composite functions and inverse functions

Composite function (f ∘ g)(x): (f ∘ g)(x) = f(g(x)). Apply g first, then f. Usually (f ∘ g)(x) ≠ (g ∘ f)(x).
f(x) = 2x + 3 and g(x) = x − 4. Find (g ∘ f)(x) and (f ∘ g)(x).: (g ∘ f)(x) = (2x + 3) − 4 = 2x − 1 (f ∘ g)(x) = 2(x − 4) + 3 = 2x − 5
How do you find f⁻¹(x)?: 1. Check f is one-to-one on its domain. 2. Write y = f(x) and make x the subject (choose the root sign the domain of f allows). 3. Swap x and y. 4. State the domain of f⁻¹ as the range of f.
Find f⁻¹(x) for f(x) = (x − 6)/4: y = (x − 6)/4, so x = 4y + 6. Swap: f⁻¹(x) = 4x + 6.
Which functions have inverses, and what is (f ∘ f⁻¹)(x)?: Only one-to-one functions (each output from exactly one input) have inverses. (f ∘ f⁻¹)(x) = (f⁻¹ ∘ f)(x) = x, the identity function.

2.6 The quadratic function

The three forms of a quadratic and what each shows: ax² + bx + c: y-intercept (0, c); axis x = −b/(2a). a(x − p)(x − q): x-intercepts (p, 0) and (q, 0). a(x − h)² + k: vertex (h, k); axis x = h.
Write x² − 10x + 18 in the form (x − h)² + k and state the vertex: x² − 10x + 18 = (x − 5)² − 25 + 18 = (x − 5)² − 7. Vertex (5, −7).
Vertex of y = −3(x − 2)² + 7: maximum or minimum?: Vertex (2, 7). It is a maximum, because a = −3 < 0.
Exam trap: the vertex of y = a(x + h)² + k: The vertex is (−h, k), not (h, k).
x-intercepts and axis of symmetry of y = 2(x + 1)(x − 5): x-intercepts (−1, 0) and (5, 0). The axis is midway between them: x = 2.

2.7 Solving quadratic equations and inequalities; the discriminant

Quadratic formula and discriminant: For ax² + bx + c = 0: x = (−b ± √(b² − 4ac))/(2a). Discriminant Δ = b² − 4ac.
What does the discriminant tell you about the roots?: Δ > 0: two distinct real roots. Δ = 0: two equal real roots (one repeated root). Δ < 0: no real roots. Δ ≥ 0 means "real roots".
How do you solve a quadratic inequality?: 1. Rearrange so one side is 0. 2. Find the critical values (roots). 3. Sketch the parabola and read off where it is above (> 0) or below (< 0) the x-axis. 4. Use ≤ or ≥ if the roots are included.
Solve x² + 2x − 15 ≤ 0: (x + 5)(x − 3) ≤ 0. The parabola is below the axis between the roots: −5 ≤ x ≤ 3.
Discriminant problems with a parameter k: method and trap: Write a, b, c in terms of k, form Δ, then set Δ > 0, = 0 or < 0 and solve for k. Trap: if a contains k, exclude the value that makes a = 0 (the equation is then linear).

2.8 The reciprocal function and rational functions

Reciprocal function f(x) = 1/x: key features: Domain x ≠ 0, range y ≠ 0. Asymptotes x = 0 and y = 0. Branches in quadrants 1 and 3. It is self-inverse, so its graph is symmetric about y = x.
Asymptotes of y = (ax + b)/(cx + d): Vertical: x = −d/c (where the denominator is 0). Horizontal: y = a/c.
Axis intercepts of y = (ax + b)/(cx + d): y-intercept (0, b/d); x-intercept (−b/a, 0), from the numerator = 0.
State the asymptotes of f(x) = (5x − 1)/(2x + 6): Vertical x = −3; horizontal y = 5/2.
Exam trap: the horizontal asymptote of (ax + b)/(cx + d): It is y = a/c, not y = b/d (b/d is the y-intercept). Give it as an equation, and do not let a branch cross it on the sketch.

2.9 Exponential and logarithmic functions

Graph of y = aˣ (a > 0, a ≠ 1): domain, range, asymptote, intercept: Domain: all real x. Range: y > 0. Asymptote y = 0. Passes through (0, 1). Increasing if a > 1, decreasing if 0 < a < 1.
Graph of y = ln x (or logₐ x): domain, range, asymptote, intercept: Domain: x > 0. Range: all real y. Asymptote x = 0. Passes through (1, 0).
Write 3ˣ in the form e^(kx): aˣ = e^(x ln a), so 3ˣ = e^(x ln 3): k = ln 3 ≈ 1.10.
How are y = aˣ and y = logₐ x related?: They are inverse functions: logₐ aˣ = x and a^(logₐ x) = x. Their graphs are reflections of each other in y = x. Example: f(x) = 5ˣ has f⁻¹(x) = log₅ x.
When does e^(kx) model growth, and when decay?: Growth if k > 0, decay if k < 0. For example y = 2⁻ˣ = (1/2)ˣ is a decay curve: the reflection of y = 2ˣ in the y-axis.

2.10 Solving equations graphically and analytically

How do you solve a disguised quadratic in eˣ or aˣ?: 1. Spot a²ˣ = (aˣ)² and substitute u = aˣ. 2. Solve the quadratic in u. 3. Reject any u ≤ 0, saying why (aˣ > 0). 4. Take logs of the positive root(s).
Solve e²ˣ − 9eˣ + 20 = 0 exactly: Let u = eˣ: u² − 9u + 20 = 0, (u − 4)(u − 5) = 0. eˣ = 4 or 5, so x = ln 4 or x = ln 5.
How do you solve an equation with technology (Paper 2)?: 1. Graph both sides (or left side minus right side). 2. Use intersect or zero. 3. Check the whole domain, especially near asymptotes, for extra roots. 4. Write the equation you solved, then the answer to 3 s.f.
P = 500e^(0.03t). Find t when P = 800.: e^(0.03t) = 1.6, so 0.03t = ln 1.6 and t = 15.7 years (3 s.f.).
"Solve" vs "find the exact value" or "show that": On Paper 2, "solve" allows a GDC. "Find the exact value" or "show that" needs algebra, with an exact answer such as x = ln 5, not 1.61.

2.11 Transformations of graphs

Effect of y = f(x) + b and y = f(x − a): f(x) + b: vertical translation by b, (x, y) → (x, y + b). f(x − a): horizontal translation by a, (x, y) → (x + a, y). Inside the bracket moves the opposite way to the sign.
Effect of y = −f(x) and y = f(−x): −f(x): reflection in the x-axis, (x, y) → (x, −y). f(−x): reflection in the y-axis, (x, y) → (−x, y).
Effect of y = p f(x) and y = f(qx): p f(x): vertical stretch, scale factor p, (x, y) → (x, py). f(qx): horizontal stretch, scale factor 1/q, (x, y) → (x/q, y). Divide x-coordinates by q.
In what order do you apply y = p f(x − a) + b?: 1. Horizontal translation by a. 2. Vertical stretch by p (reflect too if p < 0). 3. Vertical translation by b. Move key points and every asymptote at each step.
(6, −2) lies on y = f(x). Find its image on y = f(x/2) and on y = 2f(x) − 1.: f(x/2) is a horizontal stretch, factor 2: (12, −2). 2f(x) − 1: (6, 2 × (−2) − 1) = (6, −5).

2.12 Polynomial functions, zeros, roots and factors (AHL only)

Remainder theorem and factor theoremHL: Remainder: p(x) ÷ (x − r) leaves remainder p(r); p(x) ÷ (ax − b) leaves p(b/a). Factor: (x − r) is a factor of p(x) ⇔ p(r) = 0.
Sum and product of the roots of aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₀ = 0HL: Sum = −aₙ₋₁/aₙ Product = (−1)ⁿ a₀/aₙ Cubic ax³ + bx² + cx + d = 0: sum −b/a, product −d/a.
Find the remainder when x³ − 2x + 5 is divided by (x + 2)HL: p(−2) = −8 + 4 + 5 = 1
How does a repeated factor show on the graph of a polynomial?HL: (x − r)²: the graph touches the x-axis at r. (x − r)³: it crosses the x-axis at r with a point of inflexion.
Polynomial of degree n: maximum number of real zeros and of turning pointsHL: At most n real zeros and at most n − 1 turning points.
How do you factorise a cubic without a calculator?HL: 1. Try small factors of a₀/aₙ until p(r) = 0. 2. Divide by (x − r) (long division or comparing coefficients). 3. Factorise the quadratic quotient, or show its discriminant is negative.

2.13 Rational functions of higher degree (AHL only)

Rational function: how do the degrees decide the non-vertical asymptote?HL: Numerator of lower degree than denominator: y = 0. Numerator one degree higher: an oblique asymptote y = mx + k, found by division. Never both.
Find the asymptotes of y = (x² + 2x)/(x − 1)HL: x² + 2x = (x − 1)(x + 3) + 3, so y = x + 3 + 3/(x − 1). Vertical asymptote x = 1; oblique asymptote y = x + 3.
State the asymptotes of y = (x + 5)/(x² − 9)HL: Vertical x = 3 and x = −3 (denominator zero); horizontal y = 0 (numerator of lower degree).
How do you sketch a rational function of higher degree?HL: 1. Factorise the denominator: vertical asymptotes. 2. Compare degrees for y = 0 or y = mx + k. 3. Intercepts: x = 0, and numerator = 0. 4. Check signs between critical values to place each branch.

2.14 Odd and even functions; self-inverse functions (AHL only)

Odd and even functions: definitions and graph symmetryHL: Even: f(−x) = f(x); symmetric in the y-axis (for example cos x). Odd: f(−x) = −f(x); rotational symmetry of order 2 about the origin (for example sin x, tan x).
Is f(x) = x² sin x odd, even or neither?HL: f(−x) = (−x)² sin(−x) = −x² sin x = −f(x), so f is odd.
Self-inverse function: definition and standard familiesHL: f(f(x)) = x, so f⁻¹ = f; the graph is symmetric in y = x. Families: c − x, k/x, (ax + b)/(cx − a).
Find f⁻¹(x) for f(x) = x² − 4x + 1, x ≥ 2HL: f(x) = (x − 2)² − 3, range f(x) ≥ −3. Solve: x = 2 + √(y + 3) (positive root, since x ≥ 2). f⁻¹(x) = 2 + √(x + 3), x ≥ −3.
Showing a function is odd, even or neither: what counts as proof?HL: Odd or even: algebra for a general x. Substituting x = 2 and x = −2 is not enough. Neither: one counterexample is enough.

2.15 Solutions of inequalities g(x) ≥ f(x) (AHL only)

How do you solve g(x) ≥ f(x)?HL: 1. Rearrange to h(x) = g(x) − f(x) ≥ 0. Never divide by an expression of unknown sign. 2. Factorise h (or find intersections on a GDC). 3. Use a sketch or sign table. Use ≤/≥ at zeros, strict signs at asymptotes.
Solve x³ ≥ 2x² + 5x − 6HL: x³ − 2x² − 5x + 6 = (x − 1)(x + 2)(x − 3) ≥ 0. So −2 ≤ x ≤ 1 or x ≥ 3.
Solve x² ≤ 4xHL: x² − 4x ≤ 0, so x(x − 4) ≤ 0, giving 0 ≤ x ≤ 4. (Dividing by x would lose solutions.)
Solve eˣ ≥ x + 2 with a GDCHL: The graphs meet at x = −1.84 and x = 1.15 (3 s.f.), so x ≤ −1.84 or x ≥ 1.15.

2.16 Graphs of |f(x)|, f(|x|) and related functions; modulus equations (AHL only)

Graph of y = |f(x)| vs y = f(|x|)HL: |f(x)|: parts below the x-axis are reflected up (sharp corners where the graph crossed). f(|x|): keep the part for x ≥ 0 and mirror it in the y-axis; always even.
Key features of y = 1/f(x) from y = f(x)HL: Zeros of f become vertical asymptotes. A maximum (p, q) becomes a minimum (p, 1/q), q ≠ 0. Points with y = ±1 stay fixed. Signs are unchanged. Where |f| → ∞, 1/f → 0.
How do you solve |f(x)| = g(x)?HL: Solve f(x) = g(x) and f(x) = −g(x), then reject any solution where g(x) < 0. Example: |2x − 3| = x + 1 gives x = 4 or x = 2/3, both kept.
Solve |x − 2| < |2x + 1|HL: Both sides are non-negative, so square: x² − 4x + 4 < 4x² + 4x + 1, giving (3x − 1)(x + 3) > 0. x < −3 or x > 1/3
Solve |2x + 1| > 3HL: 2x + 1 > 3 or 2x + 1 < −3, so x > 1 or x < −2.
(3, −2) lies on y = f(x). Find the corresponding point on y = f(2x − 1).HL: In general (p, q) → ((p − b)/a, q) for f(ax + b). Here 2x − 1 = 3, so x = 2: the point is (2, −2).