Functions, domains, ranges, and graphs

A function is a rule that assigns exactly one output to each input. Every input (x-value) maps to one and only one output (y-value). Example: f maps every temperature in °C to a temperature in °F — one input, one output.

First mapping (1→5, 2→7, 3→5): YES, this is a function. Two inputs (1 and 3) share the same output — that is allowed. Second mapping (1→5, 1→9): NOT a function. Input 1 maps to two different outputs — that breaks the rule.

Example: "Country → Capital city." Each country has exactly one capital — every input (country) maps to exactly one output (capital). Non-example: "Person → Friend" — a person can have many friends, so one input maps to many outputs.

Yes — this is perfectly fine and does NOT stop something from being a function. What is NOT allowed: one input mapping to two different outputs. Example: f(2) = 5 and f(3) = 5 is fine. But f(2) = 5 and f(2) = 9 means it is not a function.

f(x) is the output of the function f when the input is x. Read it as "f of x." f is the name of the function. x is the input. f(x) is the corresponding output. Example: if f(x) = 2x + 1, then f(3) = 7.

f(x) = 4x − 3. Replace y with f(x). The name "f" is conventional but any letter works (g, h, p, etc.). Both y = 4x − 3 and f(x) = 4x − 3 describe the same rule.

g(t): apply the same rule but with t as the input → g(t) = t² + 1. g(a + 1): replace every x with (a + 1) → g(a + 1) = (a + 1)² + 1. The letter inside the bracket is always the input — substitute it everywhere x appears.

f(x) is not multiplication — the parentheses here mean "function of," not "times." f(x) = 4x + 2 does not mean f × x = 4x + 2. f is the function name; f(x) is the output value when the input is x.

Substitute a for every x in the function rule, then simplify. Example: f(x) = 3x + 5. Find f(4). Replace x with 4: f(4) = 3(4) + 5 = 12 + 5 = 17.

f(3) = 2(3) − 7 = 6 − 7 = −1. f(0) = 2(0) − 7 = 0 − 7 = −7. f(0) gives the y-intercept of the function.

Replace x with −2: h(−2) = (−2)² − 4(−2) + 1 = 4 + 8 + 1 = 13. Key: (−2)² = 4 (positive). −4(−2) = +8 (negative times negative = positive).

The error is in −4². When substituting a negative number, use brackets: (−4)² = +16. Without brackets: −4² = −16 (squaring only 4, then negating — wrong). Correct: f(−4) = (−4)² + 3 = 16 + 3 = 19.

The vertical line test: draw (or imagine) any vertical line through a graph. If every vertical line crosses the graph at most once → the graph represents a function. If any vertical line crosses the graph more than once → it is NOT a function (one x has two y-values).

No — a vertical line through the centre of the circle crosses it twice (two y-values for one x). Since one input (x) gives two outputs (y), the circle fails the vertical line test and is not a function.

Yes — every vertical line crosses the V-shape exactly once. Although the V looks like two lines meeting at a point, each x-value still gives exactly one y-value. y = |x| passes the vertical line test and is a function.

No — this describes a one-to-one function (injective), not just any function. The vertical line test only checks if each x gives at most one y. It is fine for two different x-values to produce the same y (many-to-one is still a function).

The domain is the set of all valid input values (x-values) for which the function is defined. Example: f(x) = √x has domain x ≥ 0 because you cannot take the square root of a negative number.

1. Division by zero — values of x that make the denominator = 0 must be excluded. Example: f(x) = 1/(x − 3) → x ≠ 3. 2. Square root of a negative — the expression inside √ must be ≥ 0. Example: f(x) = √(x + 4) → x ≥ −4.

The expression inside √ must be ≥ 0: x − 5 ≥ 0 → x ≥ 5. Domain: x ≥ 5 (or [5, ∞) in interval notation). At x = 5: f(5) = √0 = 0 ✓. At x = 4: f(4) = √(−1) — undefined ✗.

The denominator is x² − 9 = (x − 3)(x + 3). This equals zero when x = 3 or x = −3. The domain excludes x = 3 and x = −3, not x = 9. Always set the denominator equal to 0 and solve — do not guess.

The range is the set of all possible output values (y-values) that the function can produce. Example: f(x) = x² has range y ≥ 0 because squaring any real number gives a non-negative result.

Squaring any real number always gives a non-negative result: (−3)² = 9, 0² = 0. The output can never be negative. So no matter what x you input, f(x) ≥ 0. The minimum value is 0 (at x = 0); the function grows without limit as x → ±∞.

Since x² ≥ 0, we have x² + 3 ≥ 3. Range: g(x) ≥ 3 (or [3, ∞)). The minimum value is 3, reached at x = 0: g(0) = 0 + 3 = 3.

The square root function only outputs non-negative values: √x ≥ 0 for all x ≥ 0. Correct range: f(x) ≥ 0 (or [0, ∞)). The function cannot produce negative outputs — √9 = 3, not ±3.

Look at the graph horizontally — the domain is the set of x-values covered by the graph. Find the leftmost and rightmost x-values. Filled circle (●) = endpoint included. Open circle (○) = endpoint not included.

Look at the graph vertically — the range is the set of y-values covered by the graph. Find the lowest and highest y-values reached by the graph. A filled dot means that y-value is included; an open dot means it is excluded.

Domain: −2 ≤ x ≤ 6. Range: −3 ≤ y ≤ 8 (or −3 ≤ f(x) ≤ 8). IB also accepts interval notation: domain [−2, 6], range [−3, 8].

Domain → x-axis (horizontal). Range → y-axis (vertical). Memory trick: "D for domain, D for direction left-right (x-axis)." Domain = span of x-values; range = span of y-values.

A restricted domain limits the valid inputs to a practical range — not all mathematical values make sense. Examples: • Time t: must be t ≥ 0 (time cannot be negative). • Number of items n: must be a positive integer (you cannot buy half an item). • Distance d: must be d ≥ 0.

Domain: 0 ≤ t ≤ 15. t ≥ 0: time cannot be negative. t ≤ 15: V(15) = 1200 − 80(15) = 0 — the pool is empty; the model stops being valid.

No — x = 11 is outside the domain [2, 10]. The function is not defined for x = 11; the output is meaningless in this context. Always check inputs are within the stated domain before calculating.

n must be a non-negative integer (you cannot sell −3.7 units). A more appropriate domain is n ∈ {0, 1, 2, 3, ...} or n ≥ 0 with n ∈ ℤ. IB context questions often award a mark for recognising this restriction.

A composite function applies one function to the output of another. f(g(x)): first apply g to x, then apply f to the result. Notation: (f ∘ g)(x) = f(g(x)) — read "f of g of x."

(f ∘ g)(x) = f(g(x)). g is applied first (the inner function), then f is applied to the result (the outer function). Think of it like nested brackets — work from the inside out.

Step 1: g(x) = 3x (the inner function). Step 2: f(g(x)) = f(3x) = (3x) + 2 = 3x + 2. Substitute g(x) = 3x wherever x appears in f.

Composition (f ∘ g) is not multiplication. f(g(x)) means "substitute g(x) into f" — apply one function to the output of the other. f(x) × g(x) means multiply the two outputs — a completely different operation.

Step 1: Calculate the inner function first — find g(a). Step 2: Substitute that result into f — find f(g(a)). Always work inside out: inner function first, outer function second.

Step 1: g(3) = 3² = 9. Step 2: f(g(3)) = f(9) = 2(9) + 1 = 19.

Step 1: f(5) = 5 − 4 = 1. Step 2: g(f(5)) = g(1) = 3(1) + 2 = 5. Note: this asks for g(f(5)), so f is applied first, then g.

They applied the functions in the wrong order. For f(g(4)): compute the inner function g(4) first, then substitute into f. The function written on the right (inside the bracket) is always applied first.

Step 1: Write out g(x). Step 2: Substitute g(x) into f — replace every x in f(x) with the expression g(x). Step 3: Simplify if possible.

g(x) = x². f(g(x)) = f(x²) = 2(x²) + 3 = 2x² + 3.

f(x) = x − 1. g(f(x)) = g(x − 1) = 3(x − 1) = 3x − 3.

They applied the wrong rule. f(x) = (x + 1)² means: take the input, add 1, then square. f(g(x)) = (g(x) + 1)² — substitute g(x) for x throughout. Always replace every x in f with the full expression g(x), including inside brackets.

No — in general f(g(x)) ≠ g(f(x)). Counterexample: f(x) = x + 1, g(x) = x². f(g(x)) = x² + 1. g(f(x)) = (x + 1)² = x² + 2x + 1. These are different.

f(g(2)): g(2) = 5, then f(5) = 25. g(f(2)): f(2) = 4, then g(4) = 7. f(g(2)) = 25 ≠ g(f(2)) = 7. The order of composition matters.

f and g are inverse functions of each other: g = f⁻¹ (and f = g⁻¹). Each function "undoes" the other. Example: f(x) = 2x + 1 and g(x) = (x − 1)/2 satisfy f(g(x)) = x and g(f(x)) = x.

Read carefully: g(f(x)) means "f is inside g" — apply f first, then g. Memory check: the function closest to x (written on the right) is always applied first. In g(f(x)): f is closer to x → f goes first → then g.

f⁻¹ undoes the effect of f — it reverses the mapping. If f maps a → b, then f⁻¹ maps b → a. Together: f⁻¹(f(x)) = x and f(f⁻¹(x)) = x.

f(f⁻¹(x)) = x (applying f after f⁻¹ gives back x). f⁻¹(f(x)) = x (applying f⁻¹ after f gives back x). Both compositions return the original input — they cancel each other out.

f⁻¹ reverses the mapping: f⁻¹(8) = 3 and f⁻¹(12) = 5. No formula needed — just swap the input and output of f.

f⁻¹(x) is the inverse function, not the reciprocal. 1/f(x) means "1 divided by the output of f" — a completely different thing. The −1 in f⁻¹ is function notation for "inverse," not an exponent.

1. Write y = f(x). 2. Swap x and y (write x = f(y)). 3. Rearrange to make y the subject. 4. Replace y with f⁻¹(x).

y = 4x − 7. Swap: x = 4y − 7. Rearrange: x + 7 = 4y → y = (x + 7)/4. f⁻¹(x) = (x + 7)/4.

y = (2x + 1)/3. Swap: x = (2y + 1)/3. Rearrange: 3x = 2y + 1 → 2y = 3x − 1 → y = (3x − 1)/2. f⁻¹(x) = (3x − 1)/2.

They will get x = (expression in y), not y = (expression in x). The swap is essential — it converts the input-output relationship. Without swapping, the result is not expressed as f⁻¹(x).

An inverse only exists if f is one-to-one (each output comes from exactly one input). Example: f(x) = x² over all ℝ is not one-to-one — f(3) = f(−3) = 9, so the inverse would give two outputs. Restricting to x ≥ 0 makes it one-to-one: f⁻¹(x) = √x.

y = x². Swap: x = y². Rearrange: y = √x (take positive root since original domain x ≥ 0). f⁻¹(x) = √x, domain x ≥ 0.

The domain of f⁻¹ equals the range of f. The range of f⁻¹ equals the domain of f. The inverse swaps domain and range — inputs become outputs and vice versa.

Without a domain restriction, f(x) = x² is not one-to-one — the inverse is not unique. The full answer must be: f⁻¹(x) = √x for x ≥ 0. IB questions typically award a separate mark for correctly stating the domain of f⁻¹.

The graph of f⁻¹ is the reflection of the graph of f in the line y = x. Every point (a, b) on f maps to the point (b, a) on f⁻¹ — x and y coordinates are swapped.

(7, 2) and (4, −1). The inverse swaps x and y — every (a, b) on f becomes (b, a) on f⁻¹.

At any intersection point, f(x) = f⁻¹(x). These points also lie on the line y = x (since they satisfy f(x) = x at the intersection in the most common case). Note: f and f⁻¹ can intersect off the line y = x too, but they always cross y = x when they intersect.

The correct reflection is over the line y = x, not the x-axis. Reflecting over the x-axis would flip the graph vertically — that gives −f(x), not f⁻¹(x). The line y = x is the mirror that swaps x and y coordinates.