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Topic 2.2Math AA SL SL30 flashcards

Functions, domain & range

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Card 1 of 302.2.1
2.2.1
Question

What does f(x) mean?

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All Flashcards in Topic 2.2

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2.2.110 cards

Card 1concept
Question

What does f(x) mean?

Answer

The output of function f for input x. f(3) means substitute x = 3 into the rule.

Card 2concept
Question

Is f(x) the same as f × x?

Answer

No — it's 'f of x', the function applied to x. The brackets hold the input.

Card 3concept
Question

What makes a rule a function?

Answer

Each input gives exactly ONE output. (Different inputs may share an output.)

Card 4concept
Question

How do you evaluate f(a)?

Answer

Replace every x with a (in brackets for negatives/expressions), then simplify.

Card 5concept
Question

Evaluate g(x) = x² − 4x at x = −3.

Answer

(−3)² − 4(−3) = 9 + 12 = 21.

Card 6concept
Question

How do you solve f(x) = k?

Answer

Set the rule equal to k and solve for x (output → input).

Card 7concept
Question

Can two inputs give the same output?

Answer

Yes — e.g. f(x) = x² gives f(2) = f(−2) = 4. So f(x) = k may have several solutions.

Card 8concept
Question

How do you read f(a) off a graph?

Answer

Go up from x = a to the curve, then across to the y-axis.

Card 9concept
Question

How do you solve f(x) = k off a graph?

Answer

Read across from y = k to the curve, then down to the x-axis (there may be several x).

Card 10concept
Question

Find f(2a) for f(x) = 3x − 5.

Answer

Substitute the whole expression: 3(2a) − 5 = 6a − 5.

2.2.210 cards

Card 11concept
Question

What is the domain of a function?

Answer

The set of all allowed inputs (x-values).

Card 12concept
Question

What is the range of a function?

Answer

The set of all possible outputs (y-values).

Card 13concept
Question

How do you read the domain off a graph?

Answer

How far the graph extends left ↔ right (the x-extent).

Card 14concept
Question

How do you read the range off a graph?

Answer

How far the graph extends down ↕ up (the y-extent).

Card 15concept
Question

What two things restrict a domain?

Answer

No dividing by zero (denominator ≠ 0) and no even root of a negative (under √ ≥ 0). Also a log argument must be > 0.

Card 16concept
Question

Domain of 1/(x − 3)?

Answer

x ≠ 3 — the denominator can't be zero.

Card 17concept
Question

Domain of √(x − 2)?

Answer

x ≥ 2 — what's under the root must be ≥ 0 (0 is allowed).

Card 18concept
Question

Range of f(x) = (x − h)² + k opening upward?

Answer

y ≥ k — the vertex (h, k) is the minimum.

Card 19concept
Question

Range of an exponential aˣ (a > 0)?

Answer

y > 0 — always positive, approaching but never reaching 0.

Card 20concept
Question

What's the default domain if nothing restricts it?

Answer

All real numbers, x ∈ ℝ.

2.2.310 cards

Card 21concept
Question

What does an inverse function do?

Answer

It undoes f: if f(a) = b then f⁻¹(b) = a. Inputs and outputs swap.

Card 22concept
Question

Is f⁻¹(x) the same as 1/f(x)?

Answer

No — f⁻¹ is the inverse function (reverses f), not the reciprocal.

Card 23concept
Question

The graph of f⁻¹ is f reflected in which line?

Answer

y = x. Each point (a, b) on f becomes (b, a) on f⁻¹.

Card 24concept
Question

How do you find f⁻¹ algebraically?

Answer

Write y = f(x), swap x and y, then solve for y — that's f⁻¹(x).

Card 25concept
Question

Find the inverse of f(x) = 2x + 3.

Answer

Swap: x = 2y + 3 ⇒ y = (x − 3)/2, so f⁻¹(x) = (x − 3)/2.

Card 26concept
Question

How do domain and range change for f⁻¹?

Answer

They swap: domain of f⁻¹ = range of f; range of f⁻¹ = domain of f.

Card 27concept
Question

Where do f and f⁻¹ intersect?

Answer

On the line y = x — solve f(x) = x to find where.

Card 28concept
Question

How can you check an inverse?

Answer

Pick a point: f(a) = b should give f⁻¹(b) = a. Or check f(f⁻¹(x)) = x.

Card 29concept
Question

Why might f⁻¹ need a restricted domain?

Answer

Its domain is f's range, which can be limited (e.g. √x has range y ≥ 0, so its inverse x² is restricted to x ≥ 0).

Card 30concept
Question

What happens to a point already on y = x under reflection?

Answer

It maps to itself — which is why f and f⁻¹ meet on y = x.

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