Learn first: Linear equations
The function is one of the central ideas of mathematics: equations, graphs, derivatives and integrals all rest on it. Exams almost always include questions on the domain, the range, composite and inverse functions, and graph transformations. Below we go through each of them and explain why the rules work.
A function is a rule that assigns to every number x in a set X exactly one number y in a set Y. We write y = f(x); x is the independent variable (argument) and y is the dependent variable (value of the function).
The key word is exactly one. If some x corresponds to two different y, it is not a function.
Vertical line test: if every vertical line meets a curve at most once, the curve is the graph of a function. That is why a circle is not the graph of a function.
A function can be given in three ways:
f(a) means "the value at x = a": if f(x) = x² − 3x, then f(−2) = 4 + 6 = 10.
The domain D(f) is the set of all x for which the formula makes sense. To find it, remember the "forbidden" operations:
Denominator ≠ 0; under an even root ≥ 0; inside a logarithm > 0; log base > 0 and ≠ 1
For tg x we need cos x ≠ 0; for arcsin x, arccos x we need −1 ≤ x ≤ 1
If a formula contains several conditions, all of them must hold at once — we set up a system and take the intersection of the solutions.
A polynomial (such as x³ − 2x + 7) and an odd root (∛x) are defined for every real number.
How many integers lie in the domain of f(x) = √(x + 3)/(x − 2) + lg(5 − x)?
The range E(f) is the set of all values y the function takes. Main methods:
Find the range of y = 3 − 2sin x.
Find the range of y = (2x + 1)/(x − 1).
In the composite function f(g(x)) we first compute g(x) and then feed the result into f. Order matters: usually f(g(x)) ≠ g(f(x)).
If f(x) = x² − 1 and g(x) = 3 − 2x, then:
f(g(x)) = (3 − 2x)² − 1 = 4x² − 12x + 8, g(f(x)) = 3 − 2(x² − 1) = 5 − 2x²
Recovering `f`. Given f(x + 1) = x² + 3x, put t = x + 1, so x = t − 1:
f(t) = (t − 1)² + 3(t − 1) = t² + t − 2, hence f(x) = x² + x − 2.
The inverse function f⁻¹ "undoes" f: if f(a) = b, then f⁻¹(b) = a. It exists only when different x give different y (for example, for a monotonic function).
Algorithm:
D(f⁻¹) = E(f), E(f⁻¹) = D(f); the graphs are symmetric about the line y = x
For example, y = 2x + 3 → x = (y − 3)/2 → f⁻¹(x) = (x − 3)/2.
To build a new graph from a known one, use these rules:
f(x) + a — up by a; f(x) − a — down
f(x − a) — right by a; f(x + a) — left
−f(x) — reflection in Ox; f(−x) — reflection in Oy
k·f(x) — vertical stretch by factor k; f(kx) — horizontal compression by factor k (k > 1)
|f(x)| — the part below Ox is reflected upwards; f(|x|) — the right half is mirrored to the left in Oy
Why does f(x − 2) move to the right? The value the old graph had at x = 0 is now obtained when x − 2 = 0, i.e. at x = 2 — everything shifts 2 units right.
How is y = (x − 2)² + 1 obtained from y = x², and where is its vertex?