Mathematics · Intermediate · 30 min · Chapter 5 of 28

Functions, exponentials, and logarithms

A function is a rule with one output per input. Composition, inverses, and the exponential–logarithm pair are the machinery calculus will differentiate.

Read first: Algebra

The idea

A function f assigns to each input x in its domain exactly one output f(x). The set of outputs that actually occur is the range. Equality of functions means they agree on the whole domain, not merely that their formulas look similar before simplification. Piecewise functions are still functions: the absolute value |x| is one rule described by two formulas.

Composition is substitution. (g ∘ f)(x) = g(f(x)). It is associative and not commutative. An inverse f⁻¹ satisfies f⁻¹(f(x)) = x and f(f⁻¹(y)) = y on the appropriate sets. Graphically, the inverse is the reflection of the graph across the line y = x. You find a formula by writing y = f(x), swapping names, and solving for y — provided that solve step is single-valued.

The exponential function aˣ, with a > 0, converts addition in the exponent into multiplication: aˣ⁺ʸ = aˣ aʸ. The natural base e ≈ 2.71828 is the unique base whose tangent line at 0 has slope 1, which is why calculus prefers eˣ. The logarithm log_a is the inverse: log_a(aˣ) = x. Change of base says log_a x = ln x / ln a, so every logarithm is a scaled natural logarithm. Continuous compounding A = P e^{rt} is the limit of compounding n times a year as n grows.

Hold onto this

The graph of a function meets each vertical line at most once.
Composition g∘f means do f first, then g. Order matters.
An inverse undoes a function, and it exists on a domain where the function is one-to-one.
Exponential growth is multiplication by a constant per unit time. Logarithms measure the time, or the exponent.

Formulas

Composition

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

Apply f, then feed the result to g. The domain is those x where f(x) lands in the domain of g.

f, g: Functions

Inverse

f^{- 1} (f (x)) = x

The inverse undoes f. It exists when f is one-to-one on the domain you care about.

f⁻¹: Inverse function, not 1/f

Exponent rules

a^{x + y} = a^{x} a^{y}, (a^{x})^{y} = a^{x y}

Adding exponents multiplies; multiplying exponents raises a power to a power.

a: Positive base

Change of base

lo g_{a} x = \frac{ln x}{ln a}

Compute any logarithm from the natural logarithm, or from log base 10.

ln: Logarithm base e

Continuous growth

A = P e^{r t}

Amount after time t when a principal P grows at continuous rate r. If r is negative, this is decay.

P: Initial amount
r: Continuous rate, as a decimal
t: Time in the same units as the rate

Worked examples

1. Compose, and notice the order

Let f(x) = x + 1 and g(x) = x². Compute (g ∘ f)(3) and (f ∘ g)(3).

g(f(3)) = g(4) = 16.
f(g(3)) = f(9) = 10.
The numbers differ, so composition is not commutative.

(g ∘ f)(3) = 16 and (f ∘ g)(3) = 10.

2. Doubling time

A culture grows continuously at 8% per hour. How long until it doubles?

Set P e^{0.08 t} = 2P and cancel P (assuming P ≠ 0).
e^{0.08 t} = 2.
0.08 t = ln 2 ≈ 0.693.
t ≈ 0.693 / 0.08 ≈ 8.66 hours.

About 8.66 hours. The rule of 72 estimates 72/8 = 9 hours, close enough for a check.

Try these

Find the inverse of f(x) = 3x − 6.

Solve log₂(x − 1) = 5.

Common mistakes

f⁻¹ means the inverse function. The reciprocal 1/f(x) is a different object, sometimes written [f(x)]⁻¹.
You cannot take a logarithm of zero or a negative number, in the reals.
A half-life or doubling time depends on the rate, not on the starting amount. P cancels.