Functions whose values stay within fixed bounds
A function f: D → ℝ is bounded on D if there exists M > 0 such that:
∃M: f(x) ≤ M for all x ∈ D
∃m: f(x) ≥ m for all x ∈ D
| Function | Domain | Bounded? | sup | inf |
|---|---|---|---|---|
Yes | ||||
No | ||||
Yes | ||||
Yes |
Even, odd, and asymmetric functions
f(-x) = f(x) for all x in domain
Symmetric about the y-axis
Examples:
x², cos(x), |x|, x⁴ - 2x² + 1
Replace x with -x; if f(-x) = f(x), it's even
f(-x) = -f(x) for all x in domain
Symmetric about the origin (180° rotational)
Examples:
x³, sin(x), x, tan(x)
Replace x with -x; if f(-x) = -f(x), it's odd
f(-x) ≠ f(x) and f(-x) ≠ -f(x)
No symmetry about y-axis or origin
Examples:
eˣ, x² + x, ln(x), 2ˣ + x
If neither condition holds, function has no symmetry
Functions that preserve or reverse order
Non-strict:
Strict:
Example:
Allows flat sections (plateaus)
Non-strict:
Strict:
Example:
Allows flat sections (plateaus)
If f: D → ℝ is strictly monotonic, then:
Strict monotonicity implies injectivity (one-to-one), which is exactly what we need for an inverse to exist. The graph of f⁻¹ is the reflection of f's graph across y = x.
Functions that repeat their values at regular intervals
A function f: ℝ → ℝ is periodic with period T > 0 if:
The smallest positive T with this property is called the fundamental period.
Period: 2π
Period: π
Period: 2π/|k|
If f has period T₁ and g has period T₂, then f + g has period LCM(T₁, T₂) (if it exists as a rational multiple).
Similarly, f · g has period LCM(T₁, T₂) when the LCM exists.
Properties:
Properties:
Properties:
Properties:
Problem:
Prove that f(x) = x³ is strictly increasing on ℝ
Solution:
Let x₁ < x₂. We need to show f(x₁) < f(x₂).
f(x₂) - f(x₁) = x₂³ - x₁³
= (x₂ - x₁)(x₂² + x₁x₂ + x₁²)
First factor: x₂ - x₁ > 0 since x₁ < x₂
Second factor: x₂² + x₁x₂ + x₁² = (x₂ + x₁/2)² + 3x₁²/4 > 0
Product of positive numbers is positive
So f(x₂) - f(x₁) > 0, meaning f(x₁) < f(x₂) ∎
Problem:
Find the period of f(x) = sin(2x) + cos(3x)
Solution:
Period of sin(2x) is 2π/2 = π
Period of cos(3x) is 2π/3
Period of f is LCM of π and 2π/3
Express as fractions: π = 3π/3, 2π/3
LCM of numerators: LCM(3π, 2π) = 6π
GCD of denominators: GCD(3, 3) = 3
Period = 6π/3 = 2π ∎
Problem:
Prove f(x) = (1/x)sin(1/x) is unbounded on (0,1]
Solution:
For any M > 0, we need x ∈ (0,1] with |f(x)| > M
Choose n such that 2nπ + π/2 > M (so n > (M-π/2)/(2π))
Let x = 1/(2nπ + π/2)
Then 1/x = 2nπ + π/2 and sin(1/x) = sin(2nπ + π/2) = 1
So f(x) = (2nπ + π/2) · 1 = 2nπ + π/2 > M
Since x = 1/(2nπ + π/2) < 1, we have x ∈ (0,1]
Therefore f is unbounded above. Similarly for below ∎
Problem:
Show that f(x) = x + eˣ has an inverse, and describe it
Solution:
f'(x) = 1 + eˣ > 0 for all x (since eˣ > 0)
So f is strictly increasing on ℝ
By the Inverse Function Theorem, f⁻¹ exists
Range of f: As x → -∞, f(x) → -∞; as x → +∞, f(x) → +∞
So f: ℝ → ℝ is a bijection
f⁻¹ is also strictly increasing (same direction)
Note: f⁻¹ cannot be expressed in elementary form ∎
Problem:
Classify f(x) = x³ + sin(x) as even, odd, or neither
Solution:
Compute f(-x) = (-x)³ + sin(-x)
= -x³ + (-sin(x))
= -x³ - sin(x)
= -(x³ + sin(x))
= -f(x)
Since f(-x) = -f(x), f is an odd function
The graph is symmetric about the origin ∎
Problem:
Find sup and inf of f(x) = sin²(x) + cos⁴(x) on ℝ
Solution:
Let u = cos²(x), so sin²(x) = 1 - u and 0 ≤ u ≤ 1
f = (1-u) + u² = u² - u + 1
Complete the square: (u - 1/2)² + 3/4
This is minimized when u = 1/2, giving f = 3/4
Maximized when u = 0 or u = 1, giving f = 1
Therefore inf f = 3/4 (achieved at cos²x = 1/2) and sup f = 1 ∎
Problem:
Prove that f(x) = x + sin(x) is not periodic
Solution:
Suppose f has period T > 0, so f(x+T) = f(x) for all x
Then x + T + sin(x+T) = x + sin(x)
So T + sin(x+T) - sin(x) = 0
Using sum-to-product: T + 2cos((2x+T)/2)sin(T/2) = 0
At x = 0: T + 2cos(T/2)sin(T/2) = T + sin(T) = 0
But T > 0 and |sin(T)| ≤ 1, so T + sin(T) ≥ T - 1
For T ≥ 1, this is positive. For T < 1, we need T = -sin(T), impossible
Therefore f is not periodic ∎
Problem:
Find the value of ⌊3.7⌋ + ⌈-2.3⌉ + {5.8}
Solution:
⌊3.7⌋ = greatest integer ≤ 3.7 = 3
⌈-2.3⌉ = smallest integer ≥ -2.3 = -2
{5.8} = fractional part = 5.8 - ⌊5.8⌋ = 5.8 - 5 = 0.8
Sum = 3 + (-2) + 0.8 = 1.8
Therefore the answer is 1.8 ∎
If f is strictly monotonic on interval I, then f⁻¹ exists on f(I) and has the same type of monotonicity
Guarantees that strictly monotonic functions have well-defined inverses
If f and g are strictly increasing, then f∘g is strictly increasing. If f is increasing and g is decreasing, then f∘g is decreasing.
Allows us to determine monotonicity of composite functions
If f has period T and g has period S, then f(g(x)) has period LCM(T,S) if it exists, provided the composition is well-defined
Useful for analyzing trigonometric expressions
If f is continuous on closed interval [a,b], then f is bounded and attains its bounds
This is the Extreme Value Theorem - continuous functions on closed intervals achieve max/min
Any function f can be written as f(x) = fₑ(x) + fₒ(x) where fₑ is even and fₒ is odd
fₑ(x) = (f(x)+f(-x))/2 and fₒ(x) = (f(x)-f(-x))/2
If f is monotonic and surjective onto an interval, it achieves every intermediate value
Monotonic functions on intervals have no 'jumps' that skip values
If f has period p and g has period q, then f+g has period lcm(p,q) if p/q is rational
Non-commensurate periods can produce non-periodic sums
To prove f is increasing, show that x₁ < x₂ implies f(x₁) < f(x₂) (or ≤ for non-strict)
Example: For f(x) = x², on [0,∞): if 0 ≤ x₁ < x₂, then x₁² < x₂² since both are positive
If f'(x) > 0 on interval I, then f is strictly increasing on I. If f'(x) < 0, strictly decreasing.
Example: f(x) = eˣ: f'(x) = eˣ > 0 for all x, so f is strictly increasing on ℝ
To find the period T, solve f(x+T) = f(x) for the smallest positive T
Example: For sin(ax): sin(a(x+T)) = sin(ax) ⟹ aT = 2π ⟹ T = 2π/a
To show f is unbounded above, for each M > 0, find x with f(x) > M
Example: f(x) = x² is unbounded: for any M, choose x > √M, then x² > M
Introduced the modern function notation f(x) and studied many special functions
His work on exponential, logarithmic, and trigonometric functions formed the foundation of modern analysis
Gave the modern definition of a function as a correspondence between sets
The Dirichlet function (indicator of rationals) challenged intuition about continuity and integrability
Developed Fourier series to represent periodic functions
Showed that even discontinuous periodic functions can be represented as infinite sums of sines and cosines
Periodic functions model sound waves, radio signals, and electrical currents
Example: A musical note A4 (440 Hz) is modeled as sin(880πt), with period 1/440 seconds
Monotonic functions model supply/demand curves and utility functions
Example: A strictly increasing utility function guarantees consistent consumer preferences
Floor and ceiling functions are essential in algorithm analysis
Example: Binary search takes ⌈log₂(n)⌉ comparisons in the worst case
Bounded functions describe physical quantities with natural limits
Example: Temperature in Kelvin is bounded below by 0 (absolute zero)
A function f is bounded on domain D if there exists M > 0 such that |f(x)| ≤ M for all x ∈ D. This is equivalent to having both finite supremum AND finite infimum. Having only a finite supremum means bounded above but possibly unbounded below.
No! If f is periodic with period T, then f(x) = f(x+T) for all x. But if f is strictly increasing, we'd need f(x) < f(x+T) for all x. These contradict each other. A periodic function must 'come back' to previous values, preventing strict monotonicity.
Strict monotonicity implies injectivity (one-to-one): if x₁ ≠ x₂, then f(x₁) ≠ f(x₂). This is because either x₁ < x₂ (so f(x₁) < f(x₂)) or x₁ > x₂ (so f(x₁) > f(x₂)). An injective function has an inverse on its range.
Yes, with infinitely many periods! For any rational r, D(x+r) = D(x) because x is rational iff x+r is rational. So every positive rational is a period. Interestingly, D has no smallest positive period.
Check if the supremum is attained. If ∃x₀ in the domain with f(x₀) = sup f, then max = sup. For example, f(x) = 1/(1+x²) on ℝ has sup = max = f(0) = 1. But f(x) = x on (0,1) has sup = 1 but no maximum since 1 is never reached.
Even functions f(-x) = f(x) have graphs symmetric about the y-axis. Odd functions f(-x) = -f(x) have graphs symmetric about the origin. Note: f(x) = 0 is the only function that is both even and odd.
If f and g are both increasing (or both decreasing), then f∘g is increasing. If one is increasing and one is decreasing, then f∘g is decreasing. This follows from the chain rule for monotonicity.
At x = n (integer), lim(x→n⁻)⌊x⌋ = n-1 but ⌊n⌋ = n. This jump of 1 at each integer creates the discontinuity. Between integers, the floor function is constant and continuous.
Test your understanding of function properties
1
Set Theory
2
Real Number Construction
3
Properties & Inequalities
4
Functions
You're now ready to move on to Chapter 2: Sequence Limits!