Properties of Functions

Understanding properties lets you sketch graphs, solve inequalities, and invert functions with confidence. Monotonicity comes first: on an interval II, ff is strictly increasing when x1<x2⇒f(x1)<f(x2)x_1 < x_2 \Rightarrow f(x_1) < f(x_2) and strictly decreasing when the inequality reverses, which you can often read straight off the curve or, once the function is differentiable, off the sign of f′f' — positive means increasing, negative means decreasing. Boundedness answers a different question: ff is bounded on II if ∃M\exists M with ∣f(x)∣≤M|f(x)| \le M for all x∈Ix \in I, so sin⁡x\sin x is bounded by 11 on R\mathbb{R} while x2x^2 is bounded by 11 on [−1,1][-1,1]. Extrema mark the turning points — a local maximum at x0x_0 is a point with a neighborhood where f(x)≤f(x0)f(x) \le f(x_0), a local minimum one where f(x)≥f(x0)f(x) \ge f(x_0) — and the usual derivative test finds them: f′(x0)=0f'(x_0) = 0 with the sign of f′f' switching from positive to negative gives a maximum, the reverse switch gives a minimum. Parity is a symmetry statement, and it only makes sense on a domain symmetric about the origin: ff is even when f(−x)=f(x)f(-x) = f(x), making the graph symmetric about the yy-axis as x2x^2 and cos⁡x\cos x are, and odd when f(−x)=−f(x)f(-x) = -f(x), making it symmetric about the origin as x3x^3 and sin⁡x\sin x are. Periodicity repeats the picture: ff is periodic if ∃T≠0\exists T \neq 0 with f(x+T)=f(x)f(x+T) = f(x) for all xx, and the smallest positive such TT is the fundamental period — 2π2\pi for sin⁡x\sin x, π\pi for tan⁡x\tan x. Finally, convexity tells you which way the curve bends: on II, ff is convex (concave up) when the chord between any two points on the graph lies above the curve, which analytically means f′′(x)≥0f''(x) \ge 0, and concave (concave down) when f′′(x)≤0f''(x) \le 0; the points where the curvature changes sign are the inflection points.


Exercises

Exercise 1

Determine where f(x)=x3−3x+1f(x) = x^3 - 3x + 1 is increasing or decreasing.

monotonicityderivativecubic function

Approach:

The monotonicity of a differentiable function is decided by the sign of its first derivative, so differentiate and split the real line at the zeros of the derivative.

Steps:

  1. Differentiate: f′(x)=3x2−3=3(x−1)(x+1)f'(x) = 3x^2 - 3 = 3(x-1)(x+1).
  2. The derivative vanishes at x=−1x = -1 and x=1x = 1, which split the real line into three intervals.
  3. On each interval the sign of 3(x−1)(x+1)3(x-1)(x+1) is constant, so the function is monotone there.

Answer:

f′(x)=3x2−3=3(x−1)(x+1)f'(x) = 3x^2 - 3 = 3(x-1)(x+1)

  • x<−1x < -1: f′(x)>0f'(x) > 0 ⇒ increasing
  • −1<x<1-1 < x < 1: f′(x)<0f'(x) < 0 ⇒ decreasing
  • x>1x > 1: f′(x)>0f'(x) > 0 ⇒ increasing

Exercise 2

Is f(x)=ln⁡ ⁣(1+x1−x)f(x) = \ln\!\left(\dfrac{1+x}{1-x}\right) odd or even on (−1,1)(-1,1)?

parityodd functionlogarithm function

Approach:

To classify the parity of a function, first check that its domain is symmetric about the origin, then compare f(−x)f(-x) with f(x)f(x) and with −f(x)-f(x).

Steps:

  1. The domain (−1,1)(-1, 1) is symmetric about the origin.
  2. Compute f(−x)=ln⁡ ⁣(1−x1+x)=−ln⁡ ⁣(1+x1−x)=−f(x)f(-x) = \ln\!\left(\dfrac{1 - x}{1 + x}\right) = -\ln\!\left(\dfrac{1 + x}{1 - x}\right) = -f(x).
  3. Since f(−x)=−f(x)f(-x) = -f(x), the function is odd.

Answer:

Domain is symmetric. f(−x)=ln⁡ ⁣(1−x1+x)=−ln⁡ ⁣(1+x1−x)=−f(x)f(-x) = \ln\!\left(\dfrac{1 - x}{1 + x}\right) = -\ln\!\left(\dfrac{1 + x}{1 - x}\right) = -f(x), so ff is odd.


Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
f(x)f(x)Mathematical symbolf of xThe function under study
IIMathematical symbolintervalThe interval on which the property is discussed
f′(x)f'(x)Mathematical symbolf prime of xThe first derivative of the function
f′′(x)f''(x)Mathematical symbolf double prime of xThe second derivative of the function
TTMathematical symbolTThe period
R\mathbb{R}Mathematical symbolReal numbersThe set of all real numbers

Chapters

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