Curve Sketching — Mathematics Knowledge System

Overview

Curve sketching uses the first and second derivatives of a function to determine its overall shape — where it increases or decreases, where it's concave up or down, and where it has local extrema or inflection points — without needing to plot many individual points. It systematizes graphing into a checklist derived directly from calculus.

Intuition

The first derivative tells you the direction the curve is heading (increasing where f'>0, decreasing where f'<0), and where it momentarily flattens out (f'=0, candidate for a local max or min). The second derivative tells you how the curve bends: concave up (like a cup, f''>0) or concave down (like a cap, f''<0). Where concavity switches sign, you get an inflection point — the curve changes from bending one way to bending the other.

Interactive Graph

Formal Definition

Definition

Key sign tests used in curve sketching:

f'(x) > 0 \text{ on } (a,b) \implies f \text{ increasing on } (a,b)

f'(x) < 0 \text{ on } (a,b) \implies f \text{ decreasing on } (a,b)

f''(x) > 0 \text{ on } (a,b) \implies f \text{ concave up on } (a,b)

f''(x) < 0 \text{ on } (a,b) \implies f \text{ concave down on } (a,b)

f'(c)=0 \text{ and } f'' \text{ changes sign at } c \implies (c, f(c)) \text{ is a local extremum}

Derivation

The Second Derivative Test classifies a critical point c (where f'(c)=0) using the sign of f''(c):

f''(c) > 0 \implies c \text{ is a local minimum}

Concave up at a critical point means the curve bends upward, like the bottom of a cup

f''(c) < 0 \implies c \text{ is a local maximum}

Concave down at a critical point means the curve bends downward, like the top of a cap

f''(c) = 0 \implies \text{test is inconclusive; use the First Derivative Test instead}

Fall back on checking the sign of f' on either side of c

Applications

Analyzing a profit function's concavity identifies whether marginal returns are increasing or diminishing, informing production decisions.

Worked Examples

Find critical points: f'(x) = 3x² - 6x = 3x(x-2) = 0 ⟹ x=0, x=2.
$f'(x) = 3x^2-6x$
Test intervals: f'>0 on (-∞,0) and (2,∞) [increasing]; f'<0 on (0,2) [decreasing].
$f' > 0 \text{ on } (-\infty,0)\cup(2,\infty),\ f'<0 \text{ on } (0,2)$
Second derivative: f''(x)=6x-6=0 ⟹ x=1 (inflection point candidate). f''<0 for x<1 (concave down), f''>0 for x>1 (concave up).
$f''(x) = 6x-6$
Classify: x=0 is a local max (f(0)=0), x=2 is a local min (f(2)=-4), x=1 is an inflection point (f(1)=-2).
$f(0)=0,\ f(2)=-4,\ f(1)=-2$

Answer: Local max at (0,0), local min at (2,-4), inflection point at (1,-2); increasing on (-∞,0)∪(2,∞), decreasing on (0,2).

Practice Problems

Difficulty 5/10

Find all local extrema and inflection points of f(x) = x³ - 6x² + 9x.

Difficulty 4/10

Determine the intervals where f(x) = x² - 4x + 1 is increasing/decreasing and concave up/down.

Common Mistakes

Common Mistake

Assuming every critical point (where f'=0) is a local extremum.

A critical point may be neither a max nor min — e.g. f(x)=x³ has f'(0)=0 but x=0 is an inflection point (saddle-like), not an extremum.

Common Mistake

Confusing concavity with increasing/decreasing behavior.

Concavity (from f'') describes how the SLOPE is changing, not whether the function itself is increasing or decreasing (from f'). A function can be increasing and concave down simultaneously.

Summary

f'(x) determines increasing/decreasing behavior; f''(x) determines concavity.
Critical points occur where f'(x)=0 or f' is undefined; classify them with the First or Second Derivative Test.
Inflection points occur where concavity changes sign (f'' changes sign, typically where f''=0).
A complete sketch checklist: domain, intercepts, asymptotes, f' sign chart (monotonicity/extrema), f'' sign chart (concavity/inflection).

References

WebsiteWikipedia — Curve sketching