Lesson 12: Curve Sketching with the First and Second Derivative TestsAP

Curve sketching (CED topics 5.3–5.9) is where Unit 5 comes together: the sign of f' controls where f rises and falls, the sign of f'' controls concavity and points of inflection, and the two derivative tests convert that information into classified extrema and an accurate graph. The AP exam's favourite version hands you information about the *derivatives* — a graph of f', a sign chart, a formula — and asks about the *function*. Keep the three graphs (f, f', f'') conceptually separate and these questions become bookkeeping.

The dictionary

• f'(x) > 0 on an interval ⇒ f is increasing there; f'(x) < 0 ⇒ decreasing.
• First Derivative Test: if f' changes + to − at a critical point c, f has a relative maximum at c; − to + gives a relative minimum; no sign change, no extremum.
• f''(x) > 0 ⇒ f is concave up (the graph bends upward, tangents lie below); f''(x) < 0 ⇒ concave down.
• A point of inflection is a point on the graph (so f must be defined there, with a tangent) where concavity changes — requires a sign change of f'', not merely f''(c) = 0.
• Second Derivative Test: if f'(c) = 0 and f''(c) > 0, relative minimum at c; if f''(c) < 0, relative maximum; if f''(c) = 0, the test is inconclusive — fall back on the First Derivative Test.

Worked Example 1: a complete analysis

Problem. Analyse f(x) = x^3 − 3x^2 − 9x + 5: critical points, extrema, concavity, inflection, and intercepts enough to sketch.

f'(x) = 3x^2 − 6x − 9 = 3(x − 3)(x + 1);  critical points x = −1, 3.
Sign of f':  + on (−∞, −1), − on (−1, 3), + on (3, ∞)
  → relative max at x = −1:  f(−1) = −1 − 3 + 9 + 5 = 10
  → relative min at x = 3:   f(3) = 27 − 27 − 27 + 5 = −22
f''(x) = 6x − 6;  zero at x = 1, sign − before, + after
  → concave down on (−∞, 1), concave up on (1, ∞), inflection point (1, f(1)) = (1, −6)
y-intercept 5; end behaviour: → −∞ as x → −∞, +∞ as x → +∞.

Solution summary. The graph climbs to (−1, 10), falls through the inflection point (1, −6) — changing from bending down to bending up exactly midway between the extrema, as always for a cubic — and bottoms out at (3, −22) before climbing away. A six-line sign analysis produces the entire picture; on the exam, those six lines are where the points are.

Worked Example 2: reasoning from a table of f' values (numerical/graphical)

Problem. f is differentiable everywhere, and its derivative is the polynomial with values tabulated below. Determine where f increases and decreases, classify the critical points, and find the inflection point of f given f''(x) = 2x + 1.

  x      −3    −2    −1     0     1     2
  f'(x)   4     0    −2    −2     0     4

Solution. f' vanishes at x = −2 and x = 1. Between tabulated points the polynomial f' keeps its sign except at those zeros: f' > 0 on (−∞, −2), f' < 0 on (−2, 1), f' > 0 on (1, ∞). Hence f increases, then decreases, then increases; by the First Derivative Test, relative maximum at x = −2 and relative minimum at x = 1. Concavity: f''(x) = 2x + 1 < 0 for x < −1/2 and > 0 for x > −1/2, with f defined and smooth at −1/2, so the inflection point is at x = −1/2 — note it sits *inside* the decreasing interval, a useful reminder that inflection and extrema are independent events. (The tabulated f' is (x + 2)(x − 1) = x^2 + x − 2 — check f'(−1) = 1 − 1 − 2 = −2 ✓.) Exam tables of f' work exactly this way: zeros and signs first, conclusions second.

Worked Example 3: the Second Derivative Test on a two-branch function

Problem. Analyse f(x) = x + 4/x for x ≠ 0.

f'(x) = 1 − 4/x^2 = 0 → x = ±2 (critical points; x = 0 is not in the domain).
f''(x) = 8/x^3.
At x = 2:  f''(2) = 1 > 0 → relative minimum, f(2) = 4.
At x = −2: f''(−2) = −1 < 0 → relative maximum, f(−2) = −4.
Concavity: f'' < 0 on (−∞, 0) (down), f'' > 0 on (0, ∞) (up).
x = 0 is NOT an inflection point — f is not defined there; the vertical asymptote separates the branches.

Solution summary. The paradox worth remembering: the relative maximum *value* (−4) is less than the relative minimum *value* (4), which is perfectly possible when the extrema sit on different branches separated by an asymptote. "Maximum" and "minimum" in "relative" describe local neighbourhoods only.

Real-world application: reading an epidemic curve

Public-health dashboards plot daily new cases f(t). The First Derivative Test is the headline question — is f' positive (outbreak growing) or negative (shrinking)? But the *turning point of concern* comes earlier: while cases still rise, the inflection point of f — where f'' changes from positive to negative — is the moment growth stops accelerating and starts decelerating, the first mathematical sign that interventions are biting. Officials who wait for the relative maximum (f' = 0) to declare progress are weeks behind the inflection signal. Concavity is not an abstraction; in an epidemic it is the earliest evidence of control.

Common mistakes

• "f''(c) = 0, therefore inflection." f(x) = x^4 has f''(0) = 0 with no sign change and no inflection — the point is a minimum. Inflection requires a concavity *change* at a point *on the graph*.
• "f'(c) = 0, therefore extremum." f(x) = x^3 at 0 is the standing counterexample — no sign change in f', no extremum.
• Mixing up the graphs: concluding f is increasing where *f'* is increasing. f' increasing means f is concave up; f increasing requires f' > 0. On graph-of-f' problems this confusion is the number-one point loser.
• Second Derivative Test overreach: when f''(c) = 0 the test says nothing at all — it does not say "no extremum". Switch tests.

AP Exam Experience

1. (No calculator.) If f'(x) = (x − 1)^2 (x + 3), then at x = 1 the function f has

• (A) a relative maximum
• (B) a relative minimum
• (C) neither a relative maximum nor a relative minimum
• (D) a point of inflection with horizontal tangent, which is also a relative minimum

Answer: (C). (x − 1)^2 is never negative, so f' does not change sign at x = 1 (it is positive on both sides, since (x + 3) > 0 near 1). No sign change ⇒ no extremum. The action is at x = −3, where f' changes − to + (a relative minimum) — the factor structure is designed to punish students who test only the squared factor's root.

2. The sign of f'' for a function f is: negative on (−∞, −1), positive on (−1, 4), negative on (4, ∞), with f defined everywhere. Which statement is true?

• (A) f is concave up on (−1, 4) and has inflection points at x = −1 and x = 4.
• (B) f has relative extrema at x = −1 and x = 4.
• (C) f is increasing on (−1, 4).
• (D) f has inflection points at x = −1 and x = 4 only if f'' = 0 there.

Answer: (A). Concavity follows the sign of f'' directly (positive ⇒ concave up), and concavity changes at both −1 and 4 with f defined there ⇒ inflection points at both. (B) and (C) confuse f'' information with f' information — the exact trap this lesson exists to disarm.

3. If f'(2) = 0 and f''(2) = −3, then at x = 2, f has

• (A) a relative maximum
• (B) a relative minimum
• (C) an inflection point
• (D) a vertical tangent

Answer: (A). Second Derivative Test: critical point with f'' < 0 ⇒ relative maximum. The magnitude −3 is irrelevant; only the sign speaks.

Free-response question. Let f(x) = x^3 − 6x^2 + 9x + 2.

(a) Find all intervals on which f is increasing and all on which f is decreasing. Justify. (b) Find the x-coordinates of all relative extrema and classify each. Justify. (c) Find all intervals on which f is concave up and concave down, and the x-coordinate of each inflection point. Justify.

Scoring guideline.

• (a) 2 points. 1 point for f'(x) = 3x^2 − 12x + 9 = 3(x − 1)(x − 3) and its sign analysis (+ on (−∞, 1), − on (1, 3), + on (3, ∞)); 1 point for the correct conclusion: increasing on (−∞, 1) and (3, ∞), decreasing on (1, 3), tied explicitly to the sign of f'.
• (b) 2 points. 1 point for relative maximum at x = 1 (f' changes + to −; f(1) = 6); 1 point for relative minimum at x = 3 (f' changes − to +; f(3) = 2). Classification without a sign-change justification earns half — the First Derivative Test must be visible in the work.
• (c) 2 points. 1 point for f''(x) = 6x − 12, zero at x = 2 with sign − before and + after; 1 point for concave down on (−∞, 2), concave up on (2, ∞), inflection point at x = 2 (with f(2) = 4, the point is (2, 4)). Stating "inflection because f''(2) = 0" without the sign change forfeits the justification credit.

Extension: sketching f from the graph of f' — the full workout

Suppose the graph of f' is a downward parabola with x-intercepts at −1 and 3, vertex at (1, 4), and suppose f(0) = 2. Extract everything: f' > 0 on (−1, 3) and f' < 0 outside, so f increases on (−1, 3) and decreases on (−∞, −1) and (3, ∞), with a relative minimum at x = −1 (f' changes − to +) and a relative maximum at x = 3 (+ to −). Concavity is read from the *slope of the f' graph*, because f'' is the derivative of f': f' increases on (−∞, 1) so f is concave up there; f' decreases on (1, ∞) so f is concave down there; inflection point at x = 1. Not one value of f itself was needed beyond a starting height. This inversion — treating a derivative graph as the primary data — is the highest-frequency graph task on the modern AP exam, and the translation table is always the same: sign of the plotted function → monotonicity of f; slope of the plotted function → concavity of f; zeros of the plotted function → candidate extrema of f.