Lesson 24: Curve Sketching

(a) Opening Question

A continuous function f has derivative f'(x) = (x + 2)(x − 3). You are told nothing else about f. Answer using only this information:

1. On what intervals is f increasing? Decreasing?
2. At x = −2 and x = 3, does f have a relative maximum, a relative minimum, or neither?
3. The graph of f' is an upward parabola. Where does f' have its minimum, and what does that point tell you about the shape of the graph of f there?

Try it before reading on. The key move is realizing that f'(x) = (x + 2)(x − 3) is a parabola with roots at −2 and 3 — and the roots of f' are the critical points of f, while the vertex of f' (its minimum, at x = ½) is where f' is smallest, i.e., where f is steepest downward and f'' changes sign. That vertex is an inflection point of f. This lesson is about turning that kind of derivative information into a picture — and reading the picture backward.

(b) Core Concepts

Curve sketching is the capstone of Unit 5: you combine everything — limits, intercepts, and the two derivative tests — into one coherent graph. On the AP exam this shows up two ways: (1) "sketch a graph with these features," and (2) the far more common reverse task, "here is the graph of f' — describe f." Both reward the same fluency: knowing exactly what each derivative controls.

The Curve-Sketching Checklist

Work these in order. Skipping a step is how points (and asymptotes) get lost.

• 1. Domain. Where is f defined? Exclude zeros of denominators and arguments that break a function (e.g., negatives under even roots).
• 2. Intercepts. y-intercept: evaluate f(0). x-intercepts: solve f(x) = 0 (numerator = 0 for a rational function).
• 3. Symmetry. Even (f(−x) = f(x), symmetric across the y-axis) or odd (f(−x) = −f(x), symmetric about the origin). Optional but it halves the work.
• 4. Asymptotes. Vertical asymptotes occur at zeros of the denominator that do not cancel — values where lim f = ±∞. Horizontal asymptotes come from lim_{x→±∞} f(x); for a rational function compare degrees of numerator and denominator.
• 5. First-derivative analysis (f'). Find f', locate critical points (where f' = 0 or f' is undefined). Build a sign chart for f': f is increasing where f' > 0, decreasing where f' < 0. Classify each critical point by the First Derivative Test (sign change of f').
• 6. Second-derivative analysis (f''). Find f''. Build a sign chart for f'': f is concave up where f'' > 0, concave down where f'' < 0. An inflection point occurs where f'' changes sign (and f is defined there).
• 7. Assemble and sketch. Plot intercepts, draw asymptotes as dashed lines, mark extrema and inflection points with their y-values, then connect using the increasing/decreasing and concavity information.

The relationships between f, f', and f''

This table is the engine of the whole lesson. Read it both directions.

Two traps hide in this table. First: an x-intercept of f' is a critical point of f, not necessarily an extremum — f' must change sign. Second: an x-intercept of f'' is a candidate inflection point, but f'' must change sign there. A local extremum of f' corresponds to an inflection point of f — because where f' peaks or bottoms out, f'' (its derivative) is zero and switching sign.

Fully worked sketch: a rational function

Sketch f(x) = x² / (x² − 4).

Domain. Denominator x² − 4 = (x − 2)(x + 2) is zero at x = ±2. Domain: all x ≠ ±2.

Intercepts. f(0) = 0/(−4) = 0. The only x-intercept is where the numerator x² = 0, i.e., x = 0. So the curve passes through (0, 0) and that is both intercepts.

Symmetry. f(−x) = (−x)²/((−x)² − 4) = x²/(x² − 4) = f(x). Even — symmetric across the y-axis.

Vertical asymptotes. At x = 2 and x = −2 (denominator zero, numerator nonzero). Check a side: as x → 2⁺, numerator → 4, denominator → 0⁺, so f → +∞.

Horizontal asymptote. Degrees of numerator and denominator are equal (both 2), so lim_{x→±∞} f(x) = ratio of leading coefficients = 1. Horizontal asymptote y = 1.

First derivative. By the quotient rule:

f'(x) = [2x(x² − 4) − x²(2x)] / (x² − 4)²
      = (2x³ − 8x − 2x³) / (x² − 4)²
      = −8x / (x² − 4)²

The denominator (x² − 4)² is always positive (where defined), so the sign of f' is the sign of −8x. Critical point: f' = 0 at x = 0. So f' > 0 for x < 0 and f' < 0 for x > 0. f increasing on (−∞, −2) and (−2, 0); decreasing on (0, 2) and (2, ∞). At x = 0, f' changes + to −, so (0, 0) is a relative maximum.

Second derivative. Differentiate f'(x) = −8x(x² − 4)^(−2):

f''(x) = −8(x² − 4)^(−2) + (−8x)(−2)(x² − 4)^(−3)(2x)
       = (x² − 4)^(−3) [ −8(x² − 4) + 32x² ]
       = (24x² + 32) / (x² − 4)³

The numerator 24x² + 32 is always positive, so the sign of f'' is the sign of (x² − 4)³, which matches the sign of x² − 4. So f'' > 0 (concave up) when |x| > 2, and f'' < 0 (concave down) when |x| < 2. The sign changes occur only at x = ±2, which are not in the domain — so there are no inflection points.

Sketch.

Notice how each checklist item became a feature of the curve. The asymptotes frame the picture; the single critical point gives the peak; the concavity tells you how the branches bend toward y = 1.

(c) Worked Examples

Example 1 — Full sketch of a polynomial — [NO CALC]

Problem. Sketch f(x) = x³ − 3x² − 9x + 5. State intercepts (y only — x-intercepts are irrational), extrema, intervals of increase/decrease, concavity, and inflection point.

Strategy. Polynomials have domain all reals, no asymptotes, and no symmetry here. Go straight to intercepts, then f' and f''.

Solution.

• y-intercept: f(0) = 5, point (0, 5).
• f'(x) = 3x² − 6x − 9 = 3(x² − 2x − 3) = 3(x − 3)(x + 1). Critical points x = −1, 3.

- Sign chart: f' > 0 on (−∞, −1), f' < 0 on (−1, 3), f' > 0 on (3, ∞).

- At x = −1: f' changes + to − → relative max; f(−1) = −1 − 3 + 9 + 5 = 10, point (−1, 10).

- At x = 3: f' changes − to + → relative min; f(3) = 27 − 27 − 27 + 5 = −22, point (3, −22).

• f''(x) = 6x − 6 = 6(x − 1). Zero at x = 1; f'' < 0 for x < 1 (concave down), f'' > 0 for x > 1 (concave up). Sign change → inflection point at x = 1; f(1) = 1 − 3 − 9 + 5 = −6, point (1, −6).

Justification. f has a relative maximum at x = −1 because f' changes from positive to negative there; a relative minimum at x = 3 because f' changes from negative to positive; and an inflection point at x = 1 because f'' changes sign (from negative to positive) there.

Example 2 — Full sketch of a rational function — [NO CALC]

Problem. Sketch f(x) = (x + 1)/(x − 2).

Strategy. Find domain, intercepts, both asymptotes, then f' (concavity follows quickly).

Solution.

• Domain: x ≠ 2.
• Intercepts: f(0) = 1/(−2) = −½ → (0, −½). Numerator x + 1 = 0 → x-intercept (−1, 0).
• Vertical asymptote: x = 2. As x → 2⁺, numerator → 3, denominator → 0⁺, so f → +∞; as x → 2⁻, f → −∞.
• Horizontal asymptote: equal degrees (1 and 1), ratio of leading coefficients = 1, so y = 1.
• f': quotient rule, f'(x) = [(1)(x − 2) − (x + 1)(1)]/(x − 2)² = (x − 2 − x − 1)/(x − 2)² = −3/(x − 2)².

- −3/(x − 2)² < 0 everywhere in the domain → f is decreasing on (−∞, 2) and on (2, ∞). No critical points, no extrema.

• f'': f'(x) = −3(x − 2)^(−2), so f''(x) = 6(x − 2)^(−3) = 6/(x − 2)³. Sign matches (x − 2)³: f'' < 0 for x < 2 (concave down), f'' > 0 for x > 2 (concave up). No inflection point (x = 2 not in domain).

Example 3 — Given the graph of f', deduce features of f — [NO CALC]

Problem. The graph of f' (the derivative) is shown. f' is positive on (−∞, 1), crosses zero going negative at x = 1, reaches a minimum at x = 3, crosses zero going positive at x = 5, and is positive thereafter. On (−∞, 3) the graph of f' is decreasing; on (3, ∞) it is increasing.

Strategy. Translate each feature of f' into a statement about f, using the table from Core Concepts. Sign of f' → increase/decrease of f. Zeros of f' with sign change → extrema of f. Extrema of f' → inflection points of f.

Solution.

• Increasing/decreasing of f: f is increasing where f' > 0: on (−∞, 1) and (5, ∞). f is decreasing where f' < 0: on (1, 5).
• Extrema of f: At x = 1, f' changes + to − → f has a relative maximum. At x = 5, f' changes − to + → f has a relative minimum.
• Concavity/inflection of f: f is concave up where f' is increasing and concave down where f' is decreasing. f' decreases on (−∞, 3) (so f concave down) and increases on (3, ∞) (so f concave up). Therefore f has an inflection point at x = 3 — exactly where f' reaches its minimum.

Justification. f has a relative maximum at x = 1 because f' changes from positive to negative there. f has an inflection point at x = 3 because f' (and hence the slope of f) changes from decreasing to increasing there, so f'' changes sign.

Example 4 — Match a function to its derivative's graph — [NO CALC]

Problem. The function g(x) = x³ − 3x is graphed. Which graph is g'?

(A) a line through the origin with positive slope

(B) an upward parabola with x-intercepts at x = ±1 and vertex at (0, −3)

(C) an upward parabola with vertex at (0, 3), no real roots

(D) a downward parabola with x-intercepts at x = ±1

Strategy. Compute g' directly, then match shape and roots. g'(x) = 3x² − 3 = 3(x² − 1).

Solution. g'(x) = 3x² − 3 is an upward parabola (positive leading coefficient), with roots where 3x² − 3 = 0 → x = ±1, and value at x = 0 of g'(0) = −3, so vertex (0, −3). That is choice (B). As a sanity check: g has its local max at x = −1 and local min at x = 1 (its critical points), which must be the roots of g' — and (B)'s roots are at x = ±1. ✓

(d) Common Mistakes

Confusing the graphs of f, f', and f''. The most common Unit 5 error. When a problem hands you "the graph of f'," students read its y-values as values of f. Why it's wrong: a point on the f' graph is a slope of f, not a height. Fix: before answering, write at the top of your work which function is graphed, then translate every feature through the table — sign of f' controls increase/decrease, not the value of f'.

Reading extrema of f off the wrong graph. Students mark a relative max of f where the graph of f' has a maximum. Why it's wrong: a max of f' is where the slope of f is largest — that's an inflection point of f, not an extremum. Fix: extrema of f live at the x-intercepts of f' (with a sign change); maxima/minima of f' are inflection points of f.

Missing or mis-placing asymptotes. Students forget vertical asymptotes entirely, or draw a horizontal asymptote at the wrong height. Why it's wrong: skipping the domain step hides denominator zeros; guessing the HA ignores the degree comparison. Fix: always do the domain step first (VA candidates = zeros of denominator that don't cancel), and compute lim_{x→±∞} explicitly for the HA.

Sign-chart errors — claiming an extremum or inflection without a sign change. f'(c) = 0 does not guarantee an extremum, and f''(c) = 0 does not guarantee an inflection point. Why it's wrong: f(x) = x³ has f'(0) = 0 but no extremum, and f(x) = x⁴ has f''(0) = 0 but no inflection. Fix: always test the sign on both sides and require an actual change of sign.

Forgetting that asymptote-values can't be inflection points. A concavity sign change at a vertical asymptote is not an inflection point because f is undefined there. Fix: an inflection point requires f to be defined and continuous at that x.

(e) Practice Problems

[NO CALC — justification] For f(x) = (x²)/(x − 3), find the vertical and horizontal/slant behavior and all critical points. Justify whether each critical point is a relative max or min.

The table gives the sign of f' and f'':

| Interval | x < 1 | 1 < x < 4 | x > 4 |

|---|---|---|---|

| sign of f' | + | + | − |

| sign of f'' | + | − | − |

At x = 1 and x = 4, identify what f has. (Short answer.)

[NO CALC — justification] The graph of f' is shown:

Using justification language, state the intervals where f is increasing/decreasing, classify the behavior of f at x = 0 and x = 4, and identify the x-coordinate of any inflection point of f.

(f) AP Exam Focus

Free-Response Question (No calculator). Total: 9 points.

Let f be a twice-differentiable function defined for all real numbers. The graph of f', the derivative of f, is shown below. The graph of f' consists of features described here:

(a) On what open intervals is the graph of f increasing? Give a reason for your answer. (2 points)

(b) Find the x-coordinates of all relative extrema of f on (−4, 5). For each, state whether f has a relative maximum or minimum and justify your answer. (3 points)

(c) On what open intervals is the graph of f concave down? Give a reason for your answer in terms of f'. (2 points)

(d) Find the x-coordinates of all points of inflection of the graph of f on (−4, 5). Justify your answer. (2 points)

Model Solution

(a) f is increasing on (−3, 1) and on (4, 5).

Reason: f is increasing where f' > 0, and from the graph f'(x) > 0 on (−3, 1) and on (4, 5).

(b) Relative extrema occur where f' changes sign.

• At x = −3: f' changes from negative to positive → f has a relative minimum.
• At x = 1: f' changes from positive to negative → f has a relative maximum.
• At x = 4: f' changes from negative to positive → f has a relative minimum.

(Note: x = −1 and x = 3 are not extrema — f' does not equal zero there and does not change sign; they are where f' has its own extrema.)

(c) f is concave down where f''< 0, i.e., where f' is decreasing. From the graph, f' is decreasing on (−1, 3). Therefore f is concave down on (−1, 3).

(d) Points of inflection occur where f'' changes sign, i.e., where f' changes from increasing to decreasing or vice versa — the relative extrema of f'. f' has a relative maximum at x = −1 (changes increasing → decreasing) and a relative minimum at x = 3 (changes decreasing → increasing). Therefore f has points of inflection at x = −1 and x = 3.

Scoring Commentary

Where students lose points:

• Part (b): listing x = −1 and x = 3 as extrema — the classic confusion of f''s extrema with f's. Those are inflection points, not extrema.
• Part (c)/(d): justifying concavity by the sign of f' instead of whether f' is increasing or decreasing. Concavity depends on f'', which is the slope of f'. Say "f' is decreasing," not "f' is negative."
• Across all parts: forgetting the word "because" or omitting the reason. AP graders require an explicit reason tied to f', not just a correct interval.
• Part (d): writing "where f'' = 0" without confirming a sign change. State the change explicitly.