Difficulty: Intermediate | Prerequisites: Differentiation rules (power, quotient, chain, trig), basic factoring
Curve sketching pulls together everything from the first half of the course. You are using derivatives not to compute a single answer but to describe an entire function's behaviour: where it rises and falls, where it bends, and what happens at the edges. This is the topic where calculus stops being purely algebraic and becomes visual. If you can sketch a curve from its equation, you understand what a derivative actually tells you. If you missed the material on finding derivatives, revisit that first.
Curve sketching is the process of using first and second derivatives to determine where a function increases, decreases, curves upward or downward, and where it has peaks, valleys, or asymptotes. You build a sign chart for f'(x) and f''(x), plot the critical points and inflection points, then connect the dots. Master the workflow and you can sketch any polynomial, rational, or trigonometric function the exam throws at you.
Critical point (CP)
A point where f'(x) = 0 or f'(x) is undefined. These are the only candidates for local maxima and minima.
In simple terms, critical points are where the slope of the curve is flat or where the derivative breaks down. Every peak and valley lives at a critical point.
Critical value (CV)
The x-coordinate of a critical point. When you set f'(x) = 0 and solve, the solutions are the critical values.
Think of it as the input that makes the slope zero.
Inflection point (IP)
A point where the concavity of f changes, from concave up to concave down or vice versa. Found by setting f''(x) = 0 and confirming the sign of f'' actually changes across that x-value.
In simple terms, it is where the curve switches from bending like a bowl to bending like a cap (or the other way round).
Concavity
Describes the direction a curve bends. Concave up means f''(x) > 0 (the curve holds water, like a cup). Concave down means f''(x) < 0 (the curve spills water, like an arch).
Think of it as: concave up = smiley face, concave down = frowny face.
Sign chart
A number-line diagram that tracks the sign (+/ –) of f'(x) or f''(x) across intervals. You pick test values in each interval and record whether the derivative is positive or negative.
In simple terms, a sign chart is a quick visual map showing where the function is rising, falling, curving up, or curving down.
Vertical asymptote
A vertical line x = a where the function's output grows without bound. Occurs where the denominator of a rational function equals zero (provided the numerator does not also equal zero at that point).
Think of it as: the function blows up here, the graph shoots toward infinity.
Horizontal asymptote
A horizontal line y = L that the graph approaches as x tends to positive or negative infinity. For a rational function, compare the degrees of numerator and denominator.
In simple terms, it is the long-run height the curve settles toward.
Oblique (slant) asymptote
A non-horizontal straight line y = mx + b that the graph approaches as x tends to infinity. Occurs when the degree of the numerator is exactly one more than the degree of the denominator.
Think of it as: the curve chases a tilted line instead of a flat one.
Every problem in this section follows the same sequence. Learn the sequence once and it applies to polynomials, rational functions, and trig functions alike.
Step 1: Find critical points. Compute f'(x), set it equal to zero, and solve. Also note where f'(x) is undefined. Plug the critical values back into f(x) to get the y-coordinates.
Step 2: Build a sign chart for f'(x). Mark the critical values on a number line. Pick a test value in each interval and evaluate the sign of f'(x). Positive means increasing, negative means decreasing.
Step 3: Find inflection points. Compute f''(x), set it equal to zero, and solve. Confirm concavity actually changes across each candidate. Plug back into f(x) for y-coordinates.
Step 4: Build a sign chart for f''(x). Positive means concave up, negative means concave down.
Step 5: Classify critical points. Use the second derivative test: if f''(c) > 0 at a critical point, it is a local minimum. If f''(c) < 0, it is a local maximum. If f''(c) = 0, the test is inconclusive and you fall back to the first derivative test.
Step 6: Find intercepts. Set x = 0 for the y-intercept. Set y = 0 for x-intercepts (when solvable).
Step 7: Check for asymptotes (rational and trig functions). Vertical: denominator = 0. Horizontal: compare degrees. Oblique: numerator degree is exactly one more than denominator degree.
Step 8: Assemble the sketch. Plot CPs, IPs, intercepts, and asymptotes. Use the sign charts to connect the dots with the correct rising/falling and curvature.
f'(x) = 9x² – 36 = 9(x + 2)(x – 2). Critical values at x = –2 and x = 2.
f(–2) = 45, f(2) = –51, so CPs are (–2, 45) and (2, –51).
Sign chart for f': positive on (–∞, –2), negative on (–2, 2), positive on (2, ∞). The function rises, then falls, then rises.
f''(x) = 18x. Set to zero: x = 0. f(0) = –3, so the inflection point is (0, –3).
f'' < 0 for x < 0 (concave down), f'' > 0 for x > 0 (concave up).
Second derivative test: f''(–2) = –36 < 0, so (–2, 45) is a local max. f''(2) = 36 > 0, so (2, –51) is a local min.
Simple rational (f(x) = x/(x + 2)):
Vertical asymptote at x = –2 (denominator zero).
Horizontal asymptote: degrees equal, so y = leading coefficient ratio = 1/1 = 1.
f'(x) = 2/(x + 2)². This is always positive (never zero), so the function is always increasing on each piece of its domain.
f''(x) = –4/(x + 2)³. Never zero, but changes sign at x = –2. Concave up on (–∞, –2), concave down on (–2, ∞).
Intercepts: x = 0 gives y = 0, so the origin is both the x-intercept and the y-intercept.
Rational with oblique asymptote (f(x) = (x² + 4)/x = x + 4/x):
Vertical asymptote at x = 0.
As x grows large, 4/x vanishes, so the graph approaches y = x. That is the oblique asymptote.
f'(x) = (x² – 4)/x² = (x + 2)(x – 2)/x². Critical values at x = –2 and x = 2 (and undefined at x = 0).
f(–2) = –4, f(2) = 4. CPs are (–2, –4) and (2, 4).
f''(x) = 8/x³. Never zero, undefined at x = 0. Concave down for x < 0, concave up for x > 0.
f'(x) = cos x + sin x. Set to zero: sin x = –cos x, so tan x = –1. Solutions on [0, 2π]: x = 3π/4 and x = 7π/4.
f(3π/4) = √2, f(7π/4) = –√2. These are the critical points.
f''(x) = –sin x + cos x. Set to zero: tan x = 1. Solutions: x = π/4 and x = 5π/4.
f(π/4) = 0, f(5π/4) = 0. These are the inflection points.
Second derivative test: f''(3π/4) = –√2 < 0, so (3π/4, √2) is a local max. f''(7π/4) = √2 > 0, so (7π/4, –√2) is a local min.
y-intercept: f(0) = –1. x-intercepts where sin x = cos x: x = π/4 and x = 5π/4.
The exam-review problems all end with a four-column table: interval, sign of f', sign of f'', and the behaviour of f (increasing/decreasing + concavity). Practise building this table from scratch for each function. The table is the skeleton of the sketch.
First Derivative Test
If f'(x) changes from + to – at a critical value c, then f(c) is a local maximum. If f'(x) changes from – to +, then f(c) is a local minimum. If there is no sign change, c is neither.
Second Derivative Test
At a critical value c where f'(c) = 0: if f''(c) > 0, then f(c) is a local minimum. If f''(c) < 0, then f(c) is a local maximum. If f''(c) = 0, the test is inconclusive.
Asymptote Rules for Rational Functions (f(x) = P(x)/Q(x))
Vertical asymptote: solve Q(x) = 0, provided P(x) is not also zero there.
Horizontal asymptote: if deg(P) < deg(Q), then y = 0. If deg(P) = deg(Q), then y = (leading coeff of P)/(leading coeff of Q). If deg(P) > deg(Q), there is no horizontal asymptote.
Oblique asymptote: exists when deg(P) = deg(Q) + 1. Perform polynomial long division to find y = mx + b.
Inflection Point Condition
f''(x) = 0 is necessary but not sufficient. You must verify that f'' actually changes sign at that x-value. If f'' does not change sign, the point is not an inflection point.
Students often assume that f''(x) = 0 always means there is an inflection point. It does not. You need f'' to change sign at that value. For example, f(x) = x⁴ has f''(0) = 0 but no inflection point at x = 0.
Students sometimes forget that a critical value can also occur where f'(x) is undefined, not just where f'(x) = 0. In rational functions, points where the derivative is undefined can still be critical values.
Confusing the sign of f' with the sign of f''. Positive f' means the function is increasing. Positive f'' means the function is concave up. These are independent facts. A function can be decreasing and concave up at the same time.
Forgetting to check where the function itself is undefined when building sign charts. For f(x) = x/(x + 2), the point x = –2 is not a critical point or inflection point. It is not in the domain at all.
⚠️ The combined sign table (interval, f', f'', behaviour) appears in nearly every exam problem. Practise building it from scratch, not just reading one that is already filled in.
⚠️ Expect at least one rational function where the derivative is never zero but is undefined somewhere. Know how to handle an always-positive or always-negative derivative.
⚠️ Trigonometric curve sketching on a restricted interval [0, 2π] is a favourite. You need tan x = ±1 values memorised or quickly derivable from the unit circle.
⚠️ Oblique asymptotes have appeared in the review material. Be ready to rewrite a rational function as a sum (polynomial + remainder/denominator) to identify the slant line.
True or false: if f'(c) = 0, then f has a local extremum at x = c. Answer: False. f'(c) = 0 only means c is a critical value. You need the first or second derivative test to confirm an extremum.
Fill in the blank: if f''(x) > 0 on an interval, the function is concave ______ on that interval. Answer: Up.
True or false: a function can be increasing and concave down at the same time. Answer: True. For example, f(x) = 3x³ – 36x – 3 is increasing and concave down on (–∞, –2).
Fill in the blank: for f(x) = P(x)/Q(x) with deg(P) = deg(Q), the horizontal asymptote is y = ______. Answer: (leading coefficient of P)/(leading coefficient of Q).
True or false: f''(x) = 0 is sufficient to conclude that x is an inflection point. Answer: False. f'' must also change sign at that value.
Q: Given f(x) = 3x³ – 36x – 3, find the critical points and classify each as a local max or local min.
A: f'(x) = 9x² – 36 = 9(x + 2)(x – 2). Setting f'(x) = 0 gives x = –2 and x = 2. f(–2) = 45 and f(2) = –51. f''(x) = 18x, so f''(–2) = –36 < 0 (local max at (–2, 45)) and f''(2) = 36 > 0 (local min at (2, –51)).
Q: For f(x) = x/(x + 2), identify all asymptotes and determine whether the function is always increasing, always decreasing, or neither.
A: Vertical asymptote at x = –2. Horizontal asymptote at y = 1 (degrees equal, leading coefficients both 1). f'(x) = 2/(x + 2)², which is always positive wherever defined. The function is always increasing on each piece of its domain.
Q: For f(x) = (x² + 4)/x, find the oblique asymptote and the critical points.
A: Rewrite as f(x) = x + 4/x. As x grows large, 4/x vanishes, so the oblique asymptote is y = x. f'(x) = (x² – 4)/x². Setting the numerator to zero gives x = –2 and x = 2. f(–2) = –4 and f(2) = 4. Critical points are (–2, –4) and (2, 4).
Q: For f(x) = sin x – cos x on [0, 2π], find the inflection points.
A: f''(x) = –sin x + cos x. Setting f''(x) = 0 gives sin x = cos x, so tan x = 1. On [0, 2π], x = π/4 and x = 5π/4. f(π/4) = 0 and f(5π/4) = 0. The inflection points are (π/4, 0) and (5π/4, 0).
Q: Build the combined sign table for f(x) = 3x³ – 36x – 3 on the interval (–2, 0).
A: Test value x = –1. f'(–1) = 9(1) – 36 = –27 < 0 (decreasing). f''(–1) = 18(–1) = –18 < 0 (concave down). So on (–2, 0), the function is decreasing and concave down.
Curve sketching connects directly to optimisation (the next major topic). Once you can find and classify critical points, you are one step away from solving max/min word problems. The sign-chart technique also reappears when you study antiderivatives and definite integrals, where the sign of a function determines whether area is counted as positive or negative.
Implicit differentiation extends curve sketching to curves that are not functions in the usual sense, such as circles and ellipses. Related rates problems use the same derivative tools in a time-dependent setting.
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