Difficulty: Intermediate | Prerequisites: Limits, Continuity, and Derivatives notes (Doc 1 of this set).
Once you can take derivatives, the next question is: what do they tell you about the shape of a graph? Curve sketching uses the first and second derivatives to locate peaks, valleys, and inflection points. Optimisation applies the same machinery to real-world problems with constraints. The Fundamental Theorem of Calculus then bridges differentiation and integration, showing that they are inverse operations. This trio of topics ties together everything from the first half of the course.
The first derivative tells you where a function rises and falls; the second tells you whether it bends upward or downward. Optimisation problems ask you to maximise or minimise something under a constraint, using exactly these tools. The Fundamental Theorem of Calculus says that a definite integral can be evaluated using any antiderivative, and that differentiating an integral gives back the original function.
Critical point
A value of x in the domain of f where f'(x) = 0 or f'(x) does not exist. These are the only candidates for local extrema.
Increasing / decreasing
f is increasing where f'(x) > 0 and decreasing where f'(x) < 0. Think of the derivative's sign as a direction indicator.
Concave up
The graph bends upward (holds water). f''(x) > 0. In simple terms, the slope itself is increasing.
Concave down
The graph bends downward (spills water). f''(x) < 0. The slope itself is decreasing.
Inflection point
A point where concavity changes, occurring where f''(x) = 0 or is undefined (and the concavity does change on either side). Think of it as where the curve switches from cupping upward to cupping downward, or vice versa.
Antiderivative
A function F(x) whose derivative is f(x). In simple terms, it is "un-doing" the derivative.
Fundamental Theorem of Calculus (FTC)
The theorem connecting differentiation and integration. Part 1 lets you evaluate definite integrals using antiderivatives. Part 2 says that the derivative of an integral (with a variable upper bound) gives back the integrand.
Set f'(x) = 0 to find critical points. Confirm each is in the domain of f(x).
f(x) is increasing where f'(x) > 0, decreasing where f'(x) < 0.
If f' changes from positive to negative at a critical point, that point is a local maximum.
If f' changes from negative to positive, it is a local minimum.
If f' does not change sign, the critical point is neither.
Concave up where f''(x) > 0; concave down where f''(x) < 0.
An inflection point occurs where f''(x) = 0 or is undefined, provided the concavity changes on either side.
As a shortcut for classifying critical points: if f'(c) = 0 and f''(c) > 0, then c 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.
Optimisation problems have three steps:
Identify the primary equation (the quantity you want to maximise or minimise) and the constraint equation (a relationship that limits the variables).
Use the constraint to eliminate a variable from the primary equation so it depends on a single variable. Take its derivative, set it equal to zero, and solve.
Plug the result back into the constraint to find the remaining unknowns. Verify with the first or second derivative test that you have found a maximum or minimum, not the other kind.
A common example: "Maximise the area of a rectangle given a fixed perimeter." The area is the primary equation; the perimeter is the constraint.
If F is any antiderivative of f (meaning F'(x) = f(x)), then:
integral from a to b of f(x) dx = F(b) - F(a)
This is how you evaluate definite integrals without Riemann sums. Find an antiderivative, plug in the bounds, subtract.
d/dx [integral from a to g(x) of f(t) dt] = f(g(x)) * g'(x)
When the upper limit is a function of x rather than just x itself, the chain rule kicks in: you multiply by the derivative of the upper limit. If both limits are functions of x, split the integral at a constant and apply Part 2 to each piece.
This comes up regularly on multiple-choice questions. Recognise the pattern: you are differentiating an accumulation function.
Topic | Formula |
|---|---|
First Derivative Test | f'(x) = 0 at critical points; sign of f' gives increasing/decreasing |
Second Derivative Test | f''(c) > 0: local min; f''(c) < 0: local max |
Inflection point | f''(x) = 0 (with concavity change) |
FTC Part 1 | integral a to b f(x) dx = F(b) - F(a) |
FTC Part 2 | d/dx integral a to g(x) f(t) dt = f(g(x)) * g'(x) |
Students often assume that f'(c) = 0 automatically means c is a local max or min. It does not. The derivative could be zero at an inflection point (for example, f(x) = x^3 at x = 0). You must check whether f' changes sign.
Students forget to verify the domain when finding critical points. A critical point must be in the domain of the original function.
In optimisation, students sometimes maximise when the problem asks for a minimum, or vice versa. Read the question twice.
With FTC Part 2, students frequently forget to multiply by g'(x) when the upper limit is a function of x rather than just x itself.
⚠️ Justification is everything in free-response questions on curve sketching. Stating "f'(x) changes from positive to negative" is required to earn full credit for a local maximum.
⚠️ The Second Derivative Test is faster but fails when f''(c) = 0. Know when to fall back to the First Derivative Test.
⚠️ FTC Part 2 appears almost every year on the AP exam, often embedded in a question about accumulation functions or graphs of f' or f.
⚠️ Optimisation is a reliable free-response topic. Set up the problem carefully; the calculus part is usually the easy bit.
True or false: If f''(x) > 0, the function is increasing.
Fill in the blank: An inflection point occurs where ______ changes.
True or false: FTC Part 2 says d/dx integral from a to x of f(t) dt = f(x).
Fill in the blank: In an optimisation problem, the constraint equation is used to ______.
Answers: 1. False (f'' > 0 means concave up, not increasing). 2. Concavity. 3. True (when the upper limit is simply x). 4. Eliminate a variable so the primary equation has a single variable.
Q: Find the intervals where f(x) = x^3 - 3x is increasing and decreasing.
A: f'(x) = 3x^2 - 3 = 3(x-1)(x+1). Critical points at x = -1 and x = 1. f'(x) > 0 on (-inf, -1) and (1, inf): increasing. f'(x) < 0 on (-1, 1): decreasing.
Q: Find the inflection point(s) of f(x) = x^3 - 3x.
A: f''(x) = 6x. Set 6x = 0, so x = 0. f'' changes sign (negative to positive) at x = 0, confirming an inflection point at (0, 0).
Q: A farmer has 200 m of fencing and wants to enclose a rectangular pen against a barn wall (so only three sides need fencing). What dimensions maximise the area?
A: Let width = x. Then length = 200 - 2x. Area = x(200 - 2x) = 200x - 2x^2. A'(x) = 200 - 4x = 0, so x = 50 m. Length = 100 m. Maximum area = 5000 m^2.
Q: Evaluate d/dx [integral from 0 to x^2 of sin(t) dt].
A: By FTC Part 2: sin(x^2) * d/dx(x^2) = sin(x^2) * 2x = 2x sin(x^2).
Curve sketching feeds directly into the applications of integration (Doc 3), where you use the relationship between a function and its derivative to solve motion problems.
The FTC is the bridge between this material and integration techniques. Without it, you would need Riemann sums for every definite integral.
Optimisation problems appear in related rates and area/volume applications, using the same set-up-and-differentiate strategy.
curve sketching, first derivative test, second derivative test, critical point, local maximum, local minimum, increasing, decreasing, concave up, concave down, inflection point, concavity, optimisation, optimization, constraint equation, primary equation, Fundamental Theorem of Calculus, FTC, FTC Part 1, FTC Part 2, evaluation theorem, accumulation function, antiderivative, AP Calculus BC