Study guide
This chapter is where derivatives stop being computations and start becoming tools for understanding shape: where a function rises or falls, where it curves up or down, and how to guarantee or locate its largest and smallest values. Expect this material to carry significant exam weight, since it blends conceptual reasoning with algebraic execution.
Critical Points and the First Derivative Test
A critical point of f occurs where f'(x) = 0 or where f'(x) is undefined, provided x is in the domain of f itself. Critical points are the only candidates for local maxima or minima on an open interval, but not every critical point is one — some are neither, such as an inflection point with a horizontal tangent. The first derivative test resolves this by examining the sign of f' immediately to the left and right of each critical point. If f' changes from positive to negative, f has a local maximum there; if f' changes from negative to positive, f has a local minimum; if the sign does not change, the point is neither. Consider f(x) = x^3 - 3x, with f'(x) = 3x^2 - 3 = 3(x-1)(x+1), giving critical points at x = -1 and x = 1. Testing sign intervals shows f' is positive on (-infinity, -1), negative on (-1, 1), and positive again on (1, infinity), so f has a local maximum at x = -1 and a local minimum at x = 1. Building a sign chart — marking critical points on a number line and testing a value in each resulting interval — is the most reliable way to organize this reasoning and avoid sign errors under time pressure.
Concavity, Inflection Points, and the Second Derivative Test
The second derivative describes concavity: where f''(x) is positive, the graph of f curves upward (concave up), and where f''(x) is negative, the graph curves downward (concave down). An inflection point is a location where concavity changes sign, which requires f''(x) = 0 or undefined at that point along with an actual sign change on either side — a common error is declaring an inflection point wherever f'' equals zero without checking that concavity truly switches. The second derivative test offers a shortcut for classifying critical points: if f'(c) = 0 and f''(c) is positive, f has a local minimum at c because the curve is concave up there; if f''(c) is negative, f has a local maximum at c. If f''(c) equals zero, the test is inconclusive and the first derivative test must be used instead. These tools combine to build a complete qualitative sketch of a function's graph: intervals of increase and decrease from f', intervals of concavity from f'', and the coordinates of extrema and inflection points marking where behavior changes. On free-response questions, always justify conclusions with the sign of the appropriate derivative rather than merely stating the answer, since reasoning is what earns the point.
The Mean Value Theorem and Absolute Extrema
The Mean Value Theorem (MVT) states that if f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one c in (a, b) where f'(c) equals the average rate of change, [f(b) - f(a)]/(b - a). Geometrically, this guarantees some point where the tangent line is parallel to the secant line connecting the endpoints. MVT problems typically ask you to verify the hypotheses (continuity and differentiability) before solving f'(c) = [f(b) - f(a)]/(b - a) for c, and skipping the hypothesis check costs points even when the arithmetic is correct. The Extreme Value Theorem guarantees that a function continuous on a closed interval [a, b] attains both an absolute maximum and an absolute minimum somewhere on that interval. To find these absolute extrema, evaluate f at every critical point inside (a, b) and at both endpoints a and b, then compare all the resulting values — the largest is the absolute maximum and the smallest is the absolute minimum. For f(x) = x^3 - 3x^2 - 9x + 5 on the closed interval [-2, 5], the derivative f'(x) = 3x^2 - 6x - 9 factors as 3(x-3)(x+1), giving critical points at x = -1 and x = 3. Evaluating f(-2) = 3, f(-1) = 10, f(3) = -22, and f(5) = 10 shows the absolute maximum is 10 (achieved at both x = -1 and x = 5) and the absolute minimum is -22 at x = 3.
Optimization Problems
Optimization problems ask you to maximize or minimize some quantity subject to a constraint, and they combine nearly every skill in this chapter into a single multi-step process. The strategy is: identify the quantity to optimize and write it as a function of one variable, use the given constraint to eliminate a second variable if the function initially depends on two, differentiate, find critical points, and confirm a maximum or minimum using the first or second derivative test (or by comparing endpoint values if the domain is a closed interval). Consider building a rectangular garden with 200 square feet of area against an existing wall, so fencing is needed on only three sides — the side running along the wall needs no fence since the wall replaces it, meaning the three sides needing fence are the length opposite the wall (parallel to it) and the two widths. Let x represent the side parallel to the wall and y represent each perpendicular side, so the area constraint is xy = 200, giving y = 200/x. The fence length is L(x) = x + 2y = x + 400/x. Differentiating gives L'(x) = 1 - 400/x^2, which equals zero when x^2 = 400, so x = 20 (rejecting the negative root since length must be positive). This gives y = 200/20 = 10, and the minimum total fencing required is L(20) = 20 + 2(10) = 40 feet. Always check that a critical point actually minimizes (rather than maximizes) the quantity, using the second derivative test or by confirming the function increases on both sides moving away from that point.
Key terms
- Critical point
- — A value in the domain of f where f'(x) equals zero or is undefined, and the only candidate location for a local extremum.
- First derivative test
- — A method that classifies a critical point as a local max, local min, or neither based on the sign change of f' around that point.
- Concavity
- — The direction a curve bends, described as concave up where f'' is positive and concave down where f'' is negative.
- Inflection point
- — A point where the concavity of f changes sign, requiring both f'' = 0 or undefined and an actual sign change on either side.
- Second derivative test
- — A method that classifies a critical point c as a local min if f''(c) is positive or a local max if f''(c) is negative, inconclusive if f''(c) = 0.
- Mean Value Theorem (MVT)
- — For f continuous on [a,b] and differentiable on (a,b), guarantees some c in (a,b) where f'(c) equals the average rate of change over [a,b].
- Extreme Value Theorem
- — Guarantees a function continuous on a closed interval attains both an absolute maximum and an absolute minimum on that interval.
- Absolute extrema
- — The largest (absolute maximum) and smallest (absolute minimum) values a function attains over its entire domain or a specified interval.
- Candidates test
- — The method of finding absolute extrema by comparing function values at all critical points and both endpoints of a closed interval.
- Optimization
- — The process of finding the input value that maximizes or minimizes a quantity, typically subject to a given constraint.
Exam tips
- Always confirm the hypotheses of the Mean Value Theorem (continuity on the closed interval, differentiability on the open interval) explicitly before applying it.
- Do not call every point where f'' = 0 an inflection point — verify concavity actually changes sign on both sides.
- For absolute extrema on a closed interval, systematically list and compare f at every critical point plus both endpoints; do not stop at the first local extremum found.
- In optimization problems, eliminate one variable using the constraint before differentiating, and double-check whether the domain restricts your variable to positive values only.
- When the second derivative test gives f''(c) = 0, fall back to the first derivative test rather than leaving the classification blank.