Difficulty: Intermediate | Prerequisites: Limits and Continuity, Derivatives and Differentiation Rules study notes.
This is where differentiation becomes useful. Once you can take a derivative, the natural next question is: what can you do with it? The answer is curve sketching (where is the function rising, falling, concave up, concave down?), finding the largest or smallest value a function can take (optimisation), and connecting rates of change of related quantities (related rates). The Mean Value Theorem also lives here, providing the theoretical backbone for many of these results.
The first derivative tells you where a function increases or decreases and where local extrema occur. The second derivative tells you about concavity. Optimisation problems use these tools on real-world constraints. Related rates problems connect two changing quantities through differentiation with respect to time.
Increasing / decreasing function
f is increasing on an interval when f'(x) > 0 for all x in that interval, and decreasing when f'(x) < 0. In simple terms, if the derivative is positive the graph goes uphill; if negative, downhill.
Local maximum / local minimum
A local max at x = c means f(c) >= f(x) for all x near c. A local min means f(c) <= f(x) for all x near c. These occur at critical points where the derivative changes sign. Think of a hilltop (local max) and a valley bottom (local min).
Concave up / concave down
f is concave up on an interval when f''(x) > 0, and concave down when f''(x) < 0. Concave up looks like a cup; concave down looks like a cap. In simple terms, concave up means the slope itself is increasing; concave down means the slope is decreasing.
Inflection point
A point where the concavity changes, i.e. f''(x) changes sign. This is where the curve switches from curving one way to curving the other.
Mean Value Theorem (MVT)
If f is continuous on [a, b] and differentiable on (a, b), there exists at least one c in (a, b) where f'(c) = [f(b) - f(a)] / (b - a). In simple terms, somewhere on the interval the instantaneous rate of change equals the average rate of change.
Optimisation
The process of finding the maximum or minimum value of a function subject to constraints. Set up a function to optimise, find its critical points, and verify which gives the extreme value.
Related rates
Problems where two or more quantities change with respect to time, and their rates of change are connected through an equation. Differentiate the equation with respect to t and solve for the unknown rate. Think of it as: the equation ties the quantities together; the chain rule ties their rates together.
Find f'(x). Set f'(x) = 0 to locate critical points.
Test the sign of f'(x) on each interval between critical points.
f'(x) > 0 means f is increasing. f'(x) < 0 means f is decreasing.
Example: f(x) = x^3 - 6x^2 + 9x + 2. f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3). Critical points: x = 1 and x = 3. f'(x) > 0 on (-inf, 1) and (3, inf), so f increases there. f'(x) < 0 on (1, 3), so f decreases there.
If f' changes from positive to negative at x = c, there is a local maximum at c.
If f' changes from negative to positive at x = c, there is a local minimum at c.
If f' does not change sign, there is no extremum.
Continuing the example: at x = 1, f' goes from + to -, so local max. At x = 3, f' goes from - to +, so local min.
f''(x) > 0 on an interval means f is concave up there.
f''(x) < 0 on an interval means f is concave down there.
Inflection points occur where f''(x) changes sign.
Example continued: f''(x) = 6x - 12. f''(x) = 0 at x = 2. For x < 2, f'' < 0 (concave down). For x > 2, f'' > 0 (concave up). Inflection point at x = 2.
Statement: if f is continuous on [a, b] and differentiable on (a, b), then there exists c in (a, b) with f'(c) = [f(b) - f(a)] / (b - a).
The conclusion is that the tangent line at some interior point is parallel to the secant line through the endpoints.
The MVT is a theoretical result. Exam questions typically ask you to state it or identify it from the options.
A rectangular garden against a wall uses fencing for three sides. If 120 m of fencing is available: let x = the two widths (perpendicular to the wall), and l = the length (parallel to the wall).
Constraint: 2x + l = 120, so l = 120 - 2x.
Area: A(x) = x * (120 - 2x) = 120x - 2x^2.
A'(x) = 120 - 4x. Setting A'(x) = 0: x = 30. Then l = 120 - 60 = 60.
A''(x) = -4 < 0, confirming a maximum. Dimensions: 30 m by 60 m, area 1800 m^2.
A 5 m ladder leans against a wall. The base slides away at 1 m/s. How fast does the top slide down when the base is 3 m from the wall?
Relationship: x^2 + y^2 = 25 (Pythagoras).
Differentiate with respect to t: 2x(dx/dt) + 2y(dy/dt) = 0.
When x = 3: y = sqrt(25 - 9) = 4. dx/dt = 1.
Substituting: 2(3)(1) + 2(4)(dy/dt) = 0. Solve: dy/dt = -6/8 = -3/4 m/s.
The top slides down at 3/4 m/s (the negative sign indicates y is decreasing).
Position s(t), velocity v(t) = s'(t), acceleration a(t) = v'(t) = s''(t).
v'(t) = a(t), so the derivative of velocity is acceleration.
This is the correct relationship: s'(t) = v(t), v'(t) = a(t).
First derivative test: f'(x) changes + to - at c -> local max; - to + -> local min.
Second derivative test: f''(c) > 0 -> local min at c; f''(c) < 0 -> local max at c.
MVT: f'(c) = [f(b) - f(a)] / (b - a) for some c in (a, b).
s(t) -> v(t) = s'(t) -> a(t) = v'(t).
Optimisation pattern: write the quantity to optimise as a function of one variable, differentiate, set equal to zero, solve, verify with the second derivative or endpoint check.
Optimisation is everywhere: minimising material cost for packaging, maximising the area enclosed by a given length of fence, or finding the most efficient speed for fuel consumption. Related rates problems model situations like a filling tank, a growing shadow, or the changing distance between two moving objects. Concavity and inflection points help economists identify where growth rates are accelerating or decelerating.
Students assume every critical point is a local extremum. A critical point can also be an inflection point (like x = 0 for f(x) = x^3) if f' does not change sign.
Students forget to verify their optimisation answer is a maximum rather than a minimum (or vice versa). Always use the second derivative test or check endpoints.
In related rates, students sometimes forget to differentiate implicitly with respect to t. Every variable that changes with time needs the chain rule.
Students confuse f''(x) > 0 with "increasing." f''(x) > 0 means concave up, not increasing. It is f'(x) > 0 that means increasing.
The long-answer section of this exam includes an optimisation problem (the fencing problem) and a related rates problem (the ladder). These are worth 6 marks each and require full working.
"Identify the intervals of increase/decrease, local extrema, and concavity" is a classic 6-mark question. The worked process is: find f', find critical points, sign-test f', find f'', sign-test f''.
The MVT is typically tested as a multiple-choice conceptual question. Know the precise statement.
v'(t) = a(t) is a standard multiple-choice item on the position-velocity-acceleration relationship.
True or False: If f''(x) > 0 on an interval, f is increasing on that interval. (False. f is concave up. Increasing requires f'(x) > 0.)
Fill in the blank: According to the MVT, f'(c) = ___. ([f(b) - f(a)] / (b - a).)
True or False: At a critical point, f always has a local max or min. (False. The derivative could be zero without changing sign.)
Fill in the blank: If a 5 m ladder has its base 3 m from the wall, the top is ___ m up the wall. (4 m, by Pythagoras.)
Q: Given f(x) = x^3 - 6x^2 + 9x + 2, find the intervals of increase and decrease.
A: f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3). f'(x) > 0 on (-inf, 1) and (3, inf): increasing. f'(x) < 0 on (1, 3): decreasing.
Q: For the same f(x), identify all local extrema.
A: At x = 1, f' changes from + to -, so local max at (1, 6). At x = 3, f' changes from - to +, so local min at (3, 2).
Q: For the same f(x), find the intervals of concavity.
A: f''(x) = 6x - 12. f'' = 0 at x = 2. Concave down on (-inf, 2), concave up on (2, inf). Inflection point at x = 2.
Q: If f''(x) > 0 on an interval, what can you say about f(x)?
A: f(x) is concave up on that interval.
Q: State the Mean Value Theorem.
A: If f is continuous on [a, b] and differentiable on (a, b), there exists c in (a, b) such that f'(c) = [f(b) - f(a)] / (b - a).
Q: A rectangular garden against a wall uses 120 m of fencing for three sides. Find the dimensions that maximise the area.
A: Let x = width. Then length = 120 - 2x. A(x) = 120x - 2x^2. A'(x) = 120 - 4x = 0 gives x = 30. Length = 60. Max area = 1800 m^2.
Q: A 5 m ladder slides. Base moves at 1 m/s, base is 3 m from wall. How fast does the top slide down?
A: x^2 + y^2 = 25. Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0. x = 3, y = 4, dx/dt = 1. Solving: dy/dt = -3/4 m/s. The top slides down at 0.75 m/s.
Q: Which relationship is correct: v'(t) = a(t), or s'(t) = a(t)?
A: v'(t) = a(t). Velocity is the derivative of position, and acceleration is the derivative of velocity.
Optimisation and related rates are the payoff for everything in the Derivatives notes. The second derivative test connects directly to concavity, which reappears in curve sketching. The MVT is used in the proof of the Fundamental Theorem of Calculus. Position, velocity, and acceleration tie calculus to physics, and displacement (found by integrating velocity) bridges this topic to the Integration notes.
applications of derivatives, increasing, decreasing, concave up, concave down, inflection point, first derivative test, second derivative test, Mean Value Theorem, MVT, optimisation, optimization, fencing problem, related rates, sliding ladder, position, velocity, acceleration, curve sketching, local maximum, local minimum, Calculus I, MSU, Michigan State University, Calculus 1