Difficulty: Intermediate | Prerequisites: Basic differentiation (power rule, constant rule), understanding of function notation.
This topic sits at the heart of differential calculus. Once you know how to take a derivative, the next question is: what can you do with it? Critical points and extrema give you the tools to locate where a function peaks, bottoms out, or changes direction. These ideas drive optimisation problems across business, engineering, and the sciences, so they come up repeatedly on exams and in later coursework.
Critical points are x-values where a function's derivative is zero or undefined. You use them, along with the sign of the derivative, to find where a function reaches its highest or lowest values (extrema), both locally and over a closed interval. The first derivative tells you whether the function is climbing or falling on any stretch of its domain.
Critical point
A value x = c where the function g(c) is defined and either g'(c) = 0 or g'(c) does not exist.
Think of it as a spot on the curve where the slope goes flat or breaks. The function could be peaking, bottoming out, or just pausing.
Relative maximum (local maximum)
A point where the function changes from increasing to decreasing. The function value at that point is higher than all nearby values.
In simple terms, it is a local hilltop. There may be a taller hill elsewhere, but in the immediate neighbourhood this is the highest point.
Relative minimum (local minimum)
A point where the function changes from decreasing to increasing. The function value at that point is lower than all nearby values.
In simple terms, it is a local valley. The function dips down and then starts climbing again.
Absolute maximum (global maximum)
The largest function value over the entire domain or over a specified closed interval.
Think of it as the highest point the function ever reaches on the interval you are looking at.
Absolute minimum (global minimum)
The smallest function value over the entire domain or over a specified closed interval.
Think of it as the lowest point the function ever reaches on the interval you are looking at.
Increasing function
A function is increasing on an interval when f'(x) > 0 throughout that interval. As x moves to the right, f(x) rises.
Decreasing function
A function is decreasing on an interval when f'(x) < 0 throughout that interval. As x moves to the right, f(x) falls.
First derivative test
A method for classifying a critical point by checking whether the derivative changes sign (positive to negative, or negative to positive) as you pass through the point.
In simple terms, you look at the slope just before and just after the critical point. If the slope flips from positive to negative, you have a local max; negative to positive, a local min; same sign on both sides, neither.
Compute the derivative g'(x) of the function.
Solve g'(x) = 0 for x. Each solution is a candidate critical point.
Identify any x-values in the domain where g'(x) does not exist (e.g. a cusp or a vertical tangent). These are also critical points, provided g(x) itself is defined there.
Evaluate g(x) at each critical point to get the corresponding y-value. This tells you the actual location on the curve.
Pick a test value just to the left of the critical point and another just to the right.
Plug each into g'(x) to check the sign.
Positive then negative: the function rises then falls, so the critical point is a relative maximum.
Negative then positive: the function falls then rises, so the critical point is a relative minimum.
Same sign on both sides: the critical point is neither a max nor a min (for example, an inflection point with a horizontal tangent).
Where f'(x) > 0, the graph of f is rising (increasing). Imagine walking along the curve from left to right and going uphill.
Where f'(x) < 0, the graph of f is falling (decreasing). You are walking downhill.
Where f'(x) = 0, the tangent line is horizontal. The function is momentarily flat, which is why these spots are worth investigating.
This uses the closed interval method (sometimes called the candidates test):
Find all critical points of f(x) that lie within the interval [a, b].
Evaluate f(x) at each critical point and at the two endpoints, f(a) and f(b).
Compare every value you computed. The largest is the absolute maximum on [a, b]; the smallest is the absolute minimum.
This works because a continuous function on a closed interval is guaranteed to reach a highest and lowest value somewhere on that interval (the Extreme Value Theorem). The maximum and minimum can only occur at critical points or at endpoints, so checking all of them is enough.
First Derivative Test procedure
Find f'(x).
Solve f'(x) = 0 and note where f'(x) is undefined.
For each critical point c, test the sign of f'(x) on either side of c.
Classify: sign change + to - means relative max; sign change - to + means relative min; no sign change means neither.
Closed Interval Method (absolute extrema on [a, b])
Find all critical points of f in (a, b).
Evaluate f at each critical point and at x = a and x = b.
The largest value is the absolute maximum; the smallest is the absolute minimum.
Businesses use critical points to find the price that maximises revenue or the production level that minimises cost. Engineers use the same ideas to find the dimensions that minimise material usage or maximise structural strength.
Students often assume every critical point is a max or a min. It is not. A critical point can be neither if the derivative does not change sign (e.g. f(x) = x^3 at x = 0, where f'(0) = 0 but the function keeps increasing).
Students sometimes forget to check endpoints when finding absolute extrema on a closed interval. The absolute max or min often sits at an endpoint, not at a critical point.
A common error is setting f(x) = 0 instead of f'(x) = 0 when looking for critical points. You are solving for where the derivative is zero, not the function itself.
Students occasionally confuse "relative" and "absolute" extrema. A relative max is a local hilltop; an absolute max is the highest value on the entire interval. A relative max is not necessarily the absolute max.
Expect at least one question asking you to find and classify critical points. The first derivative test is the standard method in this course.
Closed interval problems ("find the absolute max and min of f on [a, b]") are a staple. You must evaluate at endpoints and critical points, then compare.
Questions may give you a derivative already and ask you to interpret what it means for the function's behaviour (increasing, decreasing, local extrema).
Watch for problems that give you a graph of f'(x) and ask about f(x). If f' is above the x-axis, f is increasing; below, f is decreasing; crossing the axis, f has a critical point.
True or false: Every point where f'(x) = 0 is a relative maximum or minimum. (False. It could be neither, e.g. an inflection point.)
Fill in the blank: When f'(x) > 0 on an interval, the function is ______ on that interval. (Increasing.)
True or false: To find absolute extrema on [a, b], you only need to check the critical points inside the interval. (False. You also need to evaluate f at the endpoints a and b.)
Fill in the blank: A critical point occurs where the derivative equals ______ or ______. (Zero; does not exist.)
True or false: A relative maximum is always the absolute maximum. (False. The absolute maximum may occur at an endpoint or at a different critical point.)
Q: Find the critical points of f(x) = x^3 - 6x^2 + 9x + 1.
A: f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3). Setting f'(x) = 0 gives x = 1 and x = 3. Both are critical points.
Q: Using the first derivative test, classify the critical points of f(x) = x^3 - 6x^2 + 9x + 1.
A: For x = 1, test x = 0 (f'(0) = 9 > 0) and x = 2 (f'(2) = -3 < 0). Sign changes + to -, so x = 1 is a relative maximum. For x = 3, test x = 2 (f'(2) = -3 < 0) and x = 4 (f'(4) = 9 > 0). Sign changes - to +, so x = 3 is a relative minimum.
Q: Find the absolute maximum and minimum of f(x) = x^2 - 4x + 3 on the interval [0, 5].
A: f'(x) = 2x - 4 = 0, so x = 2 is the only critical point in [0, 5]. Evaluate: f(0) = 3, f(2) = -1, f(5) = 8. The absolute maximum is 8 at x = 5. The absolute minimum is -1 at x = 2.
Q: On what intervals is g(x) = 2x^3 - 3x^2 - 12x increasing?
A: g'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x - 2)(x + 1). The derivative is positive when x < -1 or x > 2. So g is increasing on (-infinity, -1) and (2, infinity).
Q: If you are given that f'(x) < 0 for all x in (3, 7), what can you say about f on that interval?
A: f is strictly decreasing on (3, 7). The function values fall as x increases through that interval.
This connects to revenue optimisation because finding the price that maximises revenue is a direct application of critical points: you differentiate the revenue function, set the derivative to zero, and classify the result.
It also connects to the product rule and chain rule, since real-world functions (cost, volume, investment growth) often require these rules before you can find and classify critical points.
The second derivative test (covered separately) offers an alternative way to classify critical points, and concavity analysis builds directly on the increasing/decreasing framework here.
Critical point, critical number, stationary point, turning point, relative maximum, relative minimum, local max, local min, absolute maximum, absolute minimum, global max, global min, first derivative test, closed interval method, candidates test, increasing function, decreasing function, sign of derivative, derivative sign chart, optimisation, optimization, extrema, extreme values, Extreme Value Theorem, function behaviour, function behavior, MATH 101, brief calculus, applied calculus