Difficulty: Intermediate | Prerequisites: Calculus I differentiation (power rule, factoring polynomials), basic understanding of intervals and sign analysis.
Finding critical points is one of the central skills in single-variable calculus and carries directly into multivariable work in Calculus III. The idea is straightforward: locate where a function's derivative equals zero (or is undefined), then decide whether each of those points is a local maximum, a local minimum, or neither. You need a solid handle on taking derivatives with the power rule and on factoring, because the algebra of setting f'(x) = 0 is where most of the work happens. This topic is also the gateway to optimisation problems, which appear in nearly every applied field.
To find critical points, take the derivative, set it equal to zero and solve. Then use the first derivative test (check the sign of f'(x) on each side) to classify each point as a local max, local min, or neither. Factoring the derivative cleanly is the make-or-break step.
Critical point
A value x = c where f'(c) = 0 or f'(c) is undefined, provided c is in the domain of f. In simple terms, it is any spot where the slope of the curve is flat or the derivative does not exist.
Local maximum (relative maximum)
A point where f(c) is greater than f(x) for all x sufficiently close to c. Think of it as the top of a hill on the graph.
Local minimum (relative minimum)
A point where f(c) is less than f(x) for all x sufficiently close to c. Think of it as the bottom of a valley.
First derivative test
A method for classifying a critical point by examining the sign of f'(x) on either side. If the sign changes from positive to negative, the point is a local max; from negative to positive, a local min; no sign change means neither.
Sign chart (sign diagram)
A number-line sketch that marks the critical points and records whether f'(x) is positive or negative on each interval between them. In simple terms, it is the quick visual tool you use to read off the first derivative test results.
Apply the power rule term by term. For f(x) = 5x^3 - 3x^5 + 3:
Derivative of 5x^3 is 15x^2
Derivative of -3x^5 is -15x^4
Derivative of 3 is 0
So f'(x) = 15x^2 - 15x^4.
Factor out the greatest common factor first:
15x^2 - 15x^4 = 0
15x^2(1 - x^2) = 0
15x^2(1 - x)(1 + x) = 0
This gives three critical values: x = 0, x = 1, x = -1.
Test the sign of f'(x) on each interval created by the critical values:
x < -1 (e.g. x = -2): 15(4)(1 - 4) = negative, so f is decreasing
-1 < x < 0 (e.g. x = -0.5): 15(0.25)(1 - 0.25) = positive, so f is increasing
0 < x < 1 (e.g. x = 0.5): 15(0.25)(1 - 0.25) = positive, so f is increasing
x > 1 (e.g. x = 2): 15(4)(1 - 4) = negative, so f is decreasing
Summary: -, +, +, -
At x = -1: sign changes from negative to positive, so this is a local minimum. f(-1) = 5(-1) - 3(-1) + 3 = -5 + 3 + 3 = 1. The point is (-1, 1).
At x = 0: sign does not change (positive on both sides), so this is neither a max nor a min. It is an inflection-like flat spot. f(0) = 3. The point is (0, 3).
At x = 1: sign changes from positive to negative, so this is a local maximum. f(1) = 5(1) - 3(1) + 3 = 5 - 3 + 3 = 5. The point is (1, 5).
The handwritten solution correctly identifies all three critical values and factors the derivative properly. However, it lists the y-coordinates of the critical points as 0, which is not correct: you must substitute each x-value back into the original f(x), not f'(x). It also labels x = 0 as a minimum, but the sign chart shows no sign change there, so x = 0 is neither a max nor a min.
Power rule:
\frac{d}{dx} x^n = n x^{n-1}Critical point condition:
f'(c) = 0 \quad \text{or} \quad f'(c) \text{ undefined}First derivative test summary:
f' changes from + to - at c → local maximum at c
f' changes from - to + at c → local minimum at c
f' does not change sign at c → neither
Optimisation problems use exactly this technique. Engineers finding the dimensions that minimise material cost, economists locating the output level that maximises profit, and data scientists tuning a loss function all begin by setting a derivative to zero and classifying the result. The sign chart is the same tool whether the function describes revenue, temperature, or signal strength.
Students often evaluate f'(x) instead of f(x) to get the y-coordinate of a critical point. The derivative tells you where the critical point is (the x-value); the original function tells you the height (the y-value).
A critical point where f'(c) = 0 is not automatically a max or min. If the sign of f' does not change on either side, the point is neither. The classic example is f(x) = x^3 at x = 0.
Students sometimes forget to factor completely. Leaving 15x^2(1 - x^2) without recognising the difference of squares means missing the clean roots x = 1 and x = -1.
Labelling a point "min" just because the derivative equals zero and the y-value seems small is not valid. Classification requires the first derivative test (or second derivative test), not a guess from the y-value.
⚠️ "Find and classify the critical points" is one of the most common exam prompts in Calculus I and II, and it resurfaces in Calculus III for partial derivatives.
⚠️ You will almost certainly need to show the sign chart or the second derivative test to earn full marks. Stating the answer without justification typically loses points.
⚠️ Always substitute back into f(x) for y-values. Exams frequently ask for the coordinates of the critical point, not just the x-value.
True or false: Every point where f'(x) = 0 is a local extremum.
False. The derivative can equal zero without a sign change (e.g. f(x) = x^3 at x = 0).
Fill in the blank: To find the y-coordinate of a critical point, substitute the x-value into ____.
The original function f(x), not the derivative.
True or false: If f'(x) changes from negative to positive at x = c, then x = c is a local maximum.
False. That pattern indicates a local minimum.
Fill in the blank: 15x^2 - 15x^4 factors as 15x^2()()
(1 - x)(1 + x)
Q: Given f(x) = 5x^3 - 3x^5 + 3, what are the critical points and their classifications?
A: The critical points are x = -1, x = 0, and x = 1. By the first derivative test, x = -1 is a local minimum at (-1, 1), x = 0 is neither a max nor a min at (0, 3), and x = 1 is a local maximum at (1, 5).
Q: Why is x = 0 not a local minimum even though f'(0) = 0?
A: Because f'(x) does not change sign at x = 0. It is positive on both (-1, 0) and (0, 1). A local minimum requires f' to change from negative to positive.
Q: Suppose g(x) = x^4. Find the critical points and classify them.
A: g'(x) = 4x^3, so the only critical point is x = 0. The sign of g' changes from negative (for x < 0) to positive (for x > 0), so x = 0 is a local minimum.
Q: What is the difference between a critical point and an inflection point?
A: A critical point is where f'(x) = 0 or is undefined (flat or undefined slope). An inflection point is where f''(x) changes sign (the concavity flips). They can overlap but are different concepts.
This connects to the second derivative test, which offers an alternative way to classify critical points when f''(c) is easy to compute. In Calculus III (MATH 215), the same logic extends to functions of two variables: you find where both partial derivatives equal zero, then use the second-order partial derivatives (the Hessian determinant) to classify the critical point as a saddle, max, or min. Optimisation with constraints (Lagrange multipliers) builds directly on this foundation as well.
Critical points, local extrema, relative maximum, relative minimum, first derivative test, sign chart, sign diagram, power rule, factoring polynomials, polynomial differentiation, f'(x) = 0, increasing and decreasing intervals, inflection point, second derivative test, optimisation, Calculus III review, MATH 215 University of Michigan, max and min of polynomial functions