AP Syllabus focus:
‘To find critical points, solve f′(x)=0 and identify x-values in the domain where f′(x) does not exist; these x-values are candidates for extrema.’
Critical points reveal where a function’s behavior may change most significantly. By analyzing its derivative, we identify important locations that guide our understanding of extrema and overall graph behavior.
Finding Critical Points Using Derivatives
Critical points are fundamental to understanding how a function behaves on an interval, especially when examining where maxima or minima may occur. In this subsubtopic, we use the first derivative as our main tool for locating these important points. Because derivatives describe the instantaneous rate of change, analyzing where the derivative is zero or undefined allows us to pinpoint where a function might switch direction or flatten out. These locations become essential candidates for deeper analysis in later topics such as the First Derivative Test and optimization.
Understanding What Makes a Point “Critical”
A critical point is any point in the domain of a function where the derivative is zero or does not exist. This criterion is both necessary and precise, ensuring that only points where the function’s rate of change behaves unusually are included.
Critical Point: A point in the domain of a function where or where does not exist.
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The definition highlights that critical points must lie within the domain of the original function, not merely within the domain of the derivative. This distinction is crucial because a derivative may fail to exist at a point where the function itself is perfectly well-defined.
The Role of the Derivative in Locating Critical Points
The derivative captures how a function changes moment by moment. When the derivative equals zero, the function’s slope becomes horizontal. When the derivative is undefined, the slope becomes vertical or otherwise irregular. These conditions flag potential extremal behavior.
= instantaneous rate of change of the function
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These derivative-based indicators give us a systematic approach: compute the derivative, locate where it equals zero, and determine where it does not exist. Both outcomes produce candidates for further analysis, but not every candidate is guaranteed to be an extremum.
Step-by-Step Structure for Finding Critical Points
Students should develop a reliable, repeatable framework for identifying critical points using the derivative. A structured approach ensures accuracy, especially when dealing with more complicated functions or piecewise definitions.
Procedure for Finding Critical Points
Compute the derivative of the function using appropriate rules (power rule, product rule, chain rule, and so on).
Solve the equation to find points where the slope is horizontal.
Identify where the derivative does not exist, such as at cusps, corners, vertical tangents, or points where the derivative’s formula is undefined.
Check whether each x-value lies in the domain of the original function, since points outside the domain cannot be critical points.
Compile all x-values that meet the criteria into a complete list of critical points.
Each bullet represents a necessary step tied directly to the syllabus: a student must identify all places where the derivative equals zero and all places where it does not exist, and then confirm they lie within the function’s domain.
Critical points are -values in the domain of where or where does not exist, and these are the only places where local extrema can occur.

This figure illustrates how both horizontal tangents and non-differentiable points can act as critical points. Smooth peaks and valleys occur where , while the cornered graph demonstrates a point where does not exist but the function is still defined. These are exactly the types of points examined when identifying critical points using derivatives. Source.
Why Domain Restrictions Matter
Critical points must be valid input values for the function itself. If a derivative is undefined because the function is undefined at that point, then that x-value cannot be considered critical. This restriction prevents misinterpretation of points where the behavior of the derivative is irrelevant because the function simply does not exist.
For example, a rational function may have vertical asymptotes at certain x-values. Although the derivative will also be undefined there, such x-values are not critical because the function’s domain excludes them. Understanding this subtle constraint prevents incorrect conclusions in later analyses.
Interpreting the Meaning of Critical Points
Finding critical points is not the same as classifying them. The act of locating these points provides a list of candidates for further testing. They are potential, but not guaranteed, locations of local maxima, local minima, or points where the function simply flattens out without changing direction.
The significance of a critical point depends on what the derivative does around it. This subsubtopic focuses solely on finding the points, not determining their type. Later tools, such as the First Derivative Test and Second Derivative Test, rely on these points to conclude whether they represent meaningful features of the graph.
Key Observations for AP Calculus AB Students
Bullet points help reinforce the essential ideas students must retain when identifying critical points:
A derivative equal to zero indicates a horizontal tangent, which often occurs at the top or bottom of a curve.
A derivative that does not exist may indicate a corner, cusp, or vertical tangent.
Both conditions produce critical points, as long as the point lies in the function’s domain.
Not all critical points produce extrema, but all local extrema occur at critical points.
Accurate derivative computation is essential; errors in differentiation will misidentify or overlook critical points.
This structured emphasis ensures students understand that locating critical points is a necessary early step in the broader process of analyzing a function’s behavior.
Critical points can also arise where is undefined, as long as the original function is defined at that x-value, such as at a cusp, corner, or vertical tangent.

This graph contrasts a cusp, where does not exist, with a smooth local minimum where . Both points are critical because the derivative is zero or undefined while the function remains defined. The image reinforces that multiple types of derivative behavior can produce valid critical points. Source.
FAQ
A zero derivative appears as a horizontal tangent, so the graph flattens without any sharp change in direction.
An undefined derivative occurs at a corner, cusp, or vertical tangent. These show abrupt changes in direction, or a tangent line that would be vertical rather than horizontal.
When inspecting a graph, look for smoothness: smooth turning points signal f'(x) = 0, while abrupt or steep transitions signal f'(x) does not exist.
Critical points may be integers, fractions, irrational values, or solutions that require numerical methods. Their form depends entirely on the equation f'(x) = 0 or the structure of f'(x).
If the derivative involves transcendental expressions, the solutions may not have closed forms. In such cases, approximation or technology may be necessary.
Yes. A function can have infinitely many critical points if f'(x) = 0 at infinitely many values or if the derivative fails to exist repeatedly.
Examples include functions with oscillating slopes or piecewise definitions containing repeated corners or cusps.
In practice, AP questions typically involve a manageable number of critical points, but conceptual understanding should include the possibility of infinitely many.
Yes. Apparent smoothness can be misleading, especially when the graph is not drawn to scale or has subtle features.
A derivative may be undefined if:
• left and right derivatives do not match
• the slope approaches infinity
• the function has an implicit domain issue
Carefully check the algebraic form of f'(x) rather than trusting a rough sketch.
Only x-values belonging to the domain of f can be classified as critical points. Even if f'(x) = 0 or undefined outside the domain, those values must be discarded.
A reliable process is:
• determine the domain of the original function
• find all solutions to f'(x) = 0
• identify where f'(x) does not exist
• include only those points that lie within the valid domain
This avoids treating asymptotes or undefined function values as critical points.
Practice Questions
Question 1 (1–3 marks)
A function f is defined for all real numbers, and its derivative is given by
f'(x) = (x − 3)(x + 1).
(a) Find all critical points of f.
(b) State the reason why these values are classified as critical points.
Question 1
(a)
• Correctly solves f'(x) = 0 to obtain x = 3 and x = −1. (1 mark)
(b)
• States that these are critical points because the derivative equals zero at these x-values. (1 mark)
• Any clear statement referencing the definition of a critical point earns the mark. (e.g., “Critical points occur where the derivative is zero or undefined.”) (1 mark)
Question 2 (4–6 marks)
A function g is defined on the closed interval [−4, 2]. Its derivative is
g'(x) = (x + 2) / (x − 1).
(a) Determine all values of x in the interval [−4, 2] where g has critical points.
(b) Explain carefully whether each identified point arises from g'(x) = 0 or g'(x) being undefined.
(c) For each critical point, check whether it lies in the domain of g, and hence determine the complete set of critical points of g on the interval.
Question 2
(a)
• Correctly identifies that g'(x) = 0 when x + 2 = 0, giving x = −2. (1 mark)
• Correctly identifies the derivative is undefined at x = 1. (1 mark)
(b)
• States that x = −2 is a critical point because g'(x) = 0 there. (1 mark)
• States that x = 1 is a critical point because the derivative is undefined there. (1 mark)
(c)
• Checks whether x = 1 is in the domain of g and notes that g is not defined where its denominator is zero, so x = 1 is excluded. (1 mark)
• Concludes that the only critical point of g on the interval [−4, 2] is x = −2. (1 mark)
