First derivative test for relative extrema — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Identifying critical points, classifying critical points as relative maxima or minima via sign changes of the first derivative, handling critical points with undefined derivatives, and applying the test to polynomial, rational, and contextual functions for AP Calculus AB.

You should already know: How to compute first derivatives of common elementary functions; how to apply core differentiation rules (product, quotient, chain); how to test the sign of a function on an open interval.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is First derivative test for relative extrema?

The First Derivative Test is a core analytical technique in Unit 5: Analytical Applications of Differentiation, which counts for 15–18% of the total AP Calculus AB exam score per the official College Board CED. It is tested regularly in both multiple-choice (MCQ) and free-response (FRQ) sections, often combined with curve sketching, optimization, or contextual modeling problems. The test uses the change in sign of the first derivative of a function around a critical point to classify the critical point as a relative (local) maximum, relative (local) minimum, or neither. A relative extremum is a point where the function value is higher (maximum) or lower (minimum) than all other nearby points on the function’s domain. "Relative" and "local" are used interchangeably on the AP exam. Unlike the Second Derivative Test, which only works for differentiable points with a non-zero second derivative, the First Derivative Test works for all valid critical points, including points where the first derivative is undefined, making it more broadly applicable for most AP problems.

2. Critical Points and Sign Interval Setup

Before you can apply the First Derivative Test, you must complete two non-negotiable preliminary steps: find all critical points of the function, then split the function’s domain into intervals separated by these critical points to test the sign of $f^{\prime }(x)$ on each interval. By definition, a critical point of a function $f$ defined at $x=c$ is any point where either $f^{\prime }(c)=0$ (called a stationary point) or $f^{\prime }(c)$ is undefined. Fermat’s Theorem confirms that all relative extrema must occur at critical points, so no other locations need to be tested. Once you collect all critical points, sort them in increasing order: they divide the domain of $f$ into open intervals, and $f^{\prime }(x)$ can only change sign at critical points, so it is entirely positive or entirely negative on each interval. To find the sign of $f^{\prime }(x)$ on an interval, simply pick any test value strictly inside the interval, plug it into $f^{\prime }(x)$, and record the sign of the result.

Worked Example

Problem: Find all critical points of $f(x)=2x^3-15x^2+36x+4$ and create sign intervals for the first derivative.

1. Compute the first derivative: $f^{\prime }(x)=6x^2-30x+36$. Since $f^{\prime }$ is a polynomial, it is defined for all real $x$.
2. Find critical points by setting $f^{\prime }(x)=0$: $$6x^2-30x+36=0⟹6(x^2-5x+6)=0⟹6(x-2)(x-3)=0$$ Critical points are at $x=2$ and $x=3$.
3. Sort critical points to split the domain $(-\infty ,\infty )$ into three intervals: $(-\infty ,2)$, $(2,3)$, $(3,\infty )$.
4. Test the sign of $f^{\prime }(x)$ in each interval: For $(-\infty ,2)$, test $x=0$: $f^{\prime }(0)=36>0$. For $(2,3)$, test $x=2.5$: $f^{\prime }(2.5)=6(6.25)-30(2.5)+36=37.5-75+36=-1.5<0$. For $(3,\infty )$, test $x=4$: $f^{\prime }(4)=6(16)-30(4)+36=96-120+36=12>0$.

Exam tip: Always sort your critical points from smallest to largest before creating intervals. Skipping this step often leads to testing the wrong interval and misclassifying extrema on the AP exam.

3. First Derivative Test Classification Rules

Once you have the sign of $f^{\prime }(x)$ on either side of a critical point $x=c$, you use the pattern of sign change to classify the critical point. The rules follow directly from the relationship between derivative sign and function behavior: a positive first derivative means the function is increasing, and a negative first derivative means the function is decreasing. If $f^{\prime }(x)$ changes from positive to negative when moving left to right across $x=c$, the function increases to $c$ then decreases after $c$, so $f(c)$ is a relative maximum. If $f^{\prime }(x)$ changes from negative to positive when moving left to right across $x=c$, the function decreases to $c$ then increases after $c$, so $f(c)$ is a relative minimum. If $f^{\prime }(x)$ does not change sign (stays positive or stays negative on both sides of $c$), there is no relative extremum at $x=c$: it is neither a maximum nor a minimum. This rule works for all valid critical points, regardless of whether $f^{\prime }(c)=0$ or $f^{\prime }(c)$ is undefined.

Worked Example

Problem: Using the sign data from $f(x)=2x^3-15x^2+36x+4$ with critical points at $x=2$ and $x=3$, classify each critical point and find the coordinates of all relative extrema.

1. Recall the sign pattern we found: $f^{\prime }(x)>0$ on $(-\infty ,2)$, $f^{\prime }(x)<0$ on $(2,3)$, $f^{\prime }(x)>0$ on $(3,\infty )$.
2. Classify $x=2$: Moving left to right across $x=2$, $f^{\prime }(x)$ changes from positive to negative. By the First Derivative Test, this is a relative maximum.
3. Calculate the function value at $x=2$: $f(2)=2(8)-15(4)+36(2)+4=16-60+72+4=32$. The relative maximum is at $(2,32)$.
4. Classify $x=3$: Moving left to right across $x=3$, $f^{\prime }(x)$ changes from negative to positive. By the First Derivative Test, this is a relative minimum.
5. Calculate the function value at $x=3$: $f(3)=2(27)-15(9)+36(3)+4=54-135+108+4=31$. The relative minimum is at $(3,31)$.

Exam tip: Always compute the full function value for the extremum if the question asks for the point or value. The AP exam will dock points if you only give the x-coordinate when the question asks for the full extremum.

4. Classifying Critical Points with Undefined Derivatives

A common AP exam problem type features functions with critical points where the first derivative is undefined (corner points, cusps, or vertical tangents). The First Derivative Test works exactly the same for these points as it does for stationary points, as long as the original function is defined at the critical point (a requirement for any critical point, and for any extremum to exist there). Many new students mistakenly assume only points where $f^{\prime }(c)=0$ can be extrema, but this is incorrect: for example, $f(x)=|x|$ has a relative minimum at $x=0$, where $f^{\prime }(0)$ is undefined, and the First Derivative Test correctly classifies it. The key mistake to avoid is omitting these critical points from your interval setup: you must include all critical points (both $f^{\prime }(c)=0$ and $f^{\prime }(c)$ undefined, with $f(c)$ defined) when sorting and splitting the domain.

Worked Example

Problem: Classify all relative extrema of $f(x)=x^{1/3}(x-4)$.

1. Compute the first derivative using the product rule: $f^{\prime }(x)=\frac{1}{3}{x}^{-2/3}(x-4)+x^{1/3}(1)=\frac{x-4+3x}{3x^{2/3}}=\frac{4x-4}{3x^{2/3}}=\frac{4(x-1)}{3x^{2/3}}$.
2. Find critical points: $f^{\prime }(x)=0$ when the numerator is zero, so $x=1$. $f^{\prime }(x)$ is undefined when the denominator is zero, so $x=0$. $f(0)=0^{1/3}(0-4)=0$ is defined, so $x=0$ is also a critical point.
3. Sort critical points to get intervals: $(-\infty ,0)$, $(0,1)$, $(1,\infty )$.
4. Test the sign of $f^{\prime }(x)$: Denominator $3x^{2/3}$ is always positive for $x\ne 0$, so sign depends on numerator $4(x-1)$, which is positive when $x>1$ and negative when $x<1$. So $f^{\prime }(x)<0$ on $(-\infty ,0)$, $f^{\prime }(x)<0$ on $(0,1)$, $f^{\prime }(x)>0$ on $(1,\infty )$.
5. Classify: At $x=0$, $f^{\prime }(x)$ does not change sign, so no extremum. At $x=1$, $f^{\prime }(x)$ changes from negative to positive, so relative minimum. $f(1)=1(-3)=-3$, so the relative minimum is at $(1,-3)$.

Exam tip: Never forget to add critical points where $f^{\prime }(x)$ is undefined to your list. AP exam writers regularly include these to test if you remember all types of critical points.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Calling any critical point automatically a maximum or minimum without testing sign change. Why: Students confuse "possible location of an extremum" with "guaranteed extremum" because most introductory examples have extrema at every stationary point. Correct move: Always test the sign of $f^{\prime }(x)$ on both sides of every critical point to confirm a sign change occurs.
• Wrong move: Omitting critical points where $f^{\prime }(x)$ is undefined from your interval setup. Why: Students only look for solutions to $f^{\prime }(x)=0$ and ignore zeros in the denominator of the derivative. Correct move: After computing $f^{\prime }(x)$, always check where $f^{\prime }(x)$ is undefined, and if $f(x)$ is defined at that point, add it to your list of critical points.
• Wrong move: Reversing the classification rule (calling positive-to-negative change a minimum and negative-to-positive a maximum). Why: Students mix up the direction when moving left to right across the critical point. Correct move: Sketch a tiny increasing/decreasing line segment next to each interval to visualize: increasing then decreasing is a peak (maximum), decreasing then increasing is a valley (minimum).
• Wrong move: Using a critical point as the test value for an adjacent interval. Why: Students rush and pick the interval endpoint as the test point, which gives $f^{\prime }(c)=0$ or undefined. Correct move: Always pick a test value strictly inside the open interval to get a valid sign.
• Wrong move: Classifying a domain endpoint as a relative extremum using the First Derivative Test. Why: Students forget you need derivative values on both sides of the point to check for a sign change. Correct move: Only classify interior critical points of the domain as relative extrema; endpoints can only be absolute extrema, per AP CED conventions.
• Wrong move: Rounding messy test values early and getting an incorrect sign. Why: Students approximate irrational test values and round a small negative number to zero or a positive number. Correct move: Keep test values in exact factored form to determine the sign, do not approximate until after you confirm the sign.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Let $f^{\prime }(x)=(x+3)(x-1)(x-5)$. Which of the following correctly lists all values of $x$ where $f(x)$ has a relative maximum? (A) $x=-3$ only (B) $x=1$ only (C) $x=-3$ and $x=5$ (D) $x=1$ and $x=5$

Worked Solution: The critical points of $f(x)$ are the roots of $f^{\prime }(x)$, sorted as $x=-3$, $x=1$, $x=5$. We find the sign of $f^{\prime }(x)$ on each interval by counting negative factors: on $(-\infty ,-3)$, all three factors are negative, so $f^{\prime }(x)<0$; on $(-3,1)$, $(x-1)$ and $(x-5)$ are negative, so two negatives multiply to positive, $f^{\prime }(x)>0$; on $(1,5)$, only $(x-5)$ is negative, so $f^{\prime }(x)<0$; on $(5,\infty )$, all factors are positive, so $f^{\prime }(x)>0$. A relative maximum requires a change from positive to negative, which only occurs at $x=1$. The correct answer is B.

Question 2 (Free Response)

Let $f(x)=\frac{x^2}{x-3}$ for $x\ne 3$. (a) Find all critical points of $f(x)$. Justify your answer. (b) Use the First Derivative Test to classify each critical point as a relative maximum, relative minimum, or neither. (c) Find the coordinates of all relative extrema of $f(x)$.

Worked Solution: (a) Use the quotient rule to compute $f^{\prime }(x)$: for $u=x^2$, $v=x-3$, $f^{\prime }(x)=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}=\frac{2x(x-3)-x^2(1)}{(x-3{)}^2}=\frac{x^2-6x}{(x-3{)}^2}=\frac{x(x-6)}{(x-3{)}^2}$. Critical points occur where $f^{\prime }(x)=0$ or $f^{\prime }(x)$ is undefined with $f(x)$ defined: $f^{\prime }(x)=0$ at $x=0$ and $x=6$. $f^{\prime }(x)$ is undefined at $x=3$, but $f(x)$ is also undefined at $x=3$, so it is not a critical point. The only critical points are $x=0$ and $x=6$.

(b) Sort critical points to get intervals: $(-\infty ,0)$, $(0,3)$, $(3,6)$, $(6,\infty )$. The denominator $(x-3{)}^2$ is always positive for $x\ne 3$, so sign depends on numerator $x(x-6)$. $f^{\prime }(x)>0$ on $(-\infty ,0)$, $f^{\prime }(x)<0$ on $(0,3)$, $f^{\prime }(x)<0$ on $(3,6)$, $f^{\prime }(x)>0$ on $(6,\infty )$. At $x=0$, $f^{\prime }(x)$ changes from positive to negative: relative maximum. At $x=6$, $f^{\prime }(x)$ changes from negative to positive: relative minimum.

(c) Calculate function values: $f(0)=\frac{0}{-3}=0$, so relative maximum at $(0,0)$. $f(6)=\frac{36}{3}=12$, so relative minimum at $(6,12)$.

Question 3 (Application / Real-World Style)

A city park models the height of a new roller coaster hill $h(x)=-\frac{1}{100}{x}^3+\frac{9}{20}{x}^2-2x+10$, where $h(x)$ is height in meters, and $x\ge 0$ is horizontal distance from the start of the hill in meters. Use the First Derivative Test to find the horizontal distance that corresponds to the relative maximum height on this hill, and interpret your result in context.

Worked Solution: Compute the first derivative: $h^{\prime }(x)=-\frac{3}{100}{x}^2+\frac{9}{10}x-2$. Set $h^{\prime }(x)=0$ and multiply through by 100: $-3x^2+90x-200=0⟹3x^2-90x+200=0$. Use the quadratic formula: $x=\frac{90\pm \sqrt{8100-2400}}{6}=\frac{90\pm \sqrt{5700}}{6}\approx 2.69$ and $x\approx 27.31$. Test intervals: for $0<x<2.69$, $h^{\prime }(0)=-2<0$; for $2.69<x<27.31$, $h^{\prime }(10)=-3+9-2=4>0$; for $x>27.31$, $h^{\prime }(30)=-27+27-2=-2<0$. A relative maximum occurs at $x\approx 27$ meters. In context, the roller coaster hill reaches a local peak height of ~35 meters at 27 meters horizontally from the start of the hill, after which height decreases down the other side of the hill.

7. Quick Reference Cheatsheet

8. What's Next

Mastering the First Derivative Test is a non-negotiable prerequisite for all remaining topics in Unit 5, including curve sketching, finding absolute extrema, and optimization, which make up a large portion of the AP exam score. This topic also lays the groundwork for the Second Derivative Test, a complementary technique that speeds up classification of stationary points, and for analyzing concavity and inflection points. Without correctly finding and classifying relative extrema, you will not be able to accurately solve optimization problems or sketch function curves, which are common FRQ tasks on the AP exam. This topic also connects to particle motion in Unit 4, where identifying direction changes is just an application of the First Derivative Test to position functions.

• Second derivative test for relative extrema
• Curve sketching using derivatives
• Finding absolute extrema
• Optimization applications