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

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Critical point identification, sign analysis of the first derivative around critical points, classification of interior relative (local) extrema, handling of endpoint extrema, and application of the test to polynomial, rational, and transcendental functions.

You should already know: Definition of the derivative, how to compute derivatives of common functions, definition of relative (local) extrema.

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 BC 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 method from Unit 5 (Analytical Applications of Differentiation) of the AP Calculus CED, accounting for approximately 4-7% of the total AP exam score, and appears on both multiple-choice (MCQ) and free-response (FRQ) sections. It is also called the First Derivative Test for Local Extrema, since "relative extrema" and "local extrema" are synonymous in AP Calculus.

The test leverages the geometric meaning of the first derivative: $f^{\prime }(x)$ tells us whether $f(x)$ is increasing ($f^{\prime }(x)>0$) or decreasing ($f^{\prime }(x)<0$) at any point. By analyzing how the sign of $f^{\prime }(x)$ changes around a critical point (a point where the derivative is zero or undefined), we can classify whether that point is a relative maximum, relative minimum, or neither.

Unlike the Second Derivative Test, the First Derivative Test works for all critical points, including those where the second derivative is zero or undefined, making it more broadly applicable. On the AP exam, you will be expected to use this test to justify classifications of extrema, which is required for full credit on FRQs.

2. Critical Points and Basic First Derivative Test Logic

Before applying the First Derivative Test, you must first identify all critical points of the function on the interval of interest. By AP CED definition, a critical point of a function $f$ is an interior point $c$ of the domain of $f$ where either $f^{\prime }(c)=0$ or $f^{\prime }(c)$ does not exist.

Critical points are the only possible locations for relative extrema, because a relative extremum requires the function to change from increasing to decreasing (or vice versa), which can only happen at a point where the derivative can change sign. By the Intermediate Value Theorem for derivatives, if $f^{\prime }$ is continuous on an interval around $c$, it can only change sign at $c$ if $f^{\prime }(c)=0$ or $f^{\prime }(c)$ is undefined.

The core rules of the First Derivative Test for interior critical points are:

1. If $f^{\prime }$ changes from positive to negative at $c$, then $f$ has a relative maximum at $x=c$
2. If $f^{\prime }$ changes from negative to positive at $c$, then $f$ has a relative minimum at $x=c$
3. If $f^{\prime }$ does not change sign at $c$, then $f$ has no relative extremum at $x=c$

To conduct the test, split the domain of $f^{\prime }$ into intervals separated by critical points, then test the sign of $f^{\prime }(x)$ in each interval to check for changes.

Worked Example

Find all critical points of $f(x)=x^3-6x^2+9x+2$ and classify each using the First Derivative Test.

1. Compute the first derivative: $$f^{\prime }(x)=3x^2-12x+9=3(x-1)(x-3)$$
2. Identify critical points: $f^{\prime }(x)$ is defined for all real $x$, so set $f^{\prime }(x)=0$ to get critical points $x=1$ and $x=3$, both interior to the domain of $f$.
3. Sign analysis: Split the domain into $(-\infty ,1)$, $(1,3)$, $(3,\infty )$. Test a point in each interval:
   • $x=0$: $f^{\prime }(0)=9>0$
   • $x=2$: $f^{\prime }(2)=-3<0$
   • $x=4$: $f^{\prime }(4)=9>0$
4. Classify: At $x=1$, $f^{\prime }$ changes from positive to negative → relative maximum at $(1,6)$. At $x=3$, $f^{\prime }$ changes from negative to positive → relative minimum at $(3,2)$.

Exam tip: On AP FRQs, you must explicitly mention the sign change of the first derivative to earn full justification credit for classifying extrema—stating "f'(c)=0 so it's a minimum" is not sufficient justification.

3. Classifying Critical Points Where $f^{\prime }$ is Undefined

Many students forget that critical points can occur where $f^{\prime }(c)$ does not exist, as long as $c$ is in the domain of $f$. The First Derivative Test works exactly the same for these critical points as it does for points where $f^{\prime }(c)=0$. This scenario is common for functions involving absolute values, roots, and piecewise definitions, which appear regularly on the AP exam.

The only extra step required is confirming that $c$ (the point where $f^{\prime }$ is undefined) is actually in the domain of $f$. If $c$ is not in the domain of $f$, it cannot be a critical point or a relative extremum, so you can discard it immediately. Once you confirm $c$ is a valid critical point, the sign analysis process is identical to the case where $f^{\prime }(c)=0$.

Worked Example

Let $f(x)=x^{2/3}(x-4)$. Find all critical points and classify each using the First Derivative Test.

1. Compute $f^{\prime }(x)$ with the product rule: $$f^{\prime }(x)=\frac{2}{3}{x}^{-1/3}(x-4)+x^{2/3}=\frac{5x-8}{3x^{1/3}}$$
2. Identify critical points: $f^{\prime }(x)$ is undefined at $x=0$, and $f(0)=0$ (so $x=0$ is in the domain of $f$), so $x=0$ is a critical point. Set the numerator equal to zero to get $x=8/5$, where $f^{\prime }=0$ and $f$ is defined, so $x=8/5$ is also a critical point.
3. Sign analysis: Split into $(-\infty ,0)$, $(0,8/5)$, $(8/5,\infty )$:
   • $x=-1$: $f^{\prime }(-1)=(-13)/(-3)>0$
   • $x=1$: $f^{\prime }(1)=(-3)/(3)<0$
   • $x=2$: $f^{\prime }(2)=(2)/(3\cdot 2^{1/3})>0$
4. Classify: At $x=0$, $f^{\prime }$ changes from positive to negative → relative maximum at $(0,0)$. At $x=8/5$, $f^{\prime }$ changes from negative to positive → relative minimum at $(1.6,\approx -3.28)$.

Exam tip: Always confirm that the point where f' is undefined is actually in the domain of f before calling it a critical point. If c is not in f's domain, it cannot be a relative extremum.

4. First Derivative Test for Endpoint Extrema

On a closed interval $[a,b]$, the endpoints $x=a$ and $x=b$ can be relative extrema, because we only consider function values inside the interval when defining relative extrema. The First Derivative Test extends naturally to endpoints, since we only need to check the sign of $f^{\prime }$ on the side of the endpoint that is inside the interval.

For a left endpoint $x=a$: if $f^{\prime }(x)<0$ just to the right of $a$, $f$ is decreasing away from $a$, so $f(a)$ is a relative maximum. If $f^{\prime }(x)>0$ just to the right of $a$, $f$ is increasing away from $a$, so $f(a)$ is a relative minimum.

For a right endpoint $x=b$: if $f^{\prime }(x)>0$ just to the left of $b$, $f$ is increasing towards $b$, so $f(b)$ is a relative maximum. If $f^{\prime }(x)<0$ just to the left of $b$, $f$ is decreasing towards $b$, so $f(b)$ is a relative minimum.

Worked Example

Find all relative extrema of $f(x)=2x^3-3x^2$ on the closed interval $[-1,2]$, including endpoints.

1. Find interior critical points: $f^{\prime }(x)=6x^2-6x=6x(x-1)$, so interior critical points at $x=0$ and $x=1$.
2. Sign analysis for interior points: Intervals $(-1,0)$, $(0,1)$, $(1,2)$. Test signs: $f^{\prime }(-0.5)>0$, $f^{\prime }(0.5)<0$, $f^{\prime }(1.5)>0$. So $x=0$ is a relative maximum, $x=1$ is a relative minimum.
3. Classify endpoints: Left endpoint $x=-1$: $f^{\prime }(x)$ is positive just right of $-1$, so $f$ increases away from $x=-1$ → relative minimum at $x=-1$, $f(-1)=-5$. Right endpoint $x=2$: $f^{\prime }(x)$ is positive just left of $2$, so $f$ increases towards $x=2$ → relative maximum at $x=2$, $f(2)=4$.
4. Final result: Relative minima at $(-1,-5)$ and $(1,-1)$; relative maxima at $(0,0)$ and $(2,4)$.

Exam tip: When asked to find all relative extrema on a closed interval, don’t forget to classify the endpoints—AP exam questions frequently test if you remember to include these.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Calling $x=c$ a critical point where $f^{\prime }(c)$ is undefined but $f(c)$ is also undefined. Why: Students automatically mark any point where f' is undefined as a critical point, without checking domain of f. Correct move: For any point where f' is undefined, confirm $f(c)$ exists before marking it as a critical point to test.
• Wrong move: Justifying a relative extremum only by stating $f^{\prime }(c)=0$, without mentioning the sign change of $f^{\prime }$. Why: Students confuse the condition for a critical point with the justification for an extremum. Correct move: On all FRQ problems, explicitly state the sign change of the first derivative around c to earn full justification credit.
• Wrong move: Forgetting to check for relative extrema at endpoints when working on a closed interval. Why: Students are taught critical points are interior points, so they ignore endpoints entirely when classifying relative extrema. Correct move: When the domain is given as a closed interval, always add the endpoints to your list of points to classify using the extended First Derivative Test.
• Wrong move: Concluding no sign change because the test value of $f^{\prime }$ is zero. Why: Students sometimes accidentally test a critical point instead of a point inside the interval, getting a zero derivative and incorrectly concluding no sign change. Correct move: When testing an interval between two critical points, always pick a test point strictly inside the interval, not equal to either critical point.
• Wrong move: Using the sign of $f(testpoint)$ instead of the sign of $f^{\prime }(testpoint)$ to classify extrema. Why: Students mix up function and derivative values when working quickly on exam day. Correct move: Always explicitly compute the sign of $f^{\prime }(testpoint)$, not $f(testpoint)$, during sign analysis.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Let $f(x)=\sin x+\cos x$ on the interval $[0,2\pi ]$. How many relative extrema does $f$ have on this interval, including endpoints? A) 1 B) 2 C) 3 D) 4

Worked Solution: First compute $f^{\prime }(x)=\cos x-\sin x$. Set $f^{\prime }(x)=0$ to get $\tan x=1$, so interior critical points at $x=\pi /4$ and $x=5\pi /4$ inside $(0,2\pi )$. Next, sign analysis shows $f^{\prime }$ is positive on $(0,\pi /4)$, negative on $(\pi /4,5\pi /4)$, and positive on $(5\pi /4,2\pi )$. $f^{\prime }$ changes from positive to negative at $\pi /4$ (relative max) and negative to positive at $5\pi /4$ (relative min). For endpoints: left endpoint $x=0$ has $f^{\prime }$ positive just right of 0, so it is a relative minimum; right endpoint $x=2\pi$ has $f^{\prime }$ positive just left of $2\pi$, so it is a relative maximum. Total relative extrema: 4. The correct answer is D.

Question 2 (Free Response)

Let $f(x)=\frac{x^2}{x-2}$. (a) Find all critical points of $f$. Justify your answer. (b) Classify each critical point as a relative maximum, relative minimum, or neither using the First Derivative Test. Justify your answer. (c) Identify all relative extrema of $f$ on the interval $[0,4]$, including endpoints.

Worked Solution: (a) Use the quotient rule to compute: $$f^{\prime }(x)=\frac{(2x)(x-2)-x^2(1)}{(x-2{)}^2}=\frac{x(x-4)}{(x-2{)}^2}$$ The domain of $f$ excludes $x=2$, so even though $f^{\prime }$ is undefined at $x=2$, it is not a critical point. $f^{\prime }(x)=0$ at $x=0$ and $x=4$, both in the domain of $f$. So critical points are $x=0$ and $x=4$.

(b) The denominator $(x-2{)}^2$ is always positive for $x\ne 2$, so the sign of $f^{\prime }$ matches the sign of $x(x-4)$. Testing intervals: $f^{\prime }>0$ for $x<0$, $f^{\prime }<0$ for $0<x<2$ and $2<x<4$, and $f^{\prime }>0$ for $x>4$. At $x=0$, $f^{\prime }$ changes from positive to negative → relative maximum. At $x=4$, $f^{\prime }$ changes from negative to positive → relative minimum.

(c) On $[0,4]$, endpoints are $x=0$ and $x=4$, which we already classified, and $x=2$ is not in the domain. The relative extrema are: relative maximum at $(0,0)$ and relative minimum at $(4,8)$.

Question 3 (Application / Real-World Style)

A small business models its daily profit from selling $x$ units of a product as $P(x)=-0.001x^3+0.12x^2+3x-80$, where $P(x)$ is measured in dollars, and $0\le x\le 100$. Use the First Derivative Test to find the production level that gives a relative maximum daily profit, and interpret your result.

Worked Solution:

1. Compute the first derivative: $P^{\prime }(x)=-0.003x^2+0.24x+3$.
2. Find critical points: Set $P^{\prime }(x)=0$, which simplifies to $x^2-80x-1000=0$. Solving with the quadratic formula gives one positive critical point at $x\approx 91$, which is inside the interval $[0,100]$.
3. Sign analysis: Test $x=10$: $P^{\prime }(10)=5.1>0$. Test $x=95$: $P^{\prime }(95)\approx -1.28<0$.
4. Classification: $P^{\prime }$ changes from positive to negative at $x\approx 91$, so this is a relative maximum. $P(91)\approx 482$ dollars.

Interpretation: Producing approximately 91 units per day gives the business a relative maximum daily profit of roughly $482.

7. Quick Reference Cheatsheet

8. What's Next

The First Derivative Test is a foundational tool for all further work with extrema and curve sketching in AP Calculus BC. Immediately after mastering this topic, you will move on to the Second Derivative Test for extrema, an alternative method for classifying critical points where the second derivative exists and is non-zero. You will also apply the First Derivative Test to find absolute extrema on closed intervals as part of Extreme Value Theorem applications, a common FRQ topic. Without a solid understanding of the First Derivative Test and how to justify extrema classifications, you will not be able to earn full credit on curve sketching or optimization problems, which make up a significant portion of the AP exam. This topic also forms the basis for analyzing motion in parametric and polar contexts later in the course.

Second Derivative Test for Relative Extrema Absolute Extrema and the Extreme Value Theorem Curve Sketching with Derivatives Optimization Problems