5.4 Using the First Derivative Test to Determine Relative (Local) Extrema

The First Derivative Test finds local, or relative, maximums and minimums by checking how the sign of $f'$ changes at each critical point. If $f'$ goes from positive to negative, $f$ has a local max; if $f'$ goes from negative to positive, $f$ has a local min; if the sign stays the same, there is no extremum there. For AP Calculus, justify extrema with the sign change of $f'$.

Why This Matters for the AP Calculus Exam

This topic is part of Unit 5, which carries a large share of the AP Calculus AB exam and a smaller but real share on BC. The First Derivative Test shows up when you need to locate and justify relative extrema, which is a common task on both multiple-choice and free-response questions.

A big part of the skill here is justification. Many free-response questions ask you to find an extremum and explain why it is a max or min using the sign behavior of $f'$. Writing a clean reason like "$f$ has a relative maximum at $x = c$ because $f'$ changes from positive to negative there" is exactly the kind of calculus-based reasoning the exam rewards. Sloppy phrasing like "it goes up" does not count as a justification.

Key Takeaways

• A critical point is where $f'(x) = 0$ or $f'$ does not exist. All local extrema happen at critical points, but not every critical point is an extremum.
• Local max: $f'$ changes from positive to negative (function goes from increasing to decreasing).
• Local min: $f'$ changes from negative to positive (function goes from decreasing to increasing).
• No extremum: $f'$ keeps the same sign on both sides of the critical point.
• Use test points or a sign chart for $f'$ to check the sign on each side of every critical point.
• Always justify with the sign change of $f'$, and refer to $f$, $f'$, and $f''$ by name, not "it."

First Derivative Test

The First Derivative Test uses the sign of $f'$ to find the relative (local) extrema of a function. If $f'$ changes from positive to negative at a point (so $f$ changes from increasing to decreasing), then $f$ has a relative maximum there. If $f'$ changes from negative to positive (so $f$ changes from decreasing to increasing), then $f$ has a relative minimum there.

The process is close to finding intervals where a function increases or decreases.

First, find the critical points of the function. A critical point is where the derivative equals $0$ or is not defined. Once you have those points, check the sign of $f'$ on either side of each one.

• If $f'$ is positive on the left and negative on the right, the point is a local maximum. ⬆️
• If $f'$ is negative on the left and positive on the right, the point is a local minimum. ⬇️
• If $f'$ has the same sign on both sides, the function is either increasing or decreasing through that point, and there is no relative extremum.

First Derivative Test Walkthrough

Consider the function $f(x) = x^2$. The derivative is $f'(x) = 2x$ by the power rule. At $x = 0$, the derivative equals $0$, so this is a critical point. Now evaluate $f'$ on either side.

Left side (for $x < 0$): Substitute a value less than $0$ into $f'(x) = 2x$. Using $x = -1$:

$f'(-1) = 2(-1) = -2$

So $f'$ is negative for $x < 0$.

Right side (for $x > 0$): Substitute a value greater than $0$. Using $x = 1$:

$f'(1) = 2(1) = 2$

So $f'$ is positive for $x > 0$.

Conclusion: $f'$ is negative on the left and positive on the right, so $f$ has a relative minimum at $x = 0$. ⬇️

How to Use This on the AP Calculus Exam

Problem Solving

1. Find $f'$ and set it equal to $0$. Also note any points where $f'$ does not exist. These are your critical points.
2. Build a sign chart for $f'$. Pick test points between and around the critical points.
3. Plug each test point into $f'$ and record whether the result is positive or negative.
4. Read the sign changes: positive to negative is a local max, negative to positive is a local min, no change means no extremum.

Free Response

When a question says "Justify your answer," your work needs a clear reason tied to $f'$. A strong answer names the function and the sign change, like: "$f$ has a relative maximum at $x = c$ because $f'$ changes from positive to negative at $x = c$." Start your sentence using the language from the question to keep your answer focused.

Practice Problems

Question 1: Let $h$ be a polynomial function with derivative $h'(x) = x^2(x-3)(x+4)$. Where are $h$'s relative minima?

Question 2: Let $g(x) = x^5 - 80x$. Where are $g$'s relative maxima?

Solutions

Question 1:

Find the critical points by setting $h'(x)$ equal to $0$:

$x^2 = 0$

$(x-3) = 0$

$(x+4) = 0$

The critical points are $x = -4$, $0$, and $3$. The function can have minima only at these points. Now check the sign of $h'$ on each side of each point.

So $h$ has one relative minimum at $x = 3$. Notice that $x = 0$ is a critical point but not an extremum, because $f'$ has the same sign on both sides (the factor $x^2$ does not change sign).

Question 2:

Find the critical points. Since $g'(x) = 5x^4 - 80$, set it equal to $0$ to get $x = -2$ and $x = 2$. The function can have maxima only at these points. Now check the sign of $g'$ on each side.

So $g$ has one relative maximum at $x = -2$.

Common Misconceptions

• Every critical point is an extremum. Not true. If $f'$ keeps the same sign on both sides (like the $x^2$ factor in Question 1), there is no max or min there.
• Confusing the graph of $f'$ with the graph of $f$. A point where $f'$ is increasing tells you about concavity of $f$, not about where $f$ has a max or min. For extrema, you care about where $f'$ changes sign, not where $f'$ increases.
• Saying "it goes up" as a justification. That is too vague. Tie your reason to the sign of $f'$ and name the function, such as "$f$ is increasing because $f'$ is positive."
• Forgetting where $f'$ does not exist. Critical points also include places where the derivative is undefined, like cusps or corners. A sign change there can still create an extremum.
• Mixing up the direction of the sign change. Positive to negative is a max (the function peaks); negative to positive is a min (the function bottoms out). Sketching a quick sign chart helps you keep them straight.

Related AP Calculus Guides

• Unit 5 Overview: Analytical Applications of Differentiation
• 5.1 Using the Mean Value Theorem
• 5.2 Extreme Value Theorem, Global vs Local Extrema, and Critical Points
• 5.3 Determining Intervals on Which a Function is Increasing or Decreasing
• 5.11 Solving Optimization Problems
• 5.10 Introduction to Optimization Problems