AP Calculus AB Unit 5 Notes: Using Derivatives to Understand Function Behavior

Determining Intervals of Increase and Decrease

What “increasing” and “decreasing” really mean

A function is increasing on an interval if, as you move left to right, the outputs generally go up: larger input values give larger function values. Formally, for any a < b in the interval, you have f(a) < f(b). A function is **decreasing** on an interval if outputs go down as you move left to right: for any a < b, f(a) > f(b).

The derivative connects directly to this idea because the first derivative measures instantaneous rate of change (slope of the tangent line). If the slope is positive throughout an interval, the function is rising there; if the slope is negative, the function is falling.

Why derivatives control increasing/decreasing

Think of driving along a hilly road where your position is f(t) and time is t. Your velocity is the derivative f'(t). If your velocity is positive for a stretch of time, your position increases during that time. If your velocity is negative, your position decreases. The same logic works for any differentiable function: the sign of f'(x) tells you whether f(x) is going up or down.

The key rule (and what it does not say)

  • If f'(x) > 0 on an interval, then f is increasing on that interval.
  • If f'(x) < 0 on an interval, then f is decreasing on that interval.

A common misunderstanding is thinking “f'(x) = 0 means a maximum or minimum.” Not necessarily. f'(x)=0 means the tangent slope is horizontal at that point; the function could flatten and continue increasing, flatten and continue decreasing, or change direction.

How to find intervals of increase/decrease (process)

To determine where a function increases or decreases, you typically:

  1. Find where the derivative exists and compute it. Work with f'(x).
  2. Identify critical numbers. A critical number is an x-value in the domain of f where either:
    • f'(x) = 0, or
    • f'(x) does not exist.
  3. Use critical numbers to split the number line into test intervals. The sign of f'(x) can only change at critical numbers (or discontinuities of f'), so these points partition the domain.
  4. Test the sign of f'(x) on each interval (choose a sample point in the interval and evaluate f'(x), or do a sign analysis algebraically).
  5. State the intervals where f'(x) > 0 (increasing) and where f'(x) < 0 (decreasing).

When the question is on a closed interval [a,b], also pay attention to endpoints. Endpoints don’t have two-sided neighborhoods inside the interval, but they still matter for absolute extrema questions.

Worked example 1: Using algebra and a sign chart

Find the intervals where f(x) = x^3 - 3x^2 - 9x + 1 is increasing and decreasing.

Step 1: Differentiate.
f'(x) = 3x^2 - 6x - 9

Step 2: Find critical numbers. Solve f'(x)=0.
3x^2 - 6x - 9 = 0
Divide by 3:
x^2 - 2x - 3 = 0
Factor:
(x-3)(x+1)=0
So the critical numbers are x=-1 and x=3.

Step 3: Test intervals. The critical numbers split the real line into (-\infty,-1), (-1,3), and (3,\infty).

  • Pick x=-2:
    f'(-2)=3(4)-6(-2)-9=12+12-9=15>0
    So f is increasing on (-\infty,-1).

  • Pick x=0:
    f'(0)=-9

  • Pick x=4:
    f'(4)=3(16)-24-9=48-33=15>0
    So f is increasing on (3,\infty).

Conclusion: Increasing on (-\infty,-1) and (3,\infty); decreasing on (-1,3).

Worked example 2: Increasing/decreasing from a derivative graph (conceptual)

On the AP exam, you’re often given a graph of f'(x) rather than a formula for f(x).

  • Where the graph of f'(x) is **above** the x-axis, f'(x)>0, so f is increasing.
  • Where the graph of f'(x) is **below** the x-axis, f'(x)

This is powerful because it lets you describe the behavior of f without ever finding a formula for f.

What can go wrong (common pitfalls)

  • Forgetting domain restrictions. If f isn’t defined somewhere, you can’t claim increasing/decreasing “through” that point.
  • Mixing up f and f'. A positive value of f(x) does not imply increasing; only the sign of f'(x) controls increase/decrease.
  • Not using interval notation correctly. Increasing/decreasing is stated on intervals, not “at points.” You don’t say “increasing at x=2”; you say “increasing on (1,3),” etc.
Exam Focus
  • Typical question patterns:
    • “Find the intervals on which f is increasing/decreasing” given f(x), f'(x), or a graph/table of f'(x).
    • “For what values of x is the function increasing?” often expects sign analysis with critical points.
    • Interpreting a motion context: position increasing when velocity f'(t) is positive.
  • Common mistakes:
    • Solving f(x)=0 instead of f'(x)=0.
    • Checking signs at critical points instead of in the open intervals between them.
    • Declaring increasing on [a,b] without noting derivative sign changes inside.

First Derivative Test

What the First Derivative Test says

The First Derivative Test is a method for classifying local extrema (local maxima and local minima) using how f'(x) changes sign around a critical number.

A local maximum at x=c means f(c) is higher than nearby values of the function. A **local minimum** means f(c) is lower than nearby values.

The First Derivative Test focuses on the idea that a function changes direction at a local extremum:

  • To get a local maximum, you typically go from increasing to decreasing.
  • To get a local minimum, you typically go from decreasing to increasing.

Why it matters

This test is especially valuable when:

  • The derivative is easy to analyze by sign, but solving other conditions is harder.
  • f'(c)=0 but you’re not sure whether the point is a max, min, or neither.
  • You have f'(x) as a graph or table and need to describe behavior of f.

Conceptually, it reinforces the idea that derivatives describe behavior (increasing/decreasing), and extrema happen where that behavior changes.

The test (sign-change logic)

Let c be a critical number of f.

  • If f'(x) changes from positive to negative at c, then f has a **local maximum** at x=c.
  • If f'(x) changes from negative to positive at c, then f has a **local minimum** at x=c.
  • If f'(x) does not change sign at c (positive on both sides or negative on both sides), then f has **no local extremum** at c.

Notice how this fixes a major misconception: f'(c)=0 alone is not enough; the sign change is what creates a max or min.

How to apply it (step-by-step)

  1. Find critical numbers (solve f'(x)=0 and where f'(x) is undefined, with f still defined).
  2. Create a sign chart for f'(x) on intervals around each critical number.
  3. Use the sign changes to classify each critical number.

Worked example 1: A critical point that is NOT an extremum

Consider f(x)=x^3.

Differentiate:
f'(x)=3x^2
Critical number: 3x^2=0 gives x=0.

Now check signs:

  • For x
  • For x>0, 3x^2>0.

So f'(x) is positive on both sides of 0. That means f is increasing both before and after x=0, so there is **no** local maximum or minimum at 0. Instead, x=0 is a point where the tangent is horizontal but the function keeps increasing (this is a classic “flattening” behavior).

Worked example 2: Classifying extrema from earlier derivative information

Using the derivative sign results from the earlier polynomial example:

  • f'(x)>0 on (-\infty,-1)
  • f'(x)

Apply the First Derivative Test:

  • At x=-1, f'(x) goes from positive to negative, so f has a **local maximum** at x=-1.
  • At x=3, f'(x) goes from negative to positive, so f has a **local minimum** at x=3.

If the problem asks for the points (coordinates), you must also compute f(-1) and f(3).

Graph interpretation insight

If you graph f, a local maximum looks like a hilltop and a local minimum looks like a valley. The derivative f'(x) tracks the slope:

  • Before a hilltop, slopes are positive; after, slopes are negative.
  • Before a valley, slopes are negative; after, slopes are positive.

So the First Derivative Test is essentially “read the slope behavior to detect turning points.”

What can go wrong

  • Confusing local vs absolute extrema. The First Derivative Test finds local behavior. Absolute (global) maxima/minima on a closed interval require checking endpoints as well.
  • Ignoring points where f'(x) is undefined. Cusps/vertical tangents can still create local extrema. For example, f(x)=|x| has a local minimum at 0 even though f'(0) does not exist.
  • Using only one side. You need behavior on both sides of c to claim a sign change.
Exam Focus
  • Typical question patterns:
    • “Use the First Derivative Test to find and classify local extrema of f.”
    • “Given the graph of f', identify where f has relative maxima/minima.”
    • Free-response justification: “Explain why f has a local max at x=c.”
  • Common mistakes:
    • Claiming “local max because f'(c)=0” without showing a sign change.
    • Forgetting to verify that c is in the domain of f.
    • Mixing up which sign change corresponds to max vs min.

Determining Concavity and Points of Inflection

What concavity means (shape, not height)

Concavity describes how the slope of a function changes.

  • A function is concave up on an interval if its graph bends like a cup: as you move left to right, the slope tends to increase.
  • A function is concave down on an interval if its graph bends like a cap: as you move left to right, the slope tends to decrease.

Concavity is not about whether the function is above or below the x-axis. It’s about whether tangent lines are getting steeper (concave up) or less steep (concave down).

Why the second derivative is the concavity detector

The second derivative measures how the first derivative changes. If f'(x) represents slope, then f''(x) represents “slope of the slope.”

  • If f''(x) > 0 on an interval, then f'(x) is increasing there, so f is concave up.
  • If f''(x) < 0 on an interval, then f'(x) is decreasing there, so f is concave down.

A helpful physics analogy: if f(t) is position, then f'(t) is velocity and f''(t) is acceleration.

  • Positive acceleration means velocity is increasing, which corresponds to concave up position.
  • Negative acceleration means velocity is decreasing, which corresponds to concave down position.

Points of inflection: where concavity changes

A point of inflection is a point on the graph of f where the concavity changes (from up to down or down to up).

Important: a point where f''(x)=0 is only a _candidate_ for an inflection point. To confirm an inflection point, you must show **a change in concavity**, meaning f''(x) must change sign (or concavity must change based on another valid analysis).

How to find intervals of concavity and inflection points

  1. Compute f''(x).
  2. Find possible “transition” points where f''(x)=0 or f''(x) does not exist.
  3. Use these points to split the domain into intervals.
  4. Test the sign of f''(x) on each interval.
  5. State concavity on each interval and identify inflection points where concavity changes.

Notation reference (common on AP)

MeaningNotation you might seeWhat it represents
First derivativef'(x), \frac{dy}{dx}slope / instantaneous rate of change
Second derivativef''(x), \frac{d^2y}{dx^2}how slope changes / concavity

Worked example 1: Concavity and inflection point candidates

Let f(x)=x^4 - 4x^3.

Step 1: Differentiate twice.
f'(x)=4x^3 - 12x^2
f''(x)=12x^2 - 24x
Factor:
f''(x)=12x(x-2)

Step 2: Find where f''(x)=0.
12x(x-2)=0 gives x=0 and x=2.

Step 3: Sign test for f''(x). Test intervals (-\infty,0), (0,2), and (2,\infty).

  • If x=-1:
    f''(-1)=12(-1)(-3)=36>0 so concave up on (-\infty,0).
  • If x=1:
    f''(1)=12(1)(-1)=-12

Step 4: Inflection points. Concavity changes at both x=0 (up to down) and x=2 (down to up), so both correspond to points of inflection. If the question asks for coordinates, compute f(0) and f(2):
f(0)=0
f(2)=16-32=-16
So the inflection points are (0,0) and (2,-16) .

Worked example 2: “f''(c)=0” without an inflection point

Consider f(x)=x^4.
f'(x)=4x^3
f''(x)=12x^2

Here f''(0)=0, but 12x^2 \ge 0 for all x and is positive on both sides of 0, so the function is concave up everywhere. There is no change in concavity, so no inflection point at 0.

This is a frequent AP trap: “second derivative equals zero” is a candidate condition, not a conclusion.

How concavity connects to tangent lines (a deeper interpretation)

Concavity also predicts where tangent lines sit relative to the graph:

  • If f is concave up, tangent lines tend to lie below the graph.
  • If f is concave down, tangent lines tend to lie above the graph.

This idea is useful for reasoning and for checking whether your concavity result makes sense visually.

What can go wrong

  • Treating f''(c)=0 as automatically an inflection point. Always confirm a sign change in f'' (or a verified concavity change).
  • Finding inflection points where the function isn’t defined. An inflection point must be a point on the graph of f, so f(c) must exist.
  • Confusing concavity with increasing/decreasing. A function can be increasing and concave down (rising but flattening), or decreasing and concave up (falling but flattening).
Exam Focus
  • Typical question patterns:
    • “Find intervals where f is concave up/down and find points of inflection.”
    • Given a graph of f': concave up where f' is increasing, concave down where f' is decreasing.
    • Motion interpretation: concavity of position from the sign of acceleration f''(t).
  • Common mistakes:
    • Listing x-values where f''(x)=0 as inflection points without checking sign change.
    • Mixing up “concave up” with “increasing.”
    • Forgetting to give final answers in interval notation.

Second Derivative Test

What the Second Derivative Test does

The Second Derivative Test is another method for classifying local extrema at a critical number c. Instead of tracking a sign change in f'(x) (like the First Derivative Test), it uses concavity information from f''(x) at that point.

The intuition is geometric:

  • If the graph is concave up at a point where the tangent slope is zero, you’re sitting at the bottom of a “cup,” which suggests a local minimum.
  • If the graph is concave down at a point where the tangent slope is zero, you’re at the top of a “cap,” which suggests a local maximum.

The test (conditions and conclusions)

Assume f'(c)=0 and f''(c) exists.

  • If f''(c) > 0, then f has a **local minimum** at x=c.
  • If f''(c) < 0, then f has a **local maximum** at x=c.
  • If f''(c) = 0, the test is inconclusive (you cannot decide from this test alone).

The “inconclusive” case is important: the Second Derivative Test is fast when it works, but it doesn’t always work.

Why it matters (and when to choose it)

The Second Derivative Test is usually efficient when:

  • You can compute f''(c) easily.
  • You already have critical points from f'(x)=0.

However, if f''(c)=0 or does not exist, you often switch back to the First Derivative Test (sign analysis) because it can still classify the point.

How to apply it (step-by-step)

  1. Compute f'(x) and solve f'(x)=0 to find critical numbers (also check where f' is undefined if relevant).
  2. For each critical number c, compute f''(c) if it exists.
  3. Use the sign of f''(c) to classify the point, or declare the test inconclusive.

Worked example 1: A case where the test works cleanly

Let f(x)=x^3 - 3x.

Step 1: Find critical points.
f'(x)=3x^2 - 3
Solve:
3x^2-3=0
x^2=1
So x=-1 and x=1 are critical numbers.

Step 2: Second derivative.
f''(x)=6x
Evaluate:
f''(-1)=-6

If coordinates are needed:
f(-1)=-1+3=2
f(1)=1-3=-2
So local max at (-1,2) and local min at (1,-2).

Worked example 2: Inconclusive does not mean “no extremum”

Let f(x)=x^4.

Critical point:
f'(x)=4x^3
So f'(0)=0.

Second derivative:
f''(x)=12x^2
So f''(0)=0, which makes the Second Derivative Test inconclusive.

But you can still determine what happens using other reasoning:

  • Since x^4 \ge 0 and equals 0 only at x=0, there is a local (and absolute) minimum at 0.
  • Or use the First Derivative Test: f'(x)=4x^3 is negative for x

This is a classic example showing why “inconclusive” means “use another method,” not “nothing happens.”

How the First and Second Derivative Tests connect

These tests are two lenses on the same behavior:

  • The First Derivative Test looks at direction change (increasing to decreasing or vice versa).
  • The Second Derivative Test looks at shape at the critical point (concave up or concave down).

In practice:

  • Use the Second Derivative Test when f''(c) is easy and nonzero.
  • Use the First Derivative Test when you have a sign chart anyway (or when f''(c)=0).

What can go wrong

  • Applying it when f'(c) \ne 0. The Second Derivative Test is specifically for critical points where f'(c)=0.
  • Forgetting the test requires f''(c) to exist. If f''(c) does not exist, you cannot use this test.
  • Misreading concavity as increase/decrease. f''(c)>0 indicates a min only when f'(c)=0; otherwise it just indicates concave up.
Exam Focus
  • Typical question patterns:
    • “Use the Second Derivative Test to classify the critical points of f.”
    • “Find relative extrema and justify using derivatives” (you choose the appropriate test).
    • Mixed prompts where you must explain why a test is inconclusive and then use another method.
  • Common mistakes:
    • Concluding “no extremum” when f''(c)=0.
    • Plugging endpoints of an interval into the Second Derivative Test (endpoints are handled differently for absolute extrema).
    • Not showing derivative work/justification when the question asks for reasoning.