Derivative-Based Curve Sketching for AP Calculus BC (Unit 5)

Intervals of Increase/Decrease and the First Derivative Test

What “increasing” and “decreasing” really mean

A function is increasing on an interval if, as you move left to right along the $x$-axis within that interval, the function’s output values tend to go up. Formally, for any two inputs $a$ and $b$ in the interval with $a < b$, you have $f(a) < f(b)$. A function is **decreasing** on an interval if $a < b$ implies $f(a) > f(b)$.

This matters because “increasing/decreasing” is the backbone of describing a graph’s overall shape. On AP Calculus, you’re often asked to analyze a function without graphing technology (or to justify what your calculator shows). The derivative gives a precise, efficient way to do that.

Why the first derivative controls increase/decrease

The first derivative $f'(x)$ measures the instantaneous rate of change (slope of the tangent line) of $f$ at $x$. If slopes are positive on an interval, tangent lines tilt upward, and the function increases there. If slopes are negative, tangent lines tilt downward, and the function decreases.

The key relationship is:

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

You can connect this to the Mean Value Theorem idea: if a function is differentiable and its derivative stays positive, the “average slope” between any two points is positive, so outputs must rise as inputs increase.

Critical points: where behavior can change

Intervals of increase/decrease can only “switch” at certain important $x$-values. These are called **critical numbers** (or **critical points** when paired with the function value). A **critical number** is a value $c$ in the domain of $f$ such that:

• $f'(c) = 0$, or
• $f'(c)$ does not exist (but $f(c)$ exists).

Why these? If $f'(x)$ is positive on one side of $c$ and negative on the other, the function changes from increasing to decreasing (or vice versa). Those sign changes can only happen if the derivative hits zero or becomes undefined.

A common misconception: students sometimes treat “$f'(c) = 0$” as automatically meaning “a max or min.” That’s not true. It only means the tangent is horizontal (or the derivative is undefined). You still need to check what happens to $f'(x)$ around $c$.

How to find intervals of increase/decrease (the sign chart method)

When you’re given a formula for $f(x)$, the standard process is:

1. Compute $f'(x)$.
2. Find critical numbers by solving $f'(x) = 0$ and finding where $f'(x)$ is undefined (while $f$ is defined).
3. Break the number line into test intervals using those critical numbers (and also any domain restrictions for $f$).
4. Test the sign of $f'(x)$ in each interval (pick a representative test point).
5. Conclude where $f'(x) > 0$ (increasing) and where $f'(x) < 0$ (decreasing).

This works because the derivative’s sign cannot change within an interval unless it becomes zero or undefined.

The First Derivative Test (local maxima and minima)

A local maximum at $x=c$ means $f(c)$ is larger than nearby values of $f(x)$ (within some neighborhood around $c$). A **local minimum** means $f(c)$ is smaller than nearby values.

The First Derivative Test classifies a critical number $c$ by checking the sign of $f'(x)$ on either side:

• If $f'(x)$ changes from positive to negative at $c$, then $f$ has a **local maximum** at $c$.
• If $f'(x)$ changes from negative to positive at $c$, then $f$ has a **local minimum** at $c$.
• If $f'(x)$ does not change sign at $c$ (positive on both sides or negative on both sides), then $c$ is not a local extremum (often a “flat” point like a horizontal inflection, or it may still be neither).

Why this is powerful: it doesn’t require you to compare function values directly, and it works even when the derivative is undefined at the critical point (as long as you can determine the sign on each side).

Notation reference (you should recognize all of these)

Derivatives are written multiple ways on AP problems:

Worked Example 1: intervals of increase/decrease + local extrema

Let

$f(x) = x^3 - 3x^2 - 9x + 1$

Step 1: Differentiate.

$f'(x) = 3x^2 - 6x - 9$

Factor:

$f'(x) = 3(x^2 - 2x - 3)$

$f'(x) = 3(x-3)(x+1)$

Step 2: Critical numbers.
Solve $f'(x)=0$:

$3(x-3)(x+1)=0$

So $x=3$ and $x=-1$ are critical numbers.

Step 3: Sign chart for $f'(x)$.
Test intervals: $(-\infty,-1)$, $(-1,3)$, $(3,\infty)$.

• Pick $x=-2$:

$f'(-2)=3(-5)(-1)>0$

So increasing on $(-\infty,-1)$.

• Pick $x=0$:

$f'(0)=3(-3)(1)<0$

So decreasing on $(-1,3)$.

• Pick $x=4$:

$f'(4)=3(1)(5)>0$

So increasing on $(3,\infty)$.

Step 4: First Derivative Test conclusions.

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

If you also want the coordinate points (often required), compute:

$f(-1)=(-1)^3-3(-1)^2-9(-1)+1=-1-3+9+1=6$

$f(3)=27-27-27+1=-26$

So local max at $(-1,6)$ and local min at $(3,-26)$.

Worked Example 2: a critical number where the derivative does not exist

Consider

$f(x)=\sqrt[3]{x}$

Differentiate:

$f'(x)=\frac{1}{3}x^{-2/3}$

At $x=0$, $f(0)=0$ exists, but $f'(0)$ does not exist (division by zero behavior), so $x=0$ is a critical number.

Now check signs:

• For $x<0$, $x^{-2/3}$ is positive (because the exponent makes it effectively $1/(\sqrt[3]{x^2})$), so $f'(x)>0$.
• For $x>0$, similarly $f'(x)>0$.

So $f$ is increasing on $(-\infty,0)$ and $(0,\infty)$, with **no** local max/min at $x=0$ (the function keeps increasing through it). The point $x=0$ is still important because the graph has a vertical tangent there.

This example highlights a common error: “critical number” does not mean “turning point.” It means “possible place where the behavior can change,” and you must test.

Real-world interpretation: velocity and position

If $s(t)$ is position, then $s'(t)=v(t)$ is velocity. When $v(t)>0$, position increases (moving forward); when $v(t)<0$, position decreases (moving backward). A time when $v(t)=0$ might be a “turnaround,” but not always—you could stop and continue forward. The First Derivative Test is exactly that “turnaround check.”

Exam Focus

• Typical question patterns:
  • “Find the intervals on which $f$ is increasing/decreasing and classify relative extrema.”
  • “Given a graph/table of $f'(x)$, determine where $f$ increases/decreases and where it has local maxima/minima.”
  • “A particle’s velocity is $v(t)$; determine when the particle changes direction and when position is increasing/decreasing.”
• Common mistakes:
  • Solving $f'(x)=0$ and declaring max/min without checking sign changes in $f'(x)$.
  • Forgetting critical numbers where $f'(x)$ is undefined (but $f$ exists), or forgetting domain restrictions that split intervals.
  • Confusing “increasing” with “concave up” (increase/decrease comes from $f'(x)$, not $f''(x)$).

Concavity, Inflection Points, and the Second Derivative Test

What concavity describes (and why it’s different from increasing)

Concavity describes how the slope of the function is changing. Even if a function is increasing, it can increase in two fundamentally different ways:

• Slopes getting larger (the graph bends upward).
• Slopes getting smaller (the graph bends downward).

A function is concave up on an interval if its graph bends like a cup: as $x$ increases, the tangent slopes tend to increase. A function is **concave down** if it bends like a cap: as $x$ increases, tangent slopes tend to decrease.

This matters because concavity helps you sketch accurate graphs, understand motion (acceleration), and decide whether a critical point is a max or min using the second derivative.

Why the second derivative controls concavity

The second derivative $f''(x)$ measures the rate of change of the first derivative. In other words, it tells you how the slope is changing.

• 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 analogy: if $f'(x)$ is your “current slope,” then $f''(x)$ tells you whether you’re turning the slope knob up or down.

Inflection points: where concavity changes

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

A key subtlety (and frequent exam trap):

• $f''(c)=0$ (or undefined) is necessary to check, but it is not sufficient to conclude there’s an inflection point.
• You must verify that the concavity actually changes sign around $c$.

So the practical method is similar to first-derivative sign charts:

1. Compute $f''(x)$.
2. Find candidates where $f''(x)=0$ or $f''(x)$ does not exist.
3. Test the sign of $f''(x)$ on intervals between candidates.
4. If the sign changes, you have an inflection point (provided $f$ is defined at that $x$).

The Second Derivative Test (classifying critical points)

The Second Derivative Test is a shortcut for classifying certain critical points.

Suppose $f'(c)=0$ and $f''(c)$ exists:

• If $f''(c) > 0$, then $f$ has a **local minimum** at $x=c$ (concave up, so the point is at the bottom of a “cup”).
• If $f''(c) < 0$, then $f$ has a **local maximum** at $x=c$ (concave down, so the point is at the top of a “cap”).
• If $f''(c)=0$, the test is inconclusive. You must use another method (often the First Derivative Test).

Why it works: if the function is concave up near a critical point with a horizontal tangent, the graph locally curves upward from that flat point, forcing it to be a minimum; concave down forces a maximum.

A common misconception: thinking $f''(c)=0$ implies inflection or implies neither max nor min. In reality, $f''(c)=0$ can happen at an inflection point, a flat point with no inflection, or even at a local extremum (the test just can’t decide).

Worked Example 1: concavity and inflection points

Let

$g(x)=x^4-4x^3$

Step 1: Differentiate twice.

$g'(x)=4x^3-12x^2$

$g''(x)=12x^2-24x$

Factor:

$g''(x)=12x(x-2)$

Step 2: Inflection candidates.
Solve $g''(x)=0$:

$12x(x-2)=0$

So candidates are $x=0$ and $x=2$.

Step 3: Sign of $g''(x)$.
Test intervals: $(-\infty,0)$, $(0,2)$, $(2,\infty)$.

• Pick $x=-1$:

$g''(-1)=12(-1)(-3)>0$

Concave up on $(-\infty,0)$.

• Pick $x=1$:

$g''(1)=12(1)(-1)<0$

Concave down on $(0,2)$.

• Pick $x=3$:

$g''(3)=12(3)(1)>0$

Concave up on $(2,\infty)$.

Step 4: Inflection points.
Concavity changes at both $x=0$ and $x=2$, so both are inflection points. To give coordinates, compute:

$g(0)=0$

$g(2)=16-32=-16$

Inflection points: $(0,0)$ and $(2,-16)$.

This example reinforces the correct logic: you didn’t declare inflection points just because $g''(x)=0$; you verified sign changes.

Worked Example 2: Second Derivative Test for extrema

Let

$f(x)=x^3-3x$

Step 1: Find critical numbers using $f'(x)$.

$f'(x)=3x^2-3$

Set equal to zero:

$3x^2-3=0$

$x^2=1$

So critical numbers: $x=-1$ and $x=1$.

Step 2: Compute $f''(x)$ and evaluate at critical numbers.

$f''(x)=6x$

Evaluate:

$f''(-1)=-6<0$

So $x=-1$ is a local maximum.

$f''(1)=6>0$

So $x=1$ is a local minimum.

If needed, compute the actual function values:

$f(-1)=-1+3=2$

$f(1)=1-3=-2$

Local max at $(-1,2)$ and local min at $(1,-2)$.

A classic “inconclusive” situation (what to do next)

Consider

$h(x)=x^4$

Then

$h'(x)=4x^3$

So $h'(0)=0$, meaning $x=0$ is critical. Also,

$h''(x)=12x^2$

So $h''(0)=0$, making the Second Derivative Test inconclusive. But $x=0$ is actually a local minimum.

How do you tell? Use the First Derivative Test: $h'(x)=4x^3$ is negative for $x<0$ and positive for $x>0$, so the function decreases then increases, giving a local minimum at $x=0$.

This illustrates an AP skill: when one test fails, you switch to a more reliable tool.

Real-world interpretation: acceleration and “speeding up vs slowing down”

In motion problems, if $s(t)$ is position, then $v(t)=s'(t)$ is velocity and $a(t)=s''(t)$ is acceleration.

• Concavity of $s(t)$ is controlled by $s''(t)$.
• If $s''(t)>0$, velocity is increasing (acceleration positive).
• If $s''(t)<0$, velocity is decreasing (acceleration negative).

One common AP-style reasoning task: determine when a particle is speeding up. That depends on whether velocity and acceleration have the same sign (both positive or both negative), which is conceptually tied to how $s'(t)$ changes.

Connecting first- and second-derivative information (how AP expects you to think)

These ideas are meant to work together:

• $f'(x)$ tells you direction (up or down).
• $f''(x)$ tells you bending (slopes increasing or decreasing).

So you can have:

• Increasing and concave up: rising and getting steeper.
• Increasing and concave down: rising but flattening.
• Decreasing and concave up: falling but flattening.
• Decreasing and concave down: falling and getting steeper downward.

AP questions often give partial information (like a table of derivative signs) and ask you to infer the shape.

Exam Focus

• Typical question patterns:
  • “Find intervals where $f$ is concave up/down and find the inflection points.”
  • “Use the Second Derivative Test to classify critical points of $f$.”
  • “Given a graph/table of $f'(x)$, determine where $f$ is concave up/down (because concavity depends on whether $f'(x)$ is increasing or decreasing).”
• Common mistakes:
  • Claiming an inflection point whenever $f''(x)=0$ without checking for a sign change in $f''(x)$.
  • Using the Second Derivative Test without first confirming $f'(c)=0$ (the test is for critical points).
  • Mixing up what each derivative controls: $f'(x)$ is about increase/decrease, $f''(x)$ is about concavity (slope behavior), and they answer different questions.