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)