Lesson 1: First Derivative Test

Estimated time: 30-40 minutes

Learning Objectives

Use the sign of f' to determine where f is increasing or decreasing
Apply the First Derivative Test to classify critical points
Build a sign chart for f'

Increasing and Decreasing Functions

Increasing/Decreasing Test

On an interval where f'(x) > 0, f is increasing.

On an interval where f'(x) < 0, f is decreasing.

To find intervals of increase/decrease: (1) find critical points, (2) test the sign of f' in each interval between critical points.

The First Derivative Test

First Derivative Test

Suppose c is a critical point of a continuous function f:

If f' changes from positive to negative at c, then f has a local maximum at c.
If f' changes from negative to positive at c, then f has a local minimum at c.
If f' does not change sign at c, then f has no local extremum at c.

Example 1: Classify Critical Points

Analyze f(x) = 2x³ − 9x² + 12x − 3.

Step 1: f'(x) = 6x² − 18x + 12 = 6(x − 1)(x − 2). Critical points: x = 1, x = 2.

Step 2: Sign chart for f':

Interval	(−∞, 1)	(1, 2)	(2, ∞)
Sign of f'	+	−	+
f behavior	increasing	decreasing	increasing

Conclusion: f' changes + to − at x = 1: local max f(1) = 2. f' changes − to + at x = 2: local min f(2) = 1.

Example 2: No Extremum at a Critical Point

f(x) = x³. f'(x) = 3x² = 0 at x = 0.

Sign chart: f' > 0 for x < 0, and f' > 0 for x > 0. The sign does not change.

So x = 0 is a critical point but not a local extremum (it is an inflection point).

Key Takeaways

f' > 0 means f is increasing; f' < 0 means f is decreasing.
The First Derivative Test classifies critical points by checking if f' changes sign.
A sign chart is the standard organizational tool for this analysis.
Not every critical point is an extremum: f'(c) = 0 with no sign change means no extremum.

Check Your Understanding

1. Find the intervals where f(x) = x⁴ − 4x³ is increasing.

Answer: f'(x) = 4x³ − 12x² = 4x²(x − 3). f' > 0 when x > 3 (since 4x² ≥ 0). f is increasing on (3, ∞).

2. Use the First Derivative Test on f(x) = x − 2 sin x on [0, 2π].

Answer: f'(x) = 1 − 2 cos x = 0 when cos x = 1/2, so x = π/3, 5π/3. f' < 0 on (0, π/3), f' > 0 on (π/3, 5π/3), f' < 0 on (5π/3, 2π). Local min at x = π/3, local max at x = 5π/3.

3. True or false: If f'(c) = 0 and f''(c) = 0, then c is neither a max nor a min.

Answer: False. The Second Derivative Test is inconclusive when f''(c) = 0. You must use the First Derivative Test. For example, f(x) = x⁴ has f'(0) = 0, f''(0) = 0, but x = 0 is a local min.

4. Find and classify all critical points of f(x) = x e^−x.

Answer: f'(x) = e^−x(1 − x). Critical point: x = 1. f' > 0 for x < 1, f' < 0 for x > 1. By FDT, local max at x = 1, f(1) = 1/e.

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Lesson 2 covers the Second Derivative and concavity.

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