Read it again any time, including while you work through the exercises.
Imagine walking along a curve, from left to right, like on a mountain trail. At every moment, the slope under your feet tells you whether you are climbing or descending. That slope is exactly the derivative. This is the central idea of this lesson: the sign of f′ dictates the direction of variation of f.
The theorem is simply stated. If f′(x) > 0 on an interval, then f is strictly increasing on that interval: the slope is positive, you are climbing. If f′(x) < 0, then f is strictly decreasing: you are going down. And if f′ vanishes while changing sign, the function switches from climbing to descending (or the other way): that is a peak or a dip, in other words a local extremum.
The study method is always the same, in three steps: compute f′(x), study the sign of f′(x), then build the variation table. The table gathers everything: the intervals where f increases or decreases, and the values of f at the points where f′ changes sign.
One final word of caution: f′(a) = 0 is not enough to guarantee an extremum at a. The sign of f′ must change. The function x³ gives a famous example: its derivative 3x² vanishes at 0, yet stays positive on both sides, and the function keeps climbing. No peak, just a momentary flat step.
Study the variations of f(x) = x² - 4x + 1 on ℝ.
- 1
f′(x) = 2x - 4
I differentiate term by term: x² gives 2x, -4x gives -4, and the constant 1 vanishes.
- 2
2x - 4 = 0 ⟺ x = 2
I study the sign of f′: it vanishes at x = 2, is negative before, positive after.
- 3
f(2) = 4 - 8 + 1 = -3
I compute the value of f where the sign changes: this is the minimum.
- 4
↘ sur ]-∞ ; 2], ↗ sur [2 ; +∞[
I conclude with the table: f decreases until x = 2, reaches its minimum -3, then increases.
Result: f is decreasing on (-∞, 2] and increasing on [2, +∞), with a minimum of -3 reached at x = 2.
Study the variations of f(x) = x³ - 3x on ℝ.
- 1
f′(x) = 3x² - 3 = 3(x² - 1)
I differentiate then factor to read the sign more easily: 3(x - 1)(x + 1).
- 2
f′(x) = 0 ⟺ x = -1 ou x = 1
The derivative vanishes at -1 and 1. It is positive outside the roots, negative between them.
- 3
f(-1) = -1 + 3 = 2
At x = -1, f′ switches from positive to negative: a local maximum, with value 2.
- 4
f(1) = 1 - 3 = -2
At x = 1, f′ switches from negative to positive: a local minimum, with value -2. The table reads: climb, descent, climb.
Result: f increases on (-∞, -1], decreases on [-1, 1], then increases on [1, +∞). Local maximum 2 at x = -1, local minimum -2 at x = 1.