Sign of f' and monotonicity of f: the key link

The theorem

Let $f$ be differentiable on an interval $I$.

• $f'(x) > 0$ on $I$ ⇒ $f$ strictly increasing on $I$
• $f'(x) < 0$ on $I$ ⇒ $f$ strictly decreasing on $I$
• $f'(x) = 0$ on $I$ ⇒ $f$ constant on $I$

Method: build the variation table

1. Compute $f'(x)$.
2. Solve $f'(x) = 0$ to find where $f$ changes behavior.
3. Study the sign of $f'$ (often: quadratic, quotient, product of signs).
4. Fill the table: three rows ($x$, $f'(x)$, $f(x)$).
5. Compute values at notable points (extrema, boundaries).

Example: $f(x) = x^3 - 3x + 1$

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

$f'(x) = 0 \Leftrightarrow x = -1$ or $x = 1$.

Sign: $f' > 0$ outside $[-1, 1]$, $f' < 0$ between.

Table:

• Increasing on $(-\infty, -1]$
• Decreasing on $[-1, 1]$
• Increasing on $[1, +\infty)$

Extrema: $f(-1) = 3$ (local max), $f(1) = -1$ (local min).

Spotting an extremum

If $f'$ changes sign at $x_0$:

• $f' : + \to -$ ⇒ local maximum at $x_0$.
• $f' : - \to +$ ⇒ local minimum at $x_0$.

If $f'$ does not change sign at $x_0$ (e.g., $f'(x_0) = 0$ but stays positive around): no extremum.

Watch the hypotheses

The theorem requires $f$ differentiable on an interval. It does not apply on $\mathbb{R}^*$ (two separate intervals).

Counter-example: $f(x) = 1/x$ has $f'(x) = -1/x^2 < 0$ on each interval, but $f$ is not decreasing on $\mathbb{R}^*$ (break at $0$).

Common mistakes

1. Concluding $f$ increasing without studying the whole interval.
2. Forgetting to specify “on which interval”.
3. Claiming an extremum without checking the sign change of $f'$.
4. Confusing critical point ($f'(x_0) = 0$) with extremum (the latter needs a sign change).

?À toi de jouer

f' changes from − to + at x_0. What is x_0?

A.Local minimumB.Local maximumC.Inflection pointD.Nothing special

Valider