Increasing, Decreasing, and Critical Points

By the end of this lesson, you will be able to use $f'(x)$ to find intervals where $f$ is increasing or decreasing, locate critical points, apply the First Derivative Test to classify local extrema.

The First Derivative and Monotonicity

$f$ is increasing on an interval if $f'(x) > 0$ on that interval.

$f$ is decreasing on an interval if $f'(x) < 0$ on that interval.

This follows from the MVT: if the derivative is positive on $(a,b)$, the function must increase from $a$ to $b$.

Critical Points

$c$ is a critical point of $f$ if $f'(c) = 0$ or $f'(c)$ is undefined (and $f(c)$ exists).

Why critical points matter: the only possible locations for local extrema are critical points.

Fermat's Theorem: if $f$ has a local max or min at $c$ and $f$ is differentiable at $c$, then $f'(c) = 0$.

(But $f'(c) = 0$ doesn't guarantee an extremum — see inflection points.)

First Derivative Test

At a critical point $c$:

• $f'$ changes from $+$ to $-$: local maximum
• $f'$ changes from $-$ to $+$: local minimum
• $f'$ doesn't change sign: neither (inflection point of $f'$)

Example 1 — Complete Analysis

$f(x) = 2x^3 - 9x^2 + 12x - 4$

$f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x-1)(x-2)$

Critical points: $x = 1$ and $x = 2$.

Sign chart of $f'$:

Interval	Test $x$	$f'$	$f$
$(-\infty, 1)$	$0$: $f' = 12 > 0$	$+$	Increasing
$(1, 2)$	$1.5$: $f' = 6(.5)(-.5) = -1.5 < 0$	$-$	Decreasing
$(2, \infty)$	$3$: $f' = 6(2)(1) = 12 > 0$	$+$	Increasing

Local max at $x = 1$: $f(1) = 2 - 9 + 12 - 4 = 1$

Local min at $x = 2$: $f(2) = 16 - 36 + 24 - 4 = 0$