Find the intervals of increase and decrease and the local extremum of $f(x) = x^2 - 6x + 1$.

Decreasing on $(-\infty, 3)$, increasing on $(3, \infty)$; local min $(3, -8)$.

Find and classify the critical points of $f(x) = x^3 - 12x$.

Local max $(-2, 16)$, local min $(2, -16)$.

Find the intervals of increase and decrease and the local extrema of $f(x) = x^4 - 8x^2$.

Decreasing on $(-\infty, -2)$ and $(0, 2)$, increasing on $(-2, 0)$ and $(2, \infty)$; local max $(0, 0)$, local mins $(-2, -16)$ and $(2, -16)$.

Find the intervals of increase and decrease and the local extrema of $f(x) = \frac{x^2}{x - 3}$.

Increasing on $(-\infty, 0)$ and $(6, \infty)$, decreasing on $(0, 3)$ and $(3, 6)$; local max $(0, 0)$, local min $(6, 12)$.

Find all critical numbers of $f(x) = x - 3x^{\frac{1}{3}}$ and classify each one.

Critical numbers $-1, 0, 1$: local max $(-1, 2)$, local min $(1, -2)$, and $x = 0$ is neither.

Grade 12 · MCV4U · Calculus & Vectors

MCV4U: Increasing, Decreasing and Critical Points

The sign of the first derivative tells you which way a graph is heading. Where that sign changes, the graph turns, and those turning points are the local maximums and minimums.

How It Works

Increasing and decreasing. If $f'(x) > 0$ on an interval, $f$ is increasing there. If $f'(x) < 0$ , it is decreasing.

Critical numbers. A critical number is a value $c$ in the domain of $f$ where $f'(c) = 0$ or $f'(c)$ does not exist. These are the only places the sign of $f'$ can change. A value where $f$ itself is undefined, like a vertical asymptote, is not a critical number, but it still splits the number line.

First derivative test. At a critical number:

$f'$ changes from $+$ to $-$ : local maximum.

$f'$ changes from $-$ to $+$ : local minimum.

$f'$ does not change sign: neither. The graph flattens or turns vertical for a moment and keeps going.

The method is the same every time:

1. Find $f'(x)$ and factor it as far as you can.

2. Find the critical numbers, plus any $x$ -values outside the domain.

3. Make a sign chart. Pick a test value in each interval and find only the sign of $f'$ there, not its value.

4. Read off the intervals of increase and decrease, then classify each critical number.

5. Substitute each extremum's $x$ into $f$ , not $f'$ , to get the point.

Write intervals with the critical numbers as endpoints, like "increasing on $(-\infty, -1)$ and $(2, \infty)$ ". Join separate intervals with "and", not a union with a gap filled in, because a function can be increasing on two pieces without being increasing across the gap.

A factored derivative makes the sign chart quick: each factor's sign is easy to read, and the signs multiply.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Find the intervals of increase and decrease and the local extremum of $f(x) = x^2 - 6x + 1$ .
Show solution
1. $f'(x) = 2x - 6$ , zero at $x = 3$ .
2. $f'(0) = -6 < 0$ and $f'(4) = 2 > 0$ : decreasing, then increasing.
3. Local minimum: $f(3) = 9 - 18 + 1 = -8$ .
Answer: Decreasing on $(-\infty, 3)$ , increasing on $(3, \infty)$ ; local min $(3, -8)$ .
Easy
Find and classify the critical points of $f(x) = x^3 - 12x$ .
Show solution
1. $f'(x) = 3x^2 - 12 = 3(x - 2)(x + 2)$ , zero at $x = \pm 2$ .
2. Signs of $f'$ : $+$ for $x < -2$ , $-$ on $(-2, 2)$ , $+$ for $x > 2$ .
3. $f(-2) = -8 + 24 = 16$ and $f(2) = 8 - 24 = -16$ .
Answer: Local max $(-2, 16)$ , local min $(2, -16)$ .
Medium
Find the intervals of increase and decrease and the local extrema of $f(x) = x^4 - 8x^2$ .
Show solution
1. $f'(x) = 4x^3 - 16x = 4x(x - 2)(x + 2)$ , zero at $x = -2, 0, 2$ .
2. Signs of $f'$ : $-$ for $x < -2$ , $+$ on $(-2, 0)$ , $-$ on $(0, 2)$ , $+$ for $x > 2$ .
3. $f(\pm 2) = 16 - 32 = -16$ and $f(0) = 0$ .
Answer: Decreasing on $(-\infty, -2)$ and $(0, 2)$ , increasing on $(-2, 0)$ and $(2, \infty)$ ; local max $(0, 0)$ , local mins $(-2, -16)$ and $(2, -16)$ .
Medium
Find the intervals of increase and decrease and the local extrema of $f(x) = \frac{x^2}{x - 3}$ .
Show solution
1. Quotient rule: $f'(x) = \frac{2x(x - 3) - x^2}{(x - 3)^2} = \frac{x(x - 6)}{(x - 3)^2}$ .
2. Critical numbers: $x = 0$ and $x = 6$ . At $x = 3$ the function is undefined, so $3$ is not critical, but it still goes on the sign chart.
3. The denominator is positive, so the sign follows $x(x - 6)$ : $+$ for $x < 0$ , $-$ on $(0, 3)$ and $(3, 6)$ , $+$ for $x > 6$ .
4. $f(0) = 0$ and $f(6) = \frac{36}{3} = 12$ .
Answer: Increasing on $(-\infty, 0)$ and $(6, \infty)$ , decreasing on $(0, 3)$ and $(3, 6)$ ; local max $(0, 0)$ , local min $(6, 12)$ .
Test level
Find all critical numbers of $f(x) = x - 3x^{\frac{1}{3}}$ and classify each one.
Show solution
1. $f'(x) = 1 - x^{-\frac{2}{3}} = \frac{x^{\frac{2}{3}} - 1}{x^{\frac{2}{3}}}$ .
2. $f'(x) = 0$ when $x^{\frac{2}{3}} = 1$ , so $x = \pm 1$ . $f'$ is undefined at $x = 0$ , which is in the domain, so $0$ is critical too.
3. The denominator $x^{\frac{2}{3}}$ is positive for $x \ne 0$ . Signs of $f'$ : $+$ for $x < -1$ , $-$ on $(-1, 0)$ and $(0, 1)$ , $+$ for $x > 1$ .
4. $f(-1) = -1 + 3 = 2$ and $f(1) = 1 - 3 = -2$ . At $x = 0$ the sign does not change, so it is neither: the tangent is vertical there.
Answer: Critical numbers $-1, 0, 1$ : local max $(-1, 2)$ , local min $(1, -2)$ , and $x = 0$ is neither.

Learn It on Martian Lab

Martian Lab teaches this in the Maximums and Minimums lesson of its MCV4U Curve Sketching unit. Watch it, then take the Curve Sketching practice tests to see if it stuck.

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MCV4U: Increasing, Decreasing and Critical Points

How It Works

5 Worked Examples

Learn It on Martian Lab

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