One function decides direction, and it is not f AB & BC

On an interval where $f$ is continuous and differentiable:

1. $f' > 0$ throughout $\;\Rightarrow\; f$ is increasing there.
2. $f' < 0$ throughout $\;\Rightarrow\; f$ is decreasing there.

This is a theorem, not a definition, and 5.1 is what proves it: apply the Mean Value Theorem to any two points $x_{1} < x_{2}$ in the interval and $f(x_{2}) - f(x_{1}) = f'(c)(x_{2} - x_{1})$ for some $c$ between them. The right side inherits the sign of $f'$, so the left side does too.

The substitution that ruins it is reading the sign of $f$ instead of the sign of $f'$. "The values are positive here, so the function is going up" is a sentence about two unrelated properties. Consider $f(x) = x^{2} - 4$, which manages all four combinations on one graph:

1. On $(-\infty, -2)$: $f > 0$ and decreasing.
2. On $(-2, 0)$: $f < 0$ and decreasing.
3. On $(0, 2)$: $f < 0$ and increasing.
4. On $(2, \infty)$: $f > 0$ and increasing.

Being below the axis is a statement about height. Being on the way up is a statement about slope. A single parabola separates them four different ways.

The sign of $f'$ can only change at a place where $f'$ is zero, where $f'$ fails to exist, or where $f$ itself is undefined. So:

1. Compute $f'$.
2. Mark every critical point: $f' = 0$ or $f'$ undefined, with $c$ in the domain of $f$.
3. Mark every point not in the domain of $f$. These are not critical points, and they split the chart anyway.
4. Test one convenient value of $f'$ in each resulting subinterval, and read off the direction.

For $f(x) = x^{3} - 3x^{2} - 9x + 5$, the derivative is $f'(x) = 3x^{2} - 6x - 9 = 3(x + 1)(x - 3)$. The marks are $-1$ and $3$, and testing $f'(-2) = 15$, $f'(0) = -9$, $f'(4) = 15$ gives increasing, decreasing, increasing.

Step 3 is the one that gets dropped. For $f(x) = x + \frac{1}{x}$, the derivative is $f'(x) = \frac{x^{2} - 1}{x^{2}}$, so the critical points are $\pm 1$. But $x = 0$ is not in the domain of $f$, so it splits the line as well, and the correct answer is that $f$ is decreasing on $(-1, 0)$ and on $(0, 1)$: two intervals, not one. Writing "decreasing on $(-1, 1)$" claims something about $x = 0$, where the function shoots off in both directions.

Report intervals, never isolated points, and use the domain of $f$ as the outer boundary of everything.

When the picture on the page is a graph of $f'$, three of its features have to be translated and one is a trap.

1. Above the axis means $f' > 0$, so $f$ is increasing. Below the axis means $f$ is decreasing.
2. Crossing the axis means $f'$ changes sign, so $f$ has a relative extremum there.
3. Touching the axis without crossing means $f' = 0$ with no sign change, so $f$ has a critical point and no extremum.
4. The high point of the displayed curve is where $f'$ is largest, which means $f$ is climbing fastest. It is not a maximum of $f$. Nothing about a maximum of $f$ can be read from a high point on this graph.

The trap is item 4, and it survives because the picture is a curve with a peak and the question is about a maximum. The peak belongs to whichever function was plotted. If the label says $f'$, then the peak is a fact about the rate.

Two habits make this safe. Say out loud which function the curve is, every time, before reading anything off it. And translate the whole graph into a single sign chart on the $x$-axis before answering, so what is left in front of you is a row of plus and minus signs rather than a shape.

Marks come from both clauses of the critical point definition, and forgetting the second clause quietly deletes whole intervals.

Take $f(x) = x^{2/3}(x - 5) = x^{5/3} - 5x^{2/3}$. Then

$$f'(x) = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3} = \frac{5(x - 2)}{3x^{1/3}}.$$

Solving $f' = 0$ gives $x = 2$ and nothing else. But $f'$ also fails at $x = 0$, and $f(0) = 0$ is perfectly well defined, so $x = 0$ is a critical point too. Both marks split the chart, and the signs are:

1. $x < 0$: numerator negative, denominator negative, so $f' > 0$ and $f$ increases.
2. $0 < x < 2$: numerator negative, denominator positive, so $f' < 0$ and $f$ decreases.
3. $x > 2$: both positive, so $f' > 0$ and $f$ increases.

The cusp at $x = 0$ is a relative maximum, and a search that only solved $f' = 0$ would have reported one interval of decrease running from $-\infty$ to $2$ and missed it entirely.

The converse habit is worth breaking at the same time: a mark on the chart does not have to be a turn. $f(x) = x^{4/3} - 4x^{1/3}$ is critical at $x = 0$ and decreasing on both sides of it. Marks tell you where to test. The test tells you what happened.