Increasing and Decreasing Intervals

Checking the definition means comparing every pair of points. The derivative gives a test that's easier to use.

The proof uses the Mean Value Theorem. Suppose $f'(x) > 0$ on $(a, b)$, and take any $x_1 < x_2$ in $[a, b]$. Then $f$ is continuous on $[x_1, x_2]$ and differentiable on $(x_1, x_2)$, so there's a $c$ in $(x_1, x_2)$ with

Both factors on the right are positive, so $f(x_2) > f(x_1)$. Thus $f$ is increasing on $[a, b]$. The decreasing case is the same with $f'(c) < 0$.

The derivative may be $0$ at a single point inside an interval of increase. For $f(x) = x^3$, $f'(x) = 3x^2 > 0$ for $x \ne 0$, so $f$ is increasing on $(-\infty, 0]$ and on $[0, \infty)$. Increasing pieces that share an endpoint $b$ join up: if $x_1 < b < x_2$, then $f(x_1) < f(b) < f(x_2)$. So $x^3$ is increasing on $(-\infty, \infty)$, although $f'(0) = 0$.

Since the test needs only continuity at the ends, an interval of increase may include an endpoint where $f$ is continuous. Open intervals are also correct, and AP scoring accepts either. This page reports open intervals, except where an example says otherwise.

A sign chart on its own isn't a justification, and the shape of a graph isn't one either. A justification names the sign of $f'$ on the interval, as in "$f$ is increasing on $(2, 7)$ because $f'(x) > 0$ on $(2, 7)$."

A point where $f$ isn't defined splits the chart too, and two intervals on either side of it can't be joined into one.

When the graph of $f'$ is given, read its sign: above the axis or below it. Whether the graph of $f'$ rises or falls says nothing about whether $f$ increases.

For a particle on a line, $x'(t) = v(t)$. So the test proves what the lesson on position, velocity, and acceleration stated: the position increases, and the particle travels right, while $v(t) > 0$.