Determining Intervals on Which a Function Is Increasing or Decreasing AB & BC

A function increases exactly where its derivative is positive and decreases exactly where its derivative is negative, a fact the Mean Value Theorem supplies. The sign chart is therefore built from $f'$, and values of $f$ never enter it: $f(x) = x^{2} - 4$ is positive and decreasing on $(-\infty, -2)$, negative and decreasing on $(-2, 0)$, negative and increasing on $(0, 2)$, and positive and increasing on $(2, \infty)$, so the sign of $f$ predicts nothing.

The number line is split by every critical point, from both clauses of the definition, and also by every point where $f$ itself is undefined even though those are not critical points. For $f(x) = x + \frac{1}{x}$ the answer is decreasing on $(-1, 0)$ and on $(0, 1)$, two intervals, because $x = 0$ is a gap in the domain. For $f(x) = x^{2/3}(x - 5)$ the cusp at $x = 0$ is invisible to solving $f' = 0$ and is a relative maximum. When the picture on the page is a graph of $f'$, its high point is where $f$ climbs fastest, not where $f$ is largest.