AP Calculus AB and BC

Positive Derivative but Not Increasing: The Catch

Q: Is the tangent function increasing?

On each branch, yes, and its derivative \sec^2 x is positive wherever it is defined. Across an asymptote the values drop from +\infty to -\infty, so it is not increasing on any interval containing one.

The increasing/decreasing test needs the domain to be a single interval. A positive derivative makes a function increasing there, but across a gap the conclusion fails outright: -1/x has a positive derivative everywhere it is defined and still drops from 1 down to -1 across the origin.

Increasing/Decreasing Test

f'(x) > 0 \text{ on an interval } I \implies f \text{ is increasing on } I

If the derivative is positive at every point of an interval, then the function is increasing across that whole interval.

The hypotheses, and what each one buys

A theorem is only as strong as its conditions. Below, each hypothesis is dropped on its own while every other one is kept, so you can see precisely what it was holding up.

1
The domain is a single interval, with no gaps
$I \text{ is an interval}$
The proof runs through the Mean Value Theorem, which needs a single unbroken interval to work on. Two separate pieces of domain are two separate applications of the theorem, and nothing connects the conclusion on one piece to the conclusion on the other.
Drop it and the theorem fails
$f(x) = -\frac{1}{x}, \qquad f'(x) = \frac{1}{x^{2}} > 0$
f(x) = -1/x on [-1, 1] without the origin
The derivative $1/x^2$ is positive at every single point of the domain, with no exceptions. Yet $f(-1) = 1$ and $f(1) = -1$ , so the function is emphatically lower on the right than on the left. The check finds pairs going the wrong way. The domain is two intervals wearing one formula, and the function is increasing on each of them separately.
2
The derivative is positive throughout, not merely at one point
$f'(x) > 0 \text{ for all } x \in I$
A positive derivative at a single point says the function is instantaneously rising there and nothing more. It does not even guarantee the function is increasing on any interval around that point, once the derivative is allowed to be discontinuous. The theorem is about the whole interval, and only the whole interval.
No counterexample built from an elementary formula breaks this one on its own, so none is claimed here.

The counterexample above is checked numerically on every build: each function is evaluated and the conclusion is confirmed to fail. A counterexample that stopped working would fail the build rather than sit here misleading you.

Why it is true

Take any two points $x_1 < x_2$ in the interval.
The Mean Value Theorem applies to $f$ on $[x_1, x_2]$ , since that subinterval lies inside $I$ where $f$ is differentiable and hence continuous.
So $f(x_2) - f(x_1) = f'(c)(x_2 - x_1)$ for some $c$ strictly between them.
Both factors on the right are positive, so $f(x_2) > f(x_1)$ . The step that fails for $-1/x$ is the second one: with the origin missing, $[x_1, x_2]$ is not inside the domain and the Mean Value Theorem never applies.

What it does not say

A positive derivative always means an increasing function.

On an interval, yes. Across a domain with a hole in it, no. The function $-1/x$ and the tangent function are the two examples worth remembering.

An increasing function must have a positive derivative.

The converse is weaker than you would like: $f(x) = x^3$ is increasing everywhere and has $f'(0) = 0$ . Increasing gives you $f' \ge 0$ , never $f' > 0$ .

Increasing on each of two intervals means increasing on their union.

Only if the union is itself an interval. This is the same error as above, seen from the other direction, and it is exactly why answers should list intervals separately rather than joining them with a union symbol.

Frequently asked questions

Should I write the answer as a union of intervals?

For increasing and decreasing behaviour, list the intervals separately rather than joining them. Saying a function increases on the union of two intervals claims something about points between them that is usually false.

Is the tangent function increasing?

On each branch, yes, and its derivative $\sec^2 x$ is positive wherever it is defined. Across an asymptote the values drop from $+\infty$ to $-\infty$ , so it is not increasing on any interval containing one.