4.1.28 · D5Calculus I — Limits & Derivatives
Question bank — Applications — increasing - decreasing, local extrema (first derivative test)
True or false — justify
If then is a local maximum or minimum.
False. Zero slope only means "flat here"; the slope must switch sign to make an extremum. has yet keeps rising — an inflection with a horizontal tangent.
If on an interval , then is increasing on .
If is increasing on , then for every .
False — only is guaranteed. is strictly increasing on but : a single flat instant doesn't stop overall rising.
Every local maximum occurs at a point where .
False. Extrema occur at critical points, which include places where is undefined. has a minimum at where no derivative exists.
A critical point is always an extremum.
False. or undefined is only a candidate. (slope zero, no sign change) and (vertical tangent, no sign change) are critical but flat-through, not extrema.
If changes from to at , then is a local minimum.
True — falling then rising means you bottomed out in a valley. This is exactly the case of the First Derivative Test.
If does not exist, then cannot be an extremum.
False. A corner like at is a genuine minimum precisely because the slope flips sign across the kink, even though it's undefined at the point.
The Mean Value Theorem needs differentiable on the closed interval .
False. It needs continuity on but differentiability only on the open interval — the endpoints may be corners or vertical tangents.
If just left of and just right of , then is not an extremum.
True. Same sign both sides means the function kept rising through ; no peak or valley formed — it's "same, so nothing."
Spot the error
"Solve , get the critical points — done."
Missing the points where is undefined. Cusps, corners, and vertical tangents (, ) are critical too and can hold extrema.
", and came out zero when I tested it, so has no extremum."
You tested the critical point itself, which always gives no sign. You must sample strictly left and right of , never at .
" changes at , so the maximum is ."
The maximum location is ; the maximum value is . Reporting the -coordinate as "the maximum" confuses input with output.
" is decreasing on since everywhere."
You cannot join intervals across a break in the domain. is decreasing on each piece separately, but , so it is not decreasing on the union.
" everywhere except at one point where , so is not strictly increasing."
An isolated flat instant doesn't stop strict increase. has yet always — still strictly increasing.
"I found a critical point at ; the sign of was left and right, so it's a maximum."
is a minimum (down into a valley, then up). is the maximum. The mnemonic: PND peak, NPD valley.
"Since MVT gives one , the function has exactly one point matching the average slope."
MVT guarantees at least one such , not exactly one. A wavy graph can match the average slope at several interior points.
Why questions
Why does the sign of , not its size, decide increasing vs decreasing?
In the factor , so only the sign of can flip the sign of the change. Magnitude sets how fast, not which way.
Why is one test point enough to fix the sign of on a whole interval between critical points?
On such an interval is differentiable, so exists there and is never zero (all zeros are critical points we already removed). If is also continuous there, the Intermediate Value Theorem forbids it changing sign without passing through zero — so it keeps one sign and a single sample settles the interval.
When exactly is " continuous between critical points" guaranteed, so the one-test-point rule is valid?
It is guaranteed whenever is twice differentiable (or is otherwise known continuous), which covers polynomials, rationals, and standard functions. For pathological that exists but is discontinuous, you must check the sign more carefully rather than trusting one point.
Why must be continuous at for the First Derivative Test to classify ?
The test reads off "rise then fall" around ; if jumps at , the pieces don't connect into a genuine peak or valley, so sign of alone can mislead.
Why does Fermat's theorem only say extrema occur at critical points, not that critical points are extrema?
It's a one-way filter: it narrows candidates to where the slope is zero or undefined, but the sign change (or Second Derivative Test) is still needed to confirm each candidate.
Why can't we conclude "increasing" just because for the particular endpoints we checked?
Increasing requires for all in . A dip between two chosen points still satisfies one comparison while violating the definition overall.
Why does fool people into calling a minimum?
Because and the graph flattens there, mimicking a valley's bottom. But on both sides — no sign change, so it's an inflection, not a min.
Edge cases
What is the sign behaviour of for at , and what does the test say?
near and is positive on both sides (vertical tangent). Same sign → not an extremum, despite being undefined at .
Is an extremum of ?
Yes, a local (and global) minimum. is negative for and positive for : a cusp, so it's a valley bottom.
Can a function be increasing on yet have at interior points?
Yes. is strictly increasing with ; isolated flat points are allowed as long as and is not zero on a whole subinterval.
What happens to the First Derivative Test at an endpoint of the domain, e.g. for on ?
You only have one side to test, so the sign-change rule doesn't apply. Endpoints are handled by the closed-interval method, comparing endpoint values directly.
If is positive everywhere except undefined at one point where is still continuous and rising through it, what is that point?
A critical point that is not an extremum (like at ) — a vertical-tangent inflection where rising never pauses in the -direction.
Can a function have infinitely many critical points in a bounded interval?
Yes. (with ) oscillates so its slope hits zero infinitely often near — critical points can accumulate, so "find all critical points" isn't always finite.
Does a constant function count as a local max or min?
Yes — under the usual (non-strict) definition, every point is simultaneously a local maximum and a local minimum, since and hold on any neighbourhood. It is neither strictly increasing nor decreasing.
On an interval where (a genuine plateau, not just one point), what is doing there?
is constant on that whole interval (by MVT applied to any sub-pair). So the graph has a literal flat shelf, not an isolated flat instant.
If a function rises, then runs flat along a plateau, then falls (like a mesa), is the plateau a maximum?
Yes — every point of that flat top is a local maximum (a "flat maximum"). The test still works: goes before the shelf and after it, with across the whole top. A mirror-image plateau gives a flat minimum.
Can a plateau ever be not an extremum?
Yes — if is before the flat shelf and again after it (a flat step on a rising staircase), the plateau is neither a max nor a min; the function just paused before continuing up.
Connections
- Fermat's Theorem on Stationary Points — why critical points are only candidates.
- Mean Value Theorem — the engine behind every sign-of- claim above.
- Second Derivative Test — an alternative that sidesteps the "same sign" ambiguity for smooth points.
- Concavity and Inflection Points — explains the and "flat-through" cases.
- Optimization (Closed Interval Method) — how endpoints (excluded above) get handled.
- Curve Sketching — where all these traps show up together.