Second derivative test — concavity, inflection points: Question bank

Definition Strict vs. non-strict concavity (a wording caution)

Two precise notions hide behind the words "concave up", and the traps below exploit the gap:

Strictly concave up on an interval means $f^{''} > 0$ throughout it (slope strictly increasing). This is the sense used everywhere in this bank unless an item says otherwise.
Non-strictly (weakly) concave up means $f^{''} \geq 0$ , allowing $f^{''}$ to touch zero at isolated points while never going negative.

Why the distinction matters: for $f (x) = x^{4}$ we say "it stays concave up on both sides of $0$ ", which is true only in the non-strict $f^{''} \geq 0$ sense — at the single point $x = 0$ we have $f^{''} = 0$ , so it is not strictly concave up there. Because $f^{''}$ never turns negative, concavity never flips, so there is still no inflection. Keep the two readings separate and the $x^{4}$ trap dissolves.

The three blueprint figures below are not decoration — each one settles a specific misconception the reveals keep testing. Read the caption note under each before you tackle the matching group.

Figure 1 — the sign of $f^{''}$ is the concavity. The cyan curve ( $f^{''} > 0$ ) holds water like a cup; the amber curve ( $f^{''} < 0$ ) sheds water like a cap. This is the whole content of the "True or false" sign items: whenever you read " $f^{''} > 0$ ", picture the cyan cup, and "up is a cuP → min" stops being a slogan.

Figure — Second derivative test — concavity, inflection points

Figure 2 — crossing zero vs. touching zero. The cyan line $f^{''} = 6 x$ (from $x^{3}$ ) crosses the axis, so concavity genuinely flips → inflection. The amber parabola $f^{''} = 12 x^{2}$ (from $x^{4}$ ) merely touches zero and bounces back non-negative → no flip, no inflection. Every " $f^{''} = 0$ therefore inflection" trap in the bank lives on this one picture.

Figure 3 — at a flat point, concavity alone decides. Both curves have a horizontal tangent (dashed white). The cyan cup puts that flat point at the bottom (minimum); the amber cap puts it at the top (maximum). This is why the second derivative test needs $f^{'} (c) = 0$ first — the "Spot the error" sign confusions all come from forgetting which cup the flat point sits in.

True or false — justify

Each is a claim. The reveal states True/False and the reasoning that settles it.

f^{''} (c) = 0

then

c

is an inflection point.

False.

f^{''}

must change sign at

c

;

f (x) = x^{4}

has

f^{''} (0) = 0

yet

f^{''} = 12 x^{2} \geq 0

on both sides, so it stays concave up — no flip, no inflection.

f

is concave up on an interval, then

f

is increasing there.

False. Concave up means the slope

f^{'}

is increasing, not

f

itself.

f (x) = x^{2}

(- \infty, 0)

is concave up but decreasing.

At a local minimum found by the test,

f^{''} (c)

must be strictly positive.

False. A min can have

f^{''} (c) = 0

; e.g.

x^{4}

0

is a genuine minimum where

f^{''} (0) = 0

and the test is only inconclusive, not wrong.

f^{''} > 0

everywhere AND

f^{'} (c) = 0

at some

c

, then that critical point is a minimum, never a maximum.

True. You need a critical point to have a local extremum at all; given one,

f^{''} > 0

means globally concave up (

\cup

), so the flat point is the bottom of the cup — a minimum. A local maximum would require concave-down behaviour somewhere, which

f^{''} > 0

forbids.

An inflection point must occur at a critical point.

False. Inflection is about

f^{''}

flipping, critical is about

f^{'} = 0

(or undefined).

sin x

inflects at

x = π

, where

f^{'} = cos π = - 1 \neq = 0

— flat slope not required.

f^{''} (c)

does not exist,

c

cannot be an inflection point.

False. Concavity can still switch where

f^{''}

is undefined; e.g.

f (x) = x^{1/3}

inflects at

0

where

f^{''}

blows up, and the graph goes

\cup \to \cap

across it.

A function can be concave up and concave down at the very same point.

False. At a single point concavity is one sign (or undefined); "changes concavity" is a statement about the two sides of the point, not the point itself.

f^{''} (c) > 0

then

f

has a minimum at

c

False as stated — you must first know

f^{'} (c) = 0

f^{''} > 0

at a non-critical point just means the graph curves like a cup there, with any slope.

Between two consecutive inflection points, the concavity is constant.

True. Inflections are exactly the sign changes of

f^{''}

; between two of them

f^{''}

keeps one sign, so concavity does not switch.

A straight line

f (x) = m x + b

has an inflection point everywhere.

False.

f^{''} \equiv 0

but it never changes sign (it is always zero), so concavity never flips — there are no inflection points.

Spot the error

Each line contains a flawed argument. The reveal names the exact broken step.

f^{''} (0) = 0

for

x^{4}

, so by the second derivative test

x = 0

is neither a max nor a min."

The error:

f^{''} = 0

makes the test inconclusive, it does not declare "neither". The first-derivative test shows

f^{'}

goes

- \to +

, so

0

is a genuine minimum. (Also note: "saddle" is reserved for higher dimensions — in 1D say "neither max nor min".)

f^{'} (c) = 0

and

f^{''} (c) < 0

, so

c

is where the graph is lowest."

Sign confusion:

f^{''} (c) < 0

is concave down (

\cap

), giving a maximum (highest nearby), not lowest. Remember "up is a cuP → min".

"Concavity changes at

x = c

, and

f

has a jump there, so it's an inflection point."

Inflection requires

f

continuous at

c

. A jump disqualifies it even though the concavity technically differs on the two sides.

f^{''}

is positive at

c

, so the tangent line lies above the curve near

c

Backwards: concave up (

\cup

) means the curve bends away from the tangent upward, so the tangent lies below the curve nearby.

"We found

f^{''} (c) = 0

, so we skip checking sign changes and conclude no inflection."

Skipping the sign check is the whole trap.

f^{''} (c) = 0

is only a candidate; some candidates (like

x^{3}

0

) do flip sign and are inflections.

"The second derivative test failed at

c

, so

c

is not an extremum."

"Test failed" only means it gave no verdict. You must fall back to the First derivative test; the point may still be a max, min, or neither.

Why questions

Explain the reasoning, not just the fact.

Why does the sign of

(x - c)^{2}

never affect the conclusion of the Taylor argument?

Because

(x - c)^{2} \geq 0

always, so the sign of

f (x) - f (c) \approx \frac{1}{2} f^{''} (c) (x - c)^{2}

is fixed by

f^{''} (c)

alone — that's the whole reason

f^{''}

decides min vs max (see Taylor series).

Why can the second derivative test never classify a critical point of

x^{4}

All lower Taylor terms vanish:

f^{'} (0) = f^{''} (0) = f^{'''} (0) = 0

and the first nonzero term is

x^{4}

. The test only reads the

f^{''}

(quadratic) term, which is

0

here, so it is blind to the quartic that actually decides.

Why is "

f^{''}

changes sign" a stronger requirement than "

f^{''} = 0

A function can touch zero without crossing it (like

12 x^{2}

0

). Only an actual crossing flips concavity from

\cup

\cap

, which is what an inflection geometrically is.

Why does concave up guarantee a critical point is a minimum, with no need to test neighbours?

Concave up means slope

f^{'}

is increasing; at

c

it passes through

0

, so

f^{'} < 0

just left and

f^{'} > 0

just right — downhill then uphill = a valley, decided instantly.

Why does

f^{''} = - sin x

tell us

sin x

inflects at

π

but not at

0

2 π

(interior sense)?

- sin x

changes sign as

sin x

crosses zero going from positive to negative through

π

(interior), producing a concavity flip; at the endpoints

0, 2 π

there is no two-sided interval to compare, so no interior inflection is claimed.

Why is the second derivative test only "local"?

The Taylor expansion is an approximation valid near

c

; it says nothing about far-away behaviour, so it certifies a local extremum, never a global one (that needs Maxima and minima — optimization comparison over the whole domain).

Edge cases

Boundary and degenerate scenarios the test invites you to mishandle.

f (x) = x^{3}

x = 0

: is it a max, min, or inflection?

Neither extremum:

f^{'} (0) = 0

but

f^{''} = 6 x

flips

- \to +

across

0

and

f

is continuous, so it's an inflection point with a horizontal tangent.

f (x) = ∣ x ∣

x = 0

: what does the second derivative test say?

It cannot be applied —

f^{''} (0)

doesn't exist (corner), so

0

is a critical point of the "undefined-derivative" kind. The first-derivative test still gives a minimum since slope goes

- 1 \to + 1

f (x) = x^{1/3}

x = 0

: inflection despite

f^{''}

undefined?

Yes.

f^{''} = - \frac{2}{9} x^{- 5/3}

is undefined at

0

but changes sign (concave up for

x < 0

, concave down for

x > 0

) while

f

stays continuous, so concavity flips — a valid inflection.

A constant function

f (x) = 5

: any concavity or inflection?

f^{''} \equiv 0

with no sign change anywhere, so it is neither strictly concave up nor down and has no inflection points — flatness is not a switch.

f (x) = x^{2}

[1, 3]

: does the second derivative test find its minimum?

No — the minimum sits at the endpoint

x = 1

where

f^{'} \neq = 0

. The test only inspects interior critical points; endpoints need direct comparison (Curve sketching / boundary check).

f^{''} > 0

on all of

R

(a convex function), how many inflection points can it have?

Zero. Constant-sign

f^{''}

means concavity never flips, which is the defining feature of a convex function — see Concavity and convex functions.

f^{''} (c) = 0

and

f^{'''} (c) \neq = 0

: what can you conclude at

c

The nonzero odd-order term forces

f^{''}

to actually change sign across

c

, so (with continuity)

c

is an inflection point — a clean sufficient condition.

Connections

Second derivative test — concavity, inflection points
First derivative test
Critical points
Taylor series
Curve sketching
Maxima and minima — optimization
Concavity and convex functions