A zero second derivative is the test declining AB & BC

Suppose $f'(c) = 0$ and $f''$ exists at $c$. Then:

1. $f''(c) > 0$: the graph is concave up at a horizontal tangent, so $f(c)$ is a relative minimum.
2. $f''(c) < 0$: concave down at a horizontal tangent, so $f(c)$ is a relative maximum.
3. $f''(c) = 0$: inconclusive. The test tells you nothing, and the First Derivative Test has to be run instead.

The picture is the reason it works. A flat tangent inside a bowl is the bottom of the bowl; a flat tangent under a dome is the top of the dome. That is all the test is.

Note the hypothesis: $f'(c) = 0$, not "$c$ is critical". Critical points also come from $f'$ failing to exist, and at those points this test has nothing to say, because there is no horizontal tangent to sit inside anything. $f(x) = |x|$ has a relative minimum at $x = 0$ that the Second Derivative Test cannot reach, while the First Derivative Test settles it in one line.

Worked: $f(x) = x^{2}e^{-x}$ has $f'(x) = x(2 - x)e^{-x}$, so the critical points are $0$ and $2$, and $f''(x) = (x^{2} - 4x + 2)e^{-x}$. Then $f''(0) = 2 > 0$, a relative minimum, and $f''(2) = -2e^{-2} \approx -0.271 < 0$, a relative maximum. Two substitutions, no sign chart.

Case 3 is where the damage is done, because "$f''(c) = 0$" looks like data and gets treated as one. It is not. Three functions make the point as sharply as it can be made, and all three have $f'(0) = 0$ and $f''(0) = 0$:

1. $f(x) = x^{4}$ has a relative minimum at $x = 0$.
2. $f(x) = -x^{4}$ has a relative maximum at $x = 0$.
3. $f(x) = x^{3}$ has neither at $x = 0$.

Every input the Second Derivative Test consults is identical across the three, and the answers are all three possible answers. So no procedure that reads only $f'(c)$ and $f''(c)$ can distinguish them, and any conclusion drawn from $f''(c) = 0$ is drawn from nothing.

Two specific wrong moves to name and avoid:

1. "$f''(c) = 0$, so there is no extremum." Refuted by $x^{4}$.
2. "$f''(c) = 0$, so it is an inflection point." Also refuted by $x^{4}$, whose concavity never changes.

What to do instead takes one line: go back to $f'$ and check its sign just left and just right of $c$. The First Derivative Test never returns inconclusive.

An inflection point is a point on the graph of $f$ where the concavity changes. Two conditions, and neither one is "$f'' = 0$".

1. $f''$ must change sign there.
2. The point must be on the graph, so $c$ has to be in the domain of $f$.

Vanishing is neither necessary nor sufficient.

Not sufficient: $f(x) = x^{4}$ has $f''(x) = 12x^{2}$, which is zero at the origin and non-negative everywhere. Concave up on both sides, so no change, so no inflection point. Roots of $f''$ are candidates in exactly the way that critical points are candidates.

Not necessary: $f(x) = x^{1/3}$ has $f''(x) = -\frac{2}{9}x^{-5/3}$, which does not exist at $x = 0$ and is positive to the left and negative to the right. The concavity changes, $0$ is in the domain, and the origin is an inflection point with a vertical tangent.

And the domain rule: $f(x) = \frac{1}{x}$ has concavity changing across $x = 0$ and no inflection point, since there is no point of the graph there to be one.

So the procedure mirrors the extremum procedure exactly, one derivative up. Find where $f'' = 0$ or fails, discard anything outside the domain of $f$, and test the sign on both sides. Then report the point, both coordinates, with the $y$-coordinate coming from $f$ and not from $f'$ or $f''$.

Both tests are correct. They differ in cost and in coverage.

1. Second Derivative Test. Cheap: one substitution per critical point, no sign analysis. Best when $f''$ is easy to write down and the critical points come from $f' = 0$. Fails when $f''(c) = 0$, when $f''(c)$ does not exist, and when $f'(c)$ never was zero.
2. First Derivative Test. More work: a sign chart. Always applies at a critical point where $f$ is continuous, including corners and cusps, and never returns inconclusive.

A reasonable default is to try the second and fall back to the first the moment it declines. What is not reasonable is to treat the decline as an answer, or to report "the Second Derivative Test fails, so nothing can be said". Something can always be said; it just takes the other test to say it.

One more piece of bookkeeping worth being exact about, since 5.8 and 5.9 depend on it. The Second Derivative Test consults the value $f''(c)$ at a single point. An inflection point requires the sign change of $f''$ across a point. Those are different questions asked of the same function, and $x^{4}$ at the origin answers both of them in the negative while looking like it should answer both in the affirmative.