How does the sign of f''(c) classify a stationary point c of a one-variable function?
answer
- look at curvature, not slope
- sign of the second derivative decides
- what if that sign is zero
- x cubed and x^4 both give zero
basics
~10 sAt a stationary point c: positive f''(c) means a strict local minimum, negative means a strict local maximum, and f''(c) = 0 is inconclusive, since minima, maxima and inflections all produce it.
solid answer
~40 sOnce `f'(c) = 0`, the second derivative supplies the curvature. If `f''(c) > 0` the curve bends upward at `c`, so `c` is a strict local minimum; if `f''(c) < 0` it bends downward and `c` is a strict local maximum. If `f''(c) = 0` the test returns nothing at all, and you must look further. Three functions make that concrete: `x^4` has `f''(0) = 0` and a strict minimum at the origin; `-x^4` has `f''(0) = 0` and a strict maximum; `x^3` has `f''(0) = 0` and neither, since it increases straight through. All three look identical to the test. When it stalls you fall back on inspecting the sign of `f'` on each side of `c`, comparing values directly, or continuing to higher derivatives.
go deeper
Memorise the three cases and which sign gives which extremum, and be able to say out loud that a zero second derivative decides nothing. Getting the sign direction backwards is the failure that costs the point here.
Explain the mechanism, that the second derivative is the rate of change of the slope, and produce at least one worked function where the test stalls. Naming a concrete fallback is expected at this level.
Show judgment about when a degenerate classification actually matters in practice, and how you would resolve it numerically when derivatives are only estimated rather than computed exactly.
Own the tradeoff between cheap first-order checks and the cost of curvature information at scale, and decide when a team should stop classifying points at all and rely on direct value comparisons instead.
## What the test says Suppose `f` is twice differentiable near an interior point `c` and `f'(c) = 0`, so `c` is already known to be stationary. The **second-derivative test** classifies it from the sign of `f''(c)`: - `f''(c) > 0` implies `c` is a **strict local minimum**. - `f''(c) < 0` implies `c` is a **strict local maximum**. - `f''(c) = 0` implies **nothing**. The test is inconclusive. The intuition is that `f'` is the slope and `f''` is the rate at which the slope changes. If the slope is zero at `c` and increasing there, it was negative just left of `c` and positive just right of it: the function fell into `c` and rises out of it, which is exactly a minimum. A decreasing slope gives the mirror picture and a maximum. ## Why zero curvature tells you nothing The failure case is the part interviewers press on, because it is where a memorised rule stops working. Three functions all have a stationary point at the origin with `f''(0) = 0`, and all three are classified differently: - `f(x) = x^4`. Then `f'(x) = 4x^3` and `f''(x) = 12x^2`, so `f'(0) = 0` and `f''(0) = 0`. Yet `x^4 > 0` for every `x != 0`, so the origin is a **strict global minimum**. The curve is simply flatter than a parabola at the bottom. - `f(x) = -x^4`. By the same computation `f''(0) = 0`, and the origin is a **strict global maximum**. - `f(x) = x^3`. Here `f'(x) = 3x^2` and `f''(x) = 6x`, so again both vanish at 0. But `f` is strictly increasing everywhere, so the origin is **neither** a maximum nor a minimum; it is a flat inflection point. Since the test sees the same input, `f''(0) = 0`, in all three cases, it cannot possibly separate them. That is what "inconclusive" means: not "the point is degenerate and therefore uninteresting", but "this particular instrument has no resolution here". ## What to do when it stalls Three fallbacks, in rough order of how often they are used: **Sign of the first derivative on either side.** This is the first-derivative test and it is fully general for one variable. If `f'` changes from negative to positive as `x` passes through `c`, you have a local minimum; positive to negative gives a local maximum; no sign change gives neither. For `x^3` the derivative `3x^2` is positive on both sides, so no sign change, so no extremum — exactly right. **Direct comparison.** Evaluate `f` at a few points either side of `c`. This is crude but it settles the question locally and is often the fastest thing to do when the function is only available numerically. **Higher derivatives.** If the derivatives vanish up to some order, look at the first one that does not. If the first nonvanishing derivative at `c` is of **even** order and positive, `c` is a local minimum; even order and negative, a local maximum; **odd** order, neither. Check it against the examples: for `x^4` the first nonvanishing derivative at 0 is the fourth, which is 24 — even order, positive, so a minimum. For `x^3` it is the third derivative, 6 — odd order, so neither. The rule reproduces both answers. ## Strictness and the converse The implications run one way only. `f''(c) > 0` guarantees a strict local minimum, but a strict local minimum does not guarantee `f''(c) > 0` — `x^4` at the origin is the counterexample. What is true in the other direction is the weaker statement: if `c` is a local minimum of a twice-differentiable function then `f''(c) >= 0`. Mixing up which inequality is strict is a common and very visible slip. ## The multivariable analogue In several variables the single number `f''(c)` is replaced by curvature measured along every direction through `c`. If the curvature at a stationary point is strictly positive along every direction, the point is a strict local minimum; strictly negative along every direction, a strict local maximum; positive along some directions and negative along others, a **saddle**. And when the curvature is zero along some direction, the test is inconclusive for exactly the same reason it is inconclusive at `x = 0` on `x^4` — the second-order information has run out and you have to look at what lies beyond it.
- If c is a local minimum, must f''(c) be strictly positive?No. The implication only runs the other way. `f''(c) > 0` guarantees a strict local minimum, but a strict local minimum only guarantees `f''(c) >= 0`. The origin on `f(x) = x^4` is a strict global minimum with `f''(0) = 0`, so the weak inequality is the best you can claim.
- How does the first-derivative test resolve a case the second-derivative test cannot?It checks the sign of `f'` immediately either side of `c` rather than at `c`. Negative then positive means a local minimum, positive then negative a local maximum, and no sign change means neither. On `f(x) = x^3` the derivative `3x^2` is positive on both sides, so there is no sign change and the origin is correctly classified as neither.
- What does the higher-derivative rule say when several derivatives vanish at c?Find the first derivative at `c` that is not zero and look at its order. Even order and positive gives a local minimum, even order and negative a local maximum, and odd order gives neither. On `x^4` the fourth derivative is 24, so even and positive, a minimum; on `x^3` the third derivative is 6, so odd order, neither.
saying these in an interview costs you the question
- Says f''(c) = 0 proves the point is an inflection
- Claims a local minimum must have strictly positive second derivative
- Flips the signs, calling positive curvature a maximum
- Applies the test without first checking f'(c) = 0
- Has no fallback when the test is inconclusive