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Multivariable Second Derivative Test Intuition

Step-by-step mathematics solution: Multivariable Second Derivative Test Intuition

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1. What the problem is asking

We are asked to justify the classic second‑derivative (Hessian) test for a function of two variables

[ f:\mathbb R^{2}\to\mathbb R ,\qquad (x,y)\mapsto f(x,y) ]

and to explain why, when the first‑order partial derivatives vanish at a point ((a,b)),

  • the point is a local minimum if the quadratic form defined by the Hessian matrix is positive‑definite, i.e.

    [ \mathbf u^{T}H_f(a,b)\,\mathbf u>0\quad\text{for every non‑zero direction }\mathbf u\in\mathbb R^{2}, ]

  • and that this condition is equivalent to the two scalar inequalities

    [ f_{xx}(a,b)>0 ,\qquad \det H_f(a,b)=f_{xx}(a,b)f_{yy}(a,b)-f_{xy}(a,b)^{2}>0 . ]

The same reasoning (with “( <0)” instead of “( >0)”) yields the criterion for a local maximum, and if the quadratic form is indefinite the point is a saddle.

2. Step‑by‑step justification

2.1. Critical points – why (\nabla f(a,b)=\mathbf 0)

For a differentiable function (f) the directional derivative in direction (\mathbf u) (with (|\mathbf u|=1)) is

[ D_{\mathbf u}f(a,b)=\nabla f(a,b)\cdot\mathbf u . ]

If ((a,b)) is a local extremum (maximum or minimum) then the slope in every direction must be zero; otherwise we could move a tiny amount in a direction where the slope is non‑zero and obtain a larger (or smaller) value of (f).

Hence at an extremum

[ D_{\mathbf u}f(a,b)=0\quad\text{for all }\mathbf u\Longrightarrow \nabla f(a,b)=\mathbf 0 . ]

The equations (\partial f/\partial x=0,\;\partial f/\partial y=0) are called the first‑order conditions; their solutions are the critical points.

2.2. The second‑order (Hessian) expansion

Assume that (f) has continuous second partial derivatives near ((a,b)).
Taylor’s theorem with remainder (in vector form) gives, for a small displacement (\mathbf h=(h_1,h_2)^{T}),

[ f(a+\mathbf h)=f(a,b)+\underbrace{\nabla f(a,b)^{T}\mathbf h}_{=0} +\tfrac12\,\mathbf h^{T} H_f(a,b)\,\mathbf h +o(|\mathbf h|^{2}), ]

where

[ H_f(a,b)= \begin{bmatrix} f_{xx}(a,b) & f_{xy}(a,b)\[2pt] f_{yx}(a,b) & f_{yy}(a,b) \end{bmatrix} \quad\text{(the Hessian matrix).} ]

Because the first‑order term vanishes at a critical point, the quadratic term

[ Q(\mathbf h)=\tfrac12\,\mathbf h^{T} H_f(a,b)\,\mathbf h ]

dominates the behaviour of (f) near ((a,b)).
If (Q(\mathbf h)>0) for every non‑zero (\mathbf h), then for sufficiently small (|\mathbf h|) we have

[ f(a+\mathbf h)-f(a,b)=Q(\mathbf h)+o(|\mathbf h|^{2})>0, ]

so the function is larger in every nearby direction → local minimum.
Similarly, (Q(\mathbf h)<0) for all non‑zero (\mathbf h) gives a local maximum.
If (Q) takes both positive and negative values, the point is a saddle.

Thus the problem reduces to deciding when the quadratic form

[ \mathbf h^{T} H_f(a,b)\,\mathbf h ]

is positive‑definite, negative‑definite, or indefinite.

2.3. Positive‑definiteness of a (2\times 2) symmetric matrix

The Hessian is symmetric ((f_{xy}=f_{yx})).
For a symmetric (2\times 2) matrix

[ A= \begin{bmatrix} \alpha & \beta
\beta & \gamma \end{bmatrix}, ]

the following are equivalent:

Condition Meaning
(i) (\displaystyle \mathbf v^{T}A\mathbf v>0) for all non‑zero (\mathbf v\in\mathbb R^{2}) (A) is positive‑definite
(ii) (\alpha>0) and (\det A>0) Simple scalar test (Sylvester’s criterion)
(iii) Both eigenvalues of (A) are positive Spectral view

Proof of the equivalence (ii) ⇔ (i):

Take (\mathbf v=(v_1,0)^{T}). Then (\mathbf v^{T}A\mathbf v=\alpha v_1^{2}>0) for all (v_1\neq0) forces (\alpha>0).

Now write the quadratic form in “completed‑square” form:

[ \mathbf v^{T}A\mathbf v = \alpha\Bigl(v_1+\frac{\beta}{\alpha}v_2\Bigr)^{2} +\Bigl(\gamma-\frac{\beta^{2}}{\alpha}\Bigr)v_2^{2} = \alpha\bigl(\cdots\bigr)^{2} +\frac{\det A}{\alpha}v_2^{2}. ]

If (\alpha>0) then the sign of the whole expression is governed by the second term.
Thus the form is always positive iff (\displaystyle\frac{\det A}{\alpha}>0), i.e. (\det A>0).

Consequently, for the Hessian

[ H_f(a,b)=\begin{bmatrix} f_{xx} & f_{xy}
f_{xy} & f_{yy} \end{bmatrix}, ]

the condition

[ \mathbf h^{T}H_f(a,b)\,\mathbf h>0\quad\forall\ \mathbf h\neq\mathbf0 ]

is exactly

[ \boxed{\,f_{xx}(a,b)>0\quad\text{and}\quad f_{xx}(a,b)f_{yy}(a,b)-f_{xy}(a,b)^{2}>0\, } . ]

These are the familiar second‑derivative test inequalities for a local minimum.

The analogous condition for a local maximum is

[ f_{xx}(a,b)<0\quad\text{and}\quad f_{xx}(a,b)f_{yy}(a,b)-f_{xy}(a,b)^{2}>0, ]

i.e. the Hessian is negative‑definite (both eigenvalues negative).

If the determinant is negative, the quadratic form takes both signs and the critical point is a saddle.

2.4. Summary of the full test

  1. First‑order test – solve

    [ f_{x}(a,b)=0,\qquad f_{y}(a,b)=0 . ]

    Any solution ((a,b)) is a critical point.

  2. Second‑order test – compute the Hessian at the critical point and evaluate

    [ D = f_{xx}(a,b)f_{yy}(a,b)-f_{xy}(a,b)^{2}. ]

    • If (f_{xx}(a,b)>0) and (D>0) → local minimum.
    • If (f_{xx}(a,b)<0) and (D>0) → local maximum.
    • If (D<0) → saddle point (neither max nor min).
    • If (D=0) → the test is inconclusive; higher‑order analysis is needed.

3. Final Answer

For a twice‑continuously differentiable function (f(x,y)) :

A point ((a,b)) with (\nabla f(a,b)=\mathbf 0) is a local minimum iff the Hessian matrix at that point is positive‑definite, which for a (2\times2) Hessian is equivalent to

[ \boxed{\;f_{xx}(a,b)>0\quad\text{and}\quad f_{xx}(a,b)f_{yy}(a,b)-f_{xy}(a,b)^{2}>0\; } . ]

The analogous inequalities with “( <0)” give a local maximum; a negative determinant signals a saddle point.

4. Common Mistakes

Mistake Why it’s wrong How to avoid it
Forgetting the first‑order condition and applying the Hessian test at a point where (\nabla f\neq\mathbf0). The quadratic approximation assumes the linear term is zero; otherwise the sign of the linear term dominates. Always verify (\partial f/\partial x=\partial f/\partial y=0) before using the second‑derivative test.
Using only (f_{xx}>0) (or (f_{yy}>0)) as a test. Positivity of a single second partial does not guarantee the quadratic form is positive for all directions. Check both (f_{xx}>0) and the determinant condition (D>0).
Confusing “(D>0)” with “(f_{xx}>0)”. Either condition alone is insufficient; both are needed for definiteness. Remember Sylvester’s criterion: all leading principal minors must be positive for positive‑definiteness.
Assuming the test is decisive when (D=0). When the determinant vanishes the quadratic form is degenerate; the point could be min, max, or saddle. Treat (D=0) as “inconclusive” and examine higher‑order terms or use other methods (e.g., analyzing level curves).
Mixing up the sign of (f_{xx}) for minima vs. maxima. The sign of (f_{xx}) determines whether the Hessian is positive or negative definite, not the determinant. For minima: require (f_{xx}>0); for maxima: require (f_{xx}<0); in both cases the determinant must be positive.

Keeping these points in mind will help you apply the multivariable second‑derivative test correctly and interpret its outcome with confidence.

Original question: Multivariable Second Derivative Test Intuition on Mathematics Stack Exchange, licensed CC BY-SA.

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