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Critical Points
Outline
1. Critical Points: Simple Cases
If f : R → R is a differentiable function, a critical point for f is any value of x for which
f 0 (x) = 0. There are two simple generalizations of this definition:
• If γ : R → Rn is a differentiable curve, a critical point for γ is any value of t for which
γ̇(t) = 0.
• If f : Rn → R is a differentiable function, a critical point for f is any point p ∈ Rn for
which ∇f (p) = 0.
If t0 is a critical point for the curve γ, then the image of γ may have a cusp or bend at the
point γ(t0 ). If p is a critical point for a real-valued function f , then f may have a local
maximum, a local minimum, or a saddle point at p.
A point in the domain of a function is called a regular point if it is not a critical point.
A function is called regular if every point in its domain is a regular point. That is, a function
is regular if it has no critical points.
2. The Hessian
If f : Rn → R is a C 2 function, the Hessian of f is the matrix of second partial derivatives:


"
#
fxx fxy fxz
fxx fxy


fyx fyy fyz 
fyx fyy
fzx fzy fzz
n=2
n=3
In general, the Hessian of f is an n × n symmetric matrix. We will write Hf for the Hessian
of f , and the value of this Hessian at a point p will be denoted Hp f .
Given any vector v, the value of vT(Hp f )v can be interpreted as the directional second
derivative of f in the direction of v. Specifically,
d2
= vT(Hp f )v.
f
(p
+
tv)
2
dt
t=0
3. The Second Derivative Test
The Hessian matrix can be used to classify the critical points of a real-valued function. Specifically, let f : Rn → R be a C 2 function, let p ∈ Rn be a critical point for f , and let λ1 , . . . , λn
be the eigenvalues of the Hessian Hp f . Then:
• If the eigenvalues λ1 , . . . , λn are all positive, then f has a local minimum at p.
• If the eigenvalues λ1 , . . . , λn are all negative, then f has a local maximum at p.
• If at least one eigenvalue λi is positive and at least one eigenvalue λj is negative, then f
has a saddle point at p.
Note that the eigenvalues λ1 , . . . , λn can be interpreted as the directional second derivatives
of f in the directions of an orthonormal basis of eigenvectors for Hp f .
The second derivative test cannot always be used to classify a given critical point. For
example, if all of the eigenvalues are 0, or if some of them are 0 and the rest all have the same
sign, then none of the three cases applies.
4. Critical Points: The General Case
The definition of a critical point for a function f : Rm → Rn is a bit complicated. There are
two separate cases: either m ≤ n or m ≥ n.
If m ≤ n, then a point p ∈ Rm is called a critical point for f if it satisfies any one the
following equivalent conditions:
• The rank of the derivative matrix Dp f is less than m.
• The columns of the derivative matrix are linearly dependent.
∂f
∂f
(p), . . . ,
(p) are linearly dependent.
• The partial derivatives
∂x1
∂xm
Similarly, if m ≥ n, then a point p is a critical point for f if it satisfies any one of the
following equivalent conditions:
• The rank of the derivative matrix Dp f is less than n.
• The rows of the derivative matrix are linearly dependent.
• The gradient vectors ∇f1 (p), . . . ∇fn (p) are linearly dependent.
In the special case where m = n, both of the above definitions apply. Moreover, in this case
a point p is a critical point if and only if the square matrix Dp f is singular, i.e. if and only if
the determinant of Dp f is 0.
Practice Problems
1. Let f : R2 → R be the function f (x, y) = 31 x3 y + y 2 .
(a) Find the Hessian matrix for f at the point (1, 0).
(b) Compute the directional second derivative of f at the point (1, 0) in the direction of the
unit vector (3/5, 4/5).
2. Let f : R3 → R be the function f (x, y, z) = x2 y − 2xy + x2 − 2x − yz.
(a) Find the critical point for f .
(b) Use the Second Derivative Test to determine whether this critical point is a local maximum, a local minimum, or a saddle point.
3. Let f : R2 → R2 be the function f(x, y) = (2x3 + y 3 , x2 + 2y 2 ). Then the set of critical points
for f is a union of three lines in R2 . Find the equations of these lines.
4. Let f : R3 → R2 be the function f(x, y, z) = (x, x2 + y 2 + z 2 ).
(a) Find the set of critical points for f.
(b) Describe the level surfaces for the functions f1 and f2 . How are the critical points
reflected in the geometry of the level surfaces?
Solutions
fxx fxy
2xy x2
0 1
1. (a) Hf =
=
, so H(1,0) f =
fyx fyy
x2 2
1 2
0 1 3/5
56
(b) 3/5 4/5
=
1 2 4/5
25
2. (a) We have ∇f = (2xy − 2y + 2x − 2, x2 − 2x − z, −y), so we get the equations
2xy − 2y + 2x − 2 = 0,
x2 − 2x − z = 0,
−y = 0.
The only solution to these equations is the point (1, 0, −1) .




2 0
0
2y + 2 2x − 2 0
0
−1, so H(1,0,−1) f = 0 0 −1. This matrix has
(b) We have Hf = 2x − 2
0 −1 0
0
−1
0
eigenvalues −1, 1, and 2, so this critical point is a saddle point .
6x2 3y 2
3. The derivative of f is Df =
, so
2x 4y
det(Df) = 24x2 y − 6xy 2 = 6xy(4x − y).
Setting this equal to 0, we find that the critical points for f are the lines x = 0 , y = 0 , and
y = 4x .
1 0 0
4. (a) We have Df =
. The critical points occur when the two rows of this matrix
2x 2y 2z
are linearly dependent, which is when y = z = 0. Thus the set of critical points of f is
the x-axis .
(b) The level surfaces for f1 are planes parallel to the yz-plane, while the level surfaces for
f2 are spheres centered at the origin. The f1 -planes are tangent to the f2 -spheres at the
critical points along the x-axis.