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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.