James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 14.7 Maximum and Minimum Values 967
We close this section by giving a proof of the first part of the Second Derivatives Test.
Part (b) has a similar proof.
Proof of Theorem 3, Part (a) We compute the second-order directional derivative of
f in the direction of u − kh, kl. The first-order derivative is given by Theorem 14.6.3:
D u f − f x h 1 f y k
Applying this theorem a second time, we have
D 2 u f − D u sD u f d − − −x sD u f dh 1 − −y sD u f dk
− s f xx h 1 f yx kdh 1 s f xy h 1 f yy kdk
− f xx h 2 1 2 f xy hk 1 f yy k 2 (by Clairaut’s Theorem)
If we complete the square in this expression, we obtain
10 D 2 u f − f xxSh 1 f xy
f xx
2
kD 1 k 2
s f xx f yy 2 f xyd
2
f xx
2
We are given that f xx sa, bd . 0 and Dsa, bd . 0. But f xx and D − f xx f yy 2 f xy are continuous
functions, so there is a disk B with center sa, bd and radius . 0 such that
f xx sx, yd . 0 and Dsx, yd . 0 whenever sx, yd is in B. Therefore, by looking at Equation
10, we see that D u fsx, yd . 0 whenever sx, yd is in B. This means that if C is
2
the curve obtained by intersecting the graph of f with the vertical plane through
Psa, b, f sa, bdd in the direction of u, then C is concave upward on an interval of length
2. This is true in the direction of every vector u, so if we restrict sx, yd to lie in B, the
graph of f lies above its horizontal tangent plane at P. Thus f sx, yd > f sa, bd whenever
sx, yd is in B. This shows that f sa, bd is a local minimum.
■
1.
Suppose s1, 1d is a critical point of a function f with continuous
second derivatives. In each case, what can you say
about f ?
(a) f xxs1, 1d − 4, f x ys1, 1d − 1, f yys1, 1d − 2
(b) f xxs1, 1d − 4, f x ys1, 1d − 3, f yys1, 1d − 2
reasoning. Then use the Second Derivatives Test to confirm your
predictions.
3. f sx, yd − 4 1 x 3 1 y 3 2 3xy
y
2. Suppose (0, 2) is a critical point of a function t with continuous
second derivatives. In each case, what can you say
about t?
(a) t xxs0, 2d − 21, t x ys0, 2d − 6, t yys0, 2d − 1
(b) t xxs0, 2d − 21, t x ys0, 2d − 2, t yys0, 2d − 28
(c) t xxs0, 2d − 4, t x ys0, 2d − 6, t yys0, 2d − 9
1
3.2
3.7
3–4 Use the level curves in the figure to predict the location of
the critical points of f and whether f has a saddle point or a local
maximum or minimum at each critical point. Explain your
_1
2
1
0
3.7
3.2
_1
4
4.2
5 6
1
x
7et1407x03
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