18.02 Lecture Notes - Lecture 1: Derivative Test, Saddle Point, Talking Lifestyle 1278

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amethystmuskrat145

13 Apr 2016

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Critical points points where the derivative is zero or not de ned. Second derivative test for functions with one variable. There is a test that can tell us local max or local min. Second derivative test for functions with two variables. We know that f = f = 0 at the critical point. Saddle point: x - y critical point at (0,0) Saddle point: x + y critical point at (0,0) Saddle point if a and c are different signs. Now say that b 0, try completing the square. Maximum if a is negative and ac-b is postive. Now f(x,y) f = f = 0, a = f , b=f , c=f. Example: 13. 10 #5 f = 2x + 2xy + y + 4x - 2y +1 f = 4x + 2y + 4 and f = 2x + 2y - 2 with a critical point at (-3,4)

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answers the Qs? 5 stars for corect and complete answers

This exercise is intended to give you some insight into why the second derivative test works. A good pan of what I'm asking you to do has been adapted form the Discovery Project on page 821 of our text. In 11.4, we wrote down the tangent plane or linear approximation of a function f(x, y) at a point. To simplify this exercise, we're going to assume that the point is located at the origin, x=0, y=0. Then At a critical point, the two partial derivatives are 0, so the linear approximation just simplifies f(x, y) f(0,0), which is the equation of a horizontal plane. This just doesn't give enough information to determine whether we have a max, min, or saddle. So we look at the quadratic approximation at (0,0), which is defined as follows: At a critical point, this will reduce to It can be proved that if the function f(x, y) has continuous second partials, then it will have the same type of critical point as its quadratic approximation. Let f(x, y) = e-x2-2y2 Find the first partial derivatives Find the first partial derivatives fx(x, y) and fy(x, y): Evaluate fx(0,0) and fy(0, 0). Find the second partial derivatives fxx(x, y), fxy(x, y) and fyy(x, y): Evaluate fxx, fxy and fyy at the critical point. In order to recognize the shape of this graph from its equation, it will be helpful to rewrite the equation in a different form so that: Complete the square for the quadratic approximation and state the values of a, b, c, and d. What docs the graph of the quadratic approximation look like? A: paraboloid opening up. B: paraboloid opening down, C: saddle. Judging from your knowledge of the shape of the graph in exercise 16, does the quadratic approximation have a max, min or saddle? Enter MAX, MIN, or SADDLE. Recall that in the second derivative test, at a critical point (a. b). For f(x, y), state the values of D and fxx. Does the second derivative test tell you that the critical point of f is a local maximum, local minimum, or saddle? After you've completed the lab, we'll take time out to see how the quadratic approximation relates to the second derivative test.

A. Find the critical points.

B. Use the second derivative test to

classify each critical point.

f (x, y) = âx^3 + 9xy â y^3

A. (1,1)and(0,0)

B. (3,3)and(0,0)

C. (1, 1) only
D. (-3, -3) and (0, 0)

E. (-3, -3) only

F. not listed

B. Which are true?

A. The critical points generate one relative maximum and one relative minimum. B. The critical points generate one relative maximum and one saddle point.

C. The critical points generate one relative minimum.
D. The critical points generate one relative minimum and one saddle point.

E. The second derivative test was inconclusive about all critical points.
F. None of the statements is true.

18.02 Lecture Notes - Lecture 1: Derivative Test, Saddle Point, Talking Lifestyle 1278

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