MAC 2311C Lecture 6: 4.1: Minimum and maximum values

EXERCISES lin the difference between an absolute minimum and a (c) Sketch the graph of a function that has a local maxmm at 2 and is not continuous at 2 Saypose fis a continuoes function defined on a closed 12. Ya) Sketch the graph of a function on 1-1,21 that has an val [a,b). maxinwm value and an absolute minimem value for f minimun vales? absolute maximum but no local maximum (b) Sketch the graph of a function on [-1,21 that has a loca ia) What theorem guarantees the existence of an absolute b) What steps would you take to find those maximum and 34 For esxch of the numbers a, b,c,d, r, and s, state whether the maximum but no absolute maximum. 13. (a) Sketch the graph of a function on -1,21 that has an absolute maximum but no absolute minimum (b) Sketch the graph of a function on [-1,21 that is dis funscticon whoue gragh is shown has an ahsolate maximum or min akval maximem or minimum, or neither a maximum continuous but has both an absolute maximum and an absolute minimum imim, 14. (a) Sketch the graph of a function that has two local max- ima, one local minimum, and no absolute minimum (b) Sketch the graph of a function that has three local min- ima, two local maxima, and seven critical numbers. 15-28 Sketch the graph of f by hand and use your sketch to (Use the graphs and transformations of Sections 1.2 and 1.3.) 15, f(x)-jar-1),パ3 find the absolute and local maximum and minimum values of f De s-6 Use te gragh to state the absolute and local maximam einimum vaues of the function 2o.fu) _ san x. 0

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.