Math 121 Winter 2010 Review Sheet

Transcription

1 Math 121 Winter 2010 Review Sheet March 14, 2010 This review sheet contains a number of problems covering the material that we went over after the third midterm exam. These problems (in conjunction with the problems in the previous review sheets) are meant to help you prepare for the final exam which is scheduled to take place on Thursday, March 18 th, 2010 from 8:00 am to 10:00 am. The answers to these problems are given at the end of the this review sheet for your convenience. Analysis of Functions II: Relative Extrema; Graphing Polynomials. 1. Use the given derivative to find all critical points of f, and at each critical point determine whether a relative maximum, relative minimum, or neither occurs. Assume in each case that f is continuous everywhere. (a) f (x) = x 2 (x 3 5). (b) f (x) = 2 3x 3 x+2. (c) f (x) = xe 1 x2. (d) f (x) = ln( 2 1+x 2 ). 2. (a) Show that f(x) = 1 x 5 and g(x) = 3x 4 8x 3 both have stationary points at x = 0. (b) What does the second derivative test tell you about the nature of these stationary points? (c) What does the first derivative test tell you about the nature of these stationary points? 3. Is the following statement true or false. If p(x) is a polynomial such that p (x) has a simple root at x = 1, then p has a relative extremum at x = Find the relative extrema for f(x) = sin 2x where 0 < x < π using both first and second derivative tests. 5. Use any method to find the relative extrema of the function f(x) = x. 1

2 6. Give a graph of the polynomial p(x) = x 4 6x and label the coordinates of the intercepts, stationary points, and inflection points. Absolute Maxima and Minima. 1. Determine whether the statement is true or false. Explain your answer. (a) If a function f is continuous on [a, b], then f has an absolute maximum on [a, b]. (b) If a function f is continuous on (a, b), then f has an absolute minimum on (a, b). (c) If a function f has an absolute minimum on (a, b), then there is a critical point of f in (a, b). (d) If a function f is continuous on [a, b] and f has no relative extreme values in (a, b), then the absolute maximum value of f exists and occurs either at x = a or at x = b. 2. Find the absolute maximum and minimum values of f, if any, on the given interval, and state where those values occur. (a) f(x) = x 2 x 2; (, + ). (b) f(x) = 4x 3 3x 4 ; (, + ). (c) f(x) = 2x 3 6x + 2; (, + ). (d) f(x) = x 2 x+1 ; ( 1, 5]. (e) f(x) = x 2/3 (20 x); [ 1, 20]. (f) f(x) = 1 + 1/x; (0, + ). (g) f(x) = 2x2 3x 3 x 2 2x+2 ; [1, + ). (h) f(x) = 2 cos x sin x ; [π/4, 3π/4]. { 4x 2, x < 1 (i) f(x) = (x 2)(x 3), x 1 ; [1/2, 7/2]. 3. Let f(x) = x 2 + px + q. Find the values of p and q such that f(1) = 3 is an absolute extremum of f on [0, 2]. Is this value an absolute maximum or an absolute minimum? 2

3 Solutions Analysis of Functions II: Relative Extrema; Graphing Polynomials. 1. (a) Critical Points x = 0, 5 1/3. x = 0 : neither; x = 5 1/3 relative minimum. (b) Critical Points x = 2, 2/3. x = 2 : relative minimum; x = 2/3 relative maximum. (c) Critical Points x = 0. x = 0 : relative minimum. (d) Critical Points x = 1, 1. x = 1 : relative minimum; x = 1 relative maximum. 2. (a) f (x) = 5x 4 and g (x) = 12x 3 24x 2 f (0) = g (0) = 0. (b) f (x) = 20x 3 and g (x) = 36x 2 48x f (0) = g (0) = 0, which yields no information. (c) There is no relative extremum at 0 for the function f nor the function g since there is no change in the signs of the derivatives f (x) and g (x) before and after True. By Theorem 4.2.5(c), the graph of p (x) crosses the x-axis at x = 1. By either case (a) or case (b) of Theorem 4.2.3, f has either a relative maximum or a relative minimum at x = x = 1: relative minimum. Absolute Maxima and Minima. 1. (a) True, by Theorem (b) False. A( counter example is the function f(x) = tan x on the open interval π 2, ) π 2. This function has no absolute minimum (and no absolute maximum) on this open interval. (c) True, by Theorem (d) True. The absolute maximum of f on [a, b] exists, by Theorem If it occurred in (a, b), then it would also be a relative maximum. Since f has no relative maximum in (a, b), the absolute maximum must occur at either x = a or x = b. 2. Find the absolute maximum and minimum values of f, if any, on the given interval, and state where those values occur. (a) The function f has no absolute maximum but it has an absolute minimum of -9/4 that occurs at the critical point x = 1/2. 3

4 5. (b) The function f has no absolute minimum but it has an absolute maximum of 1 that occurs at the critical point x = 1. (c) The function f has neither an absolute maximum nor an absolute minimum since lim x ± f(x) = ±. (d) The function f has no absolute minimum since lim x 1 + f(x) =. However, f has an absolute maximum of 1/2 at x = 5 since f is increasing on the interval ( 1, 5] because f (x) = 3/(x + 1) 2 > 0 for all x in ( 1, 5]. (e) f (x) = 5(8 x) 3x 1/3, f (x) = 0 when x = 8 and f (x) does not exist when x = 0; f( 1) = 21, f(0) = 0, f(8) = 48, f(20) = 0. So, the absolute maximum is 48 at x = 8 and the absolute minimum is 0 at x = 0, 20. (f) f has neither an absolute maximum nor an absolute minimum on the open interval (0, + ) because it has no critical points on that interval. However, we note here that f is bounded from below because f(x) > 1 for all x > 0. (g) f (x) = x(x 2) (x 2 2x+2) 2, and for 1 x <, f (x) = 0 when x = 2. Thus, f has one relative extremum on the interval [1, + ) which is a relative maximum. Therefore, f has an absolute maximum of 5/2 at x = 2. Moreover, since f(1) = 2 = lim x + f(x), it follows that f has an abslute minimum of 2 at x = 1. (h) f 1 2 cos x (x) = sin 2 x ; f (x) = 0 on [π/4, 3π/4] only when x = π/3. Then, f(π/4) = 2 2 1, f(π/3) = 3, and f(3π/4) = So, f has an absolute maximum value of at x = 3π/4 and an absolute minimum value of 3 at x = π/3. 4

5 { 4, x < 1 (i) f (x) = 2x 5, x > 1. So, f (x) = 0 when x = 5/2, and f (x) does not exist when x = 1 because lim x 1 + f(x) f(1) x 1 = 3 4 = f(x) f(1) lim x 1 x 1 ; f(1/2) = 0, f(1) = 2, f(5/2) = 1/4, f(7/2) = 3/4. So, the absolute maximum is 2 and the absolute minimum is 1/4. 3. f (x) = 2x+p which exists throughout the interval (0, 2) for all values of p so f (1) = 0 because f(1) is an extreme value. Thus, 2+p = 0 p = 2. Moreover, f(1) = ( 2)(1) + q = 3 q = 4. Thus, f(x) = x 2 2x + 4 and f(0) = 4, f(2) = 4 so f(1) is an absolute minimum. 5