Suppose f($) is some function and you want to determine the intervals were it is increasing and decreasing. You determine the derivative is f'(:c) = 3(3: 2) (:1: + 3)2. To find the critical values, you solve f'(a:) : O and come up with the critical values a: : 3, a: : 2. You divide the number line up into three intervals: :1: 2. On each interval, pick a test value and determine if the derivative is positive or negative, and use that to determine if the original function is increasing or decreasing. m) at test value Function behavior --_ -- Suppose f(:c) is some function, and you determine the derivative is f'(:c) = 5(:c + 4) (a: 3). Find the intervals on which the function is increasing and decreasing. Write the intervals using inequalities not including the endpoints. Intervals where f(:v) is increasing: Intervals where f(:v) is decreasing: Suppose f(:r:) is some function, and you determine the second derivative is f"(:r:) = 3(:I: 1) (a: 4). Find the intervals on which the function is concave up and concave down. Write the intervals using inequalities not including the endpoints. Intervals where f(a:) is concave up: Intervals where f(:r:) is concave down: 1 5:33 g:::2 + 7:1: + 9, 0"(m) = 21:2 7:1: + 7, and C\"(:v) : 3:1: 7 The x-value(s) of inflection point(s) of C(32) are a: = [: Concave Down interval 3:] O Concave Down interval does not exist. 0 Concave Up interval does not exist. Given C(53) Given the function f(:z:) = 21:3 + 33:2 122*: + 3, determine the intervals where the function is increasing and decreasing, and the intervals of concavity. Write the intervals using inequalities. f(m) is increasing for: [: at) is decreasing for: :] at) is concave up for: [: f(3:) is concave down for: [:1 Given 3; = 3:3 108:1: + 16 Find the critical number(s) of y. a: = C] O No critical values of y Find the x-value(s) of inflection point(s) of y. a: = C] O No inflection point(s) of y Given the function g(:r:) : 8:63 12:1:2 288w, find the first derivative, g'(:c). W) _ Notice that 9"(22) = 0 when m = 4, that is, g'(4) : 0. Now, we want to know whether there is a local minimum or local maximum at a: = 4, so we will use the second derivative test. Find the second derivative, g'( (.3) a\") = Evaluate g\"(4). M4) = Based on the sign of this number, does this mean the graph of 9(37) is concave up or concave down at a: = 4? At :1: = 4 the graph of g(:1:) is Based on the concavity of g(:1:) at a: = 4, does this mean that there is a local minimum or local maximum at a: : 4? At a: = 4 there is a local The function f(a:) = 2:c3 + 30:1:2 961: + 7 has one local minimum and one local maximum. This function has a local minimum at a: = [:1 and a local maximum at :1: = [j Consider the function f(m) = 4:1: + 5m'1. For this function there are four important open intervals: (00, A), (A, B),(B, C), and (0, 00) where A, and C are the critical numbers and the function is not defined at B. For each of the following open intervals, tell whether f(a:) is increasing or decreasing. (oo, A): (A3)- (B, 0): (0,00)- Let f(x) = (4 - 4x) 4 f (@ ) has one critical value at A = For x > A, f(x) is Select an answerThe function at) 2 5.7: + 5531 has one local minimum and one local maximum. This function has a local maximum at a: = :]