Calculus : please put a box on the answer because I get confused easily and mix them up.

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1. [0.01/0.31 Points] DETAILS PREVIOUS ANSWERS SESSCALC2 3.3.001.MI.SA. MY NOTES ASK YOUR TE This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Consider the equation below. f(x) = 2x3 + 3x2 - 120x Exercise (a) Find the interval on which f is increasing. Find the interval on which f is decreasing. Click here to begin! Exercise (b) Find the local minimum and maximum values of f. Part 1 of 2 The function f(x) changes from increasing to decreasing at x = -5. Therefore, f(-5) = IS a maximum E V . Submit Skip (you cannot come back). Exercise (c) Find the inflection point. Find the interval on which fis concave up. Find the interval on which fis concave down. Part 1 of 3 We have f'(x) = 6x2 + 6x - 120, so f"(x) = which equals 0 when (x, f(x)) = Submit Skip (you cannot come back).15. [0.12/0.16 Points] DETAILS PREVIOUS ANSWERS SESSCALC2 3.1.AE.006. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER EXAMPLE 8 Find the absolute maximum and minimum values of the function below. 400 f(x) = x3 - 9x2 + 1 2SX S 12 300 SOLUTION Since f is continuous on - 3, 12 , we can use the Closed Interval Method: 200 f(x) = x3 - 9x2 + 1 100 f' ( x ) = 3x2- 18x 5 10 -100 Since f'(x) exists for all x, the only critical numbers of f occur when f'(x) = 0 , that is, x = 0 or x = 0 X . Notice that each of these critical numbers lies in (- , 12 ). The values of f at these critical numbers are f(0) = 1 and f(6) = -107 The values of f at the endpoints of the interval are f( - 2) = X and f(12) = 433 Comparing these four numbers, we see that the absolute maximum value is f(12) = 433 and the absolute minimum value is f(6) = -107 Note that in this example the absolute maximum occurs at an endpoint, whereas the absolute minimum occurs at a critical number. The graph of f is sketched in the figure.2. [0.06/0.21 Points] DETAILS PREVIOUS ANSWERS SESSCALC2 3.3.002. Consider the equation below. f(x) = 4x3 + 15x2 - 150x + 5 (a) Find the intervals on which f is increasing. (Enter your answer using interval notation.) (-00, -5) U (2.5,00) Find the interval on which f is decreasing. (Enter your answer using interval notation.) (-5, 2.5) (b) Find the local minimum and maximum values of f. local minimum value X local maximum value X (c) Find the inflection point. ( x, y) = Find the interval on which fis concave up. (Enter your answer using interval notation.) Find the interval on which fis concave down. (Enter your answer using interval notation.)Consider the equation below. f(x) = 7 sin(x) + 7 cos(x), 0 S X S 2:: (a) Find the interval on which fis increasing. (Enter your answer using interval notation.) E Find the interval on which fis decreasing. (Enter your answer using interval notation.) S (b) Find the local minimum and maximum values of f. (c) Find the inflection points. local minimum value local maximum value (X, y) = ( 3% X )(smaller x-value) (X, y) = ( 9% X )(Iarger x-value) Find the interval on which fis concave up. (Enter your answer using interval notation.) 3_7r Ln g, 4 ' 4 Find the interval on which fis concave down. (Enter your answer using interval notation.) Need Help? _' _' _' In each part state the x-coordinates of the inflection points off. Give reasons for your answers. Y (a) The curve is the graph off. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) Give the reason for your answer. f Will have an inflection point whenever f changes between concave up and concave down J (b) The curve is the graph of f'. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) Give the reason for your answer. f will have an inflection point whenever i\" has a local minimum or local maximum y (c) The curve is the graph of f\". (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) Give the reason for your answer. f will have an inflection point whenever f" changes between positive and negative J 7. [0.02/0.06 Points] DETAILS PREVIOUS ANSWERS SESSCALC2 3.2.009. MY NOTES ASK Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = 2x2 - 5x + 1, [0, 2] Yes, it does not matter if f is continuous or differentiable, every function satifies the Mean Value Theorem. Yes, f is continuous on [0, 2] and differentiable on (0, 2) since polynomials are continuous and differentiable on R. No, f is not continuous on [0, 2]. No, f is continuous on [0, 2] but not differentiable on (0, 2). There is not enough information to verify if this function satifies the Mean Value Theorem. If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem. (Enter your answers as a comma-separated list. If it does not satisify the hypotheses, enter DNE). C = X Need Help? Read It 8. [0/0.05 Points] DETAILS PREVIOUS ANSWERS SESSCALC2 3.2.023. MY NOTES ASK .If. f(2.).....5.and.f '(x) 2 3 for 2 s x S 4, how small can f(4) possibly be? EX Enter mber