Calculus :

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11. [0.03l0.09 Points] 5.5m: 3.3....001. 7 2000 7 3000 74000 7 5000 7 6000 Video Example-4 ASK YOUR TEACHER PRACTICE ANOTHER EXAMPLE 1 Find where the function f(x) = 3X 8x3 210x2 + 5 is increasing and where it is decreasing. SOLUTION f'(x) = 12x3 24x2 420x = 12x(x )(X + To use the [/0 Test, we have to know where f'(x) > O and where f'(x) 7 + + + increasing w on (7, co) The graph of fshown in the gure conrms the information in the chart. EXAMPLE 2 Find the local minimum and maximum values of the function below. f(x) = 3x4 4x3 120x2 + 6 Video Example SOLUTION f'(x) = 12x3 12x2 240x = 12x(x 5)(x + 4) From the chart Interval 12X X 5 x + 4 f'(X) f x 5 + + + + increasing on (5, 00) we see that f'(x) changes from negative to positive at 4, so f(4) = is a local minimum value by the First Derivative Test. Similarly, f'(x) changes from negative to positive at 5, so f(5) = is a local minimum value. And, f(0) = is a local maximum value because f'(x) changes from positive to negative at 0. Need Hem? 14. [-/0.06 Points] DETAILS SESSCALC2 3.1.AE.004. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER EXAMPLE 4 The graph of the function f(x) = 3x4 - 16x3 + 18x2 -1 5x 5 4 (-1, 37) is shown in the figure. You can see that f(1) = is a local maximum whereas the absolute maximum is f . (This absolute maximum is not a local maximum because it occurs at an endpoint.) Also, f(0) = is a local minimum and f(3) = is both a (1, 5) local and an absolute minimum. Note that f has neither a local nor an absolute maximum at x = 4. -2 -1 3 (3, -27) Video Example () Need Help? Read It 15. [-/0.16 Points] DETAILS 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 3 5x5 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 ) = 5 10 -100 Since f'(x) exists for all x, the only critical numbers of f occur when f'(x) = , that is, x = 0 or X = . Notice that each of these critical numbers lies in (- 3, 12 ). The values of f at these critical numbers are f(0 ) = and f(6 ) = The values of f at the endpoints of the interval are 1( - 2) = and f(12) = Comparing these four numbers, we see that the absolute maximum value is f(12) = and the absolute minimum value is f(6) = Note that in this example the absolute maximum occurs at an endpoint, whereas the absolute minimum occurs at a critical number. The graph of fis sketched in the figure.2. [-/0.21 Points] DETAILS 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.) Find the interval on which f is decreasing. (Enter your answer using interval notation.) (b) Find the local minimum and maximum values of f. local minimum value local maximum value (c) Find the inflection point. ( x, y) = ( Find the interval on which f is concave up. (Enter your answer using interval notation.) Find the interval on which f is concave down. (Enter your answer using interval notation.) Need Help? Read ItConsider 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 1'. local minimum value |:| local maximum value (c) Find the inflection points. (X, y) = (|:| ) (smaller xvalue) (x, y) = (|:| ) (larger x-value) Find the interval on which fis concave up. (Enter your answer using interval notation.) E Find the interval on which fis concave down. (Enter your answer using interval notation.) S Find the local maximum and minimum values off using both the First and Second Derivative Tests. f(x) = 9 + 6x2 4x3 Need Help? _ local maximum value local minimum value In each part state the X-coordinates of the inflection points of f. Give reasons for your answers. Y (a) 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 concave up and concave down J (b) The curve is the graph of)\". (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) X= DNE X Give the reason for your answer. f will have an inflection point whenever I\" has a local minimum or local maximum J (c) The curve is the graph of f\". (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) X= DNE X Give the reason for your answer. fwill have an inflection point whenever f" changes between positive and negative J Need Help