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Find the local maximum and minimum values of f using both the First and Second Derivative Tests. f(x) = 2 + 9x^{2} – 6x^{3}

EXPERT ANSWER f(x) = 2+9x^2-6x^3 f'(x) = 18x-18x^2 = 18x(1-x) = 0 x = 0 , 1 f”(x) = 18-36x f”(0) = 18 , at x = 0 local minimum f”(1) = -18 , at x = 1 local maximum f(0) = 2 , f(1) = 5 local maximum value = 5 . (Answer) local …

Find the local maximum and minimum values of f using both the First and Second Derivative Tests. f(x) = 2 + 9x^{2} – 6x^{3} Read More »

Hello! I need help with these questions. Please do not skip any steps because I keep getting them wrong and I want to see where I am making mistakes. thank you! use the first derivative and second derivative test to find all maximums and minimums of the function. f(x)=2x^3-9×62+12x-3. 6 use The First Derivative And Second Derivative Test To Find All Maximums And Minimums on the interval of the function 0 is less than or equal to theta less than or equal to 2pi. f(theta )=sin(theta)+cos(theta).

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18. When adding an entry to an array-based implementation of the ADT List at the end of the list a. no shift is necessary b. all of the previous elements must be shifted toward the front of the array c. only one element is shifted toward the end of the array to make room d. only one element is shifted toward the beginning of the array to make room 19. When inserting a node between two adjacent nodes in a chain you must locate a. the first node and the node before the insertion position b. the first node and the node after the insertion position c. the last node and the node after the insertion position d. the node before the insertion position and the node after the insertion position

EXPERT ANSWER When adding an entry to an array-based implementation of the ADT List at the end of the list no shift is necessary all of the previous elements must be shifted toward the front of the array only one element is shifted toward the end of the array to make room only one element …

18. When adding an entry to an array-based implementation of the ADT List at the end of the list a. no shift is necessary b. all of the previous elements must be shifted toward the front of the array c. only one element is shifted toward the end of the array to make room d. only one element is shifted toward the beginning of the array to make room 19. When inserting a node between two adjacent nodes in a chain you must locate a. the first node and the node before the insertion position b. the first node and the node after the insertion position c. the last node and the node after the insertion position d. the node before the insertion position and the node after the insertion position Read More »

Assume array size is 50 in a circular array implementation of a Queue; Assume frontIndex is 49 and backIndex is 46; After dequeueing three, ie removing 3 elements, what will the values of backIndex and FrotnIndex be: a. FrontIndex: 49 BackIndex: 46 b. FrontIndex: 47 BackIndex: 46 c. FrontIndex: 49 BackIndex: 44 d. FrontIndex: 2 BackIndex: 46 Answer: 16. In an array-based implementation of a Queue, if waitingList is the name of the array, and the index to the last entry in the queue is called backIndex, which assignment statement updates backIndex to add an entry to the queue: a. backIndex = (backIndex + 2) % waitingList.length; b. backIndex = (backIndex + 1) % waitingList.length; c. backIndex = (backIndex – 1) % waitingList.length; d. backIndex = (backIndex – 2) % waitingList.length;

EXPERT ANSWER Assume array size is 50 in a circular array implementation of a Queue; Assume frontIndex is 49 and backIndex is 46; After dequeueing three, ie removing 3 elements, what will the values of backIndex and FrotnIndex be: a. FrontIndex: 49 BackIndex: 46 b. FrontIndex: 47 BackIndex: 46 c. FrontIndex: 49 BackIndex: 44 d. …

Assume array size is 50 in a circular array implementation of a Queue; Assume frontIndex is 49 and backIndex is 46; After dequeueing three, ie removing 3 elements, what will the values of backIndex and FrotnIndex be: a. FrontIndex: 49 BackIndex: 46 b. FrontIndex: 47 BackIndex: 46 c. FrontIndex: 49 BackIndex: 44 d. FrontIndex: 2 BackIndex: 46 Answer: 16. In an array-based implementation of a Queue, if waitingList is the name of the array, and the index to the last entry in the queue is called backIndex, which assignment statement updates backIndex to add an entry to the queue: a. backIndex = (backIndex + 2) % waitingList.length; b. backIndex = (backIndex + 1) % waitingList.length; c. backIndex = (backIndex – 1) % waitingList.length; d. backIndex = (backIndex – 2) % waitingList.length; Read More »