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anonymous
 5 years ago
For each of the given functions, find the intervals on which f(x) is increasing, the intervals on which f(x) is decreasing, and the coordinates of all local extrema.
f(x)= 2x^2  16ln x
anonymous
 5 years ago
For each of the given functions, find the intervals on which f(x) is increasing, the intervals on which f(x) is decreasing, and the coordinates of all local extrema. f(x)= 2x^2  16ln x

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0according to the function, x∈(0,+∞), its derivative should be f'(x)=4x16/x. let f'(x)=4x16/x=0, x=2. the change of x, f'(x), and f(x) is as following: x (0,2) 2 (2,+∞) f'(x)  0 + f(x) decrease 816ln 2 increase from above, we can see that the interval on which f(x) is increasing is (2,+∞), and the interval on which f(x) is decreasing is (0,2).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is the local minimum

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the local mininum, naturally, is f(2)=816ln 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got the point (2,2.5) but she circled the 2.5 and i dont understand that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it's not 2.5, 816ln2≈3.090 , not 2.5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0could you help me on one more like this?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f(x)= 4x ^{3} + 3x ^{2} +2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'd love to. well, first we should remember the basic derivatives!? for example, f(x)=x^a, then f'(x)=ax^(a1). f(x)=c (c is a constant), then f'(x)=0. also the algorithm of derivatives, such as [f(x)±g(x)]'=f'(x)±g'(x). so f(x)=4x^3+3x^2+2, f'(x)=12x^2+6x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay i understand that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0great. so just now did you want me to work out the intervals that are increasing or decreasing?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes and the local extrema please

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay. first, it may be a good habit to do a factoring with the derivative (at least my teacher told me like this) because it will be convenient to calculate. f'(x)=12x^2+6x=x(12x+6). next, we are sure that the derivative at local extrema should be 0. let f'(x)=x(12x+6)=0. so you can work out the solutions of x here very fast, x1=0, x2=1/2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0notice here that the domain of definition of the function is R. split it into intervals and points according to the solutions of x, as (∞,1/2), 1/2, (1/2,0), 0 , (0,+∞)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hold on working on it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so it is decreasing at 0,0 and increasing at 0,0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the change of x, f'(x), and f(x) is as following: x (∞,1/2) 1/2 (1/2,0) 0 (0,+∞) f'(x) + 0  0 + f(x) ↗ 9/4 ↘ 2 ↗

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the +/ of the f'(x) stands for whether f'(x) is positive or negative. when f'(x)>0, increases; when f'(x)<0, decreases. so it is directly perceived that the increasing intervals are (∞,1/2) , (0,+∞); the decreasing interval is (1/2,0). the maximum value is f(1/2)=9/4 and the minimum value is f(0)=2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(^ ^) love to communicate with you.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0love to communicate with you?
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