anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
according to the function, x∈(0,+∞), its derivative should be f'(x)=4x-16/x. let f'(x)=4x-16/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 8-16ln 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
  • anonymous
what is the local minimum
anonymous
  • anonymous
the local mininum, naturally, is f(2)=8-16ln 2

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anonymous
  • anonymous
i got the point (2,-2.5) but she circled the -2.5 and i dont understand that
anonymous
  • anonymous
it's not -2.5, 8-16ln2≈-3.090 , not -2.5
anonymous
  • anonymous
duh...lol
anonymous
  • anonymous
could you help me on one more like this?
anonymous
  • anonymous
\[f(x)= 4x ^{3} + 3x ^{2} +2\]
anonymous
  • anonymous
i'd love to. well, first we should remember the basic derivatives!? for example, f(x)=x^a, then f'(x)=ax^(a-1). 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
  • anonymous
okay i understand that
anonymous
  • anonymous
great. so just now did you want me to work out the intervals that are increasing or decreasing?
anonymous
  • anonymous
yes and the local extrema please
anonymous
  • anonymous
okay. 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
  • anonymous
notice 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
  • anonymous
okay so far?
anonymous
  • anonymous
hold on working on it
anonymous
  • anonymous
so it is decreasing at 0,0 and increasing at 0,0
anonymous
  • anonymous
the 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
  • anonymous
the +/- 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
  • anonymous
oh! now i get you
anonymous
  • anonymous
(^ ^) love to communicate with you.
anonymous
  • anonymous
what do you mean?
anonymous
  • anonymous
what?
anonymous
  • anonymous
love to communicate with you?

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