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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

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  1. anonymous
    • 5 years ago
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    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).

  2. anonymous
    • 5 years ago
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    what is the local minimum

  3. anonymous
    • 5 years ago
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    the local mininum, naturally, is f(2)=8-16ln 2

  4. anonymous
    • 5 years ago
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    i got the point (2,-2.5) but she circled the -2.5 and i dont understand that

  5. anonymous
    • 5 years ago
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    it's not -2.5, 8-16ln2≈-3.090 , not -2.5

  6. anonymous
    • 5 years ago
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    duh...lol

  7. anonymous
    • 5 years ago
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    could you help me on one more like this?

  8. anonymous
    • 5 years ago
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    \[f(x)= 4x ^{3} + 3x ^{2} +2\]

  9. anonymous
    • 5 years ago
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    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

  10. anonymous
    • 5 years ago
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    okay i understand that

  11. anonymous
    • 5 years ago
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    great. so just now did you want me to work out the intervals that are increasing or decreasing?

  12. anonymous
    • 5 years ago
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    yes and the local extrema please

  13. anonymous
    • 5 years ago
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    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.

  14. anonymous
    • 5 years ago
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    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,+∞)

  15. anonymous
    • 5 years ago
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    okay so far?

  16. anonymous
    • 5 years ago
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    hold on working on it

  17. anonymous
    • 5 years ago
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    so it is decreasing at 0,0 and increasing at 0,0

  18. anonymous
    • 5 years ago
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    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 ↗

  19. anonymous
    • 5 years ago
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    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

  20. anonymous
    • 5 years ago
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    oh! now i get you

  21. anonymous
    • 5 years ago
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    (^ ^) love to communicate with you.

  22. anonymous
    • 5 years ago
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    what do you mean?

  23. anonymous
    • 5 years ago
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    what?

  24. anonymous
    • 5 years ago
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    love to communicate with you?

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