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hiramoby Group Title

Find all relative extrema of the function F(X)=16/(X^2+1). Use the Second-Derivative Test when applicable. Answer a. The relative maximum is .1,0 b. The relative minimum is .0,1 c. The relative maximum is .0,16 d. The relative minimum is .0,16 e. The relative maximum is .16,0

  • 2 years ago
  • 2 years ago

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  1. Australopithecus Group Title
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    Take the derivative of this function first

    • 2 years ago
  2. hiramoby Group Title
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    here we go

    • 2 years ago
  3. Australopithecus Group Title
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    First set f'(x) = 0 then solve for x, those will give you your critical points

    • 2 years ago
  4. hiramoby Group Title
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    I got 0

    • 2 years ago
  5. hiramoby Group Title
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    -32

    • 2 years ago
  6. Australopithecus Group Title
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    also make sure to check the domain of f'(x) and the domain of f(x) if something is not in the domain of f'(x) but is in the domain of f(x) then it is a critical point if a point is not in the domain of f(x) but is in the domain of f'(x) then it is not a critical point

    • 2 years ago
  7. hiramoby Group Title
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    so it is going to be e then

    • 2 years ago
  8. Australopithecus Group Title
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    ok make a table |dw:1351050419295:dw|

    • 2 years ago
  9. Australopithecus Group Title
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    check in between these numbers and right if they are positive or negative

    • 2 years ago
  10. hiramoby Group Title
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    no b is the answer

    • 2 years ago
  11. Australopithecus Group Title
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    you need to check to see if they are maximums or minimums

    • 2 years ago
  12. Australopithecus Group Title
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    and I'm showing you how

    • 2 years ago
  13. Australopithecus Group Title
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    Oh I made a mistake the table should be |dw:1351050692097:dw| so the general rule is 1. If f'(x) > 0 on the Interval than f(x) is increasing on the interval 2. If f'(x) < 0 on the Interval than f(x) is decreasing on the interval

    • 2 years ago
  14. Australopithecus Group Title
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    So to show you an example of this -32x/(x^4 + 2x^2 + 1) sub in a number in the interval (-infinity, 0) so I'm going with -1 -32(-1)/((-1)^4 + 2(-1)^2 + 1) = a positive number so we add in the table |dw:1351050909010:dw|

    • 2 years ago
  15. Australopithecus Group Title
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    then we do it again for the next interval (0, -32) so I will pick 1 f'(1) = -32(1)/((1)^4 + 2(1)^2 + 1) = a negative number |dw:1351051038432:dw| so we see from this table that 0 is a maximum

    • 2 years ago
  16. Australopithecus Group Title
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    I mean minimum

    • 2 years ago
  17. Australopithecus Group Title
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    we can insert 0 into f(x) to see where this minimum is f(0) = 16/(X^2+1) f(0) = 16/1 f(0) = 16 so we have a minimum at (0,16)

    • 2 years ago
  18. Australopithecus Group Title
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    I hope this was helpful :S

    • 2 years ago
  19. Australopithecus Group Title
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    oh crap made a mistake lol |dw:1351051431080:dw| it should be this you are right it is e maximum at (0,16)

    • 2 years ago
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