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hiramoby

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

  • one year ago
  • one year ago

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  1. Australopithecus
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    Take the derivative of this function first

    • one year ago
  2. hiramoby
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    here we go

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

    • one year ago
  4. hiramoby
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    I got 0

    • one year ago
  5. hiramoby
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    -32

    • one year ago
  6. Australopithecus
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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

    • one year ago
  7. hiramoby
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    so it is going to be e then

    • one year ago
  8. Australopithecus
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    ok make a table |dw:1351050419295:dw|

    • one year ago
  9. Australopithecus
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    check in between these numbers and right if they are positive or negative

    • one year ago
  10. hiramoby
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    no b is the answer

    • one year ago
  11. Australopithecus
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    you need to check to see if they are maximums or minimums

    • one year ago
  12. Australopithecus
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    and I'm showing you how

    • one year ago
  13. Australopithecus
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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

    • one year ago
  14. Australopithecus
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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|

    • one year ago
  15. Australopithecus
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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

    • one year ago
  16. Australopithecus
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    I mean minimum

    • one year ago
  17. Australopithecus
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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)

    • one year ago
  18. Australopithecus
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    I hope this was helpful :S

    • one year ago
  19. Australopithecus
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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)

    • one year ago
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