anonymous
  • anonymous
(I really need help with this) The function F has first derivative given by f'(x) = (square root) X / 1 + x +x^3. What is the x- coordinate of the inflection point of the graph of F? (use calculator) I know its the second derivate. but can't get it.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
ok is the square root only in the numerator?
anonymous
  • anonymous
yea. Square root of X so its basically. x^(1/2)
anonymous
  • anonymous
if you're doing inflection points, you should have covered the multiplication rule and chain rule by now. just treat it as \[\sqrt{x}*(1+x+x^3)^{-1}\]

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anonymous
  • anonymous
ok i get one critical number at x = .4725
anonymous
  • anonymous
also think 0 makes it undefined so it would be a critial number as well.
anonymous
  • anonymous
hmmm i'll try and do it again. every time i get a different answer. can i see the derivation process. of \[\sqrt{x} \times (1+ x+ x ^{3}) ^{-1}\]
anonymous
  • anonymous
edit critical numbers above. should be -.1393 and 1.0080...
anonymous
  • anonymous
thanks :)
anonymous
  • anonymous
ok, (1.0080, 0.3311) sould be an inflection point, curve changes from - to + concavity.
anonymous
  • anonymous
I get undef for above & below -.1393. Seems like it's been forever since I was in Calculus I, hope the answer is correct.

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