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- 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.

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- anonymous

- schrodinger

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- anonymous

ok is the square root only in the numerator?

- anonymous

yea. Square root of X so its basically. x^(1/2)

- 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

ok i get one critical number at x = .4725

- anonymous

also think 0 makes it undefined so it would be a critial number as well.

- 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

edit critical numbers above. should be -.1393 and 1.0080...

- anonymous

thanks :)

- anonymous

ok, (1.0080, 0.3311) sould be an inflection point, curve changes from - to + concavity.

- 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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