## anonymous 5 years ago (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.

1. anonymous

ok is the square root only in the numerator?

2. anonymous

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

3. anonymous

if you're doing inflection points, you should have covered the multiplication rule and chain rule by now. just treat it as $\sqrt{x}*(1+x+x^3)^{-1}$

4. anonymous

ok i get one critical number at x = .4725

5. anonymous

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

6. anonymous

hmmm i'll try and do it again. every time i get a different answer. can i see the derivation process. of $\sqrt{x} \times (1+ x+ x ^{3}) ^{-1}$

7. anonymous

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

8. anonymous

thanks :)

9. anonymous

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

10. 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.