Question

Find the coordinates of the inflection point of the function
f(x)=x(x^2+2)^-.5+3
i got the first derivative to be x(.5)(x^2+2)^-.5
and i think the second derivative is -.25x(x^2+2)^-1.5
how do i set that equal to 0?

Answer 1

1st derive is a product rule i think in part

Answer 2

f'(x)=-.5x(2x)(x^2+2)^-.5+(x^2+2)^-.5

Answer 3

forgot to subtract 1 from my expo lol

Answer 4

f'(x)=-.5x(2x)(x^2+2)^-1.5+(x^2+2)^-.5

Answer 5

just got to keep a good eye about you on those

Answer 6

i still dont see how i would set that equal to 0.

Answer 7

thats only the first derivative

Answer 8

f'(x)=-.5x(2x)(x^2+2)^-1.5+(x^2+2)^-.5

f''(x)=
-2x(x^2+2)^-1.5+3x^3(x^2+2)^-2.5-x(x^2+2)^-1.5


\[ (-3x(x^2+2)+3x^3)(x^2+2)^{-2.5}\]

\[ (-3x^3-6x+3x^3)(x^2+2)^{-2.5}\]

it looks like when x=0 to me if i did this right :)

Answer 9

http://www.wolframalpha.com/input/?i=y%3Dx%28x%5E2%2B2%29%5E-.5%2B3

looks good to me

Answer 10

by definition POI (point of inflection is when f"(x) = 0.

Answer 11

not quite

Answer 12

f''=0 is a critical point to check, but it doesnt gaurentee inflection

Answer 13

f'' = undefined are also point to check

Answer 14

oh right, thanks for reminding

Answer 15

:) it happens lol

Answer 16

f''=0 is neccessary, but not sufficient

Answer 17

thanks for your help! it took me awhile to get what you did, but it made sense. :)