anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
1st derive is a product rule i think in part
amistre64
  • amistre64
f'(x)=-.5x(2x)(x^2+2)^-.5+(x^2+2)^-.5
amistre64
  • amistre64
forgot to subtract 1 from my expo lol

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More answers

amistre64
  • amistre64
f'(x)=-.5x(2x)(x^2+2)^-1.5+(x^2+2)^-.5
amistre64
  • amistre64
just got to keep a good eye about you on those
anonymous
  • anonymous
i still dont see how i would set that equal to 0.
amistre64
  • amistre64
thats only the first derivative
amistre64
  • amistre64
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 :)
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=y%3Dx%28x%5E2%2B2%29%5E-.5%2B3 looks good to me
anonymous
  • anonymous
by definition POI (point of inflection is when f"(x) = 0.
amistre64
  • amistre64
not quite
amistre64
  • amistre64
f''=0 is a critical point to check, but it doesnt gaurentee inflection
amistre64
  • amistre64
f'' = undefined are also point to check
anonymous
  • anonymous
oh right, thanks for reminding
amistre64
  • amistre64
:) it happens lol
amistre64
  • amistre64
f''=0 is neccessary, but not sufficient
anonymous
  • anonymous
thanks for your help! it took me awhile to get what you did, but it made sense. :)

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