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anonymous

  • 5 years ago

Let f(x)=1/(8x^2+6). I need to find the intervals on which the function is concave up and where it is down. I keep getting imaginary numbers so it doesn't make sense. I need assistance

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  1. amistre64
    • 5 years ago
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    whats your 2nd derivative show you?

  2. amistre64
    • 5 years ago
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    f'(x) = -16x --------------- (8x^2 +6)^2

  3. amistre64
    • 5 years ago
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    at x = 0 there is a critical point, and probably and inflection.... better consult the 2nd derivative for concavity

  4. amistre64
    • 5 years ago
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    f''(x) = (8x^2 +6)^2 (-16) - (-16x)(2)(8x^2 +6)(16x) ---------------------------------------- [(8x^2 +6)^2]^2

  5. amistre64
    • 5 years ago
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    (0+6)^2 (-16) - 0(......) = 36(-16) = (-) it looks like x=0 is a Maximum value.....

  6. amistre64
    • 5 years ago
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    our inflections will be when that top of the 2nd derivative =0

  7. amistre64
    • 5 years ago
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    this making any sense to you?

  8. anonymous
    • 5 years ago
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    Yeah I got the x=o part and that it was a maximum. But why is the top of the second derivative equaling zero the inflection point. I thought when the second derivative equals zero is your point of inflection. That would mean that the bottom equals zero.

  9. amistre64
    • 5 years ago
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    the 2nd derivative = 0 indicates that the concavity of the graph changes at that point, it is no longer cave up or cave down at "0"; it is switching concavities. So the top of your 2nd derivative = 0 will produce the f''(x) = 0 effect... the bottom is just there for decoration ;)

  10. amistre64
    • 5 years ago
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    lets clean this top up so we can use it.. (8x^2 +6)^2 (-16) - (-16x)(2)(8x^2 +6)(16x) (-16)(64x^4 +96x^2 +36) + 512x^2(8x^2 +6)

  11. amistre64
    • 5 years ago
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    +4096x^4 -1024x^4 + 3072x^2 -1536x^2 -576

  12. amistre64
    • 5 years ago
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    3072x^4 + 1536x^2 -576 you see any common factors :)

  13. amistre64
    • 5 years ago
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    this is a disguised quadratic; a^2 = x^4; a = x^2 3072a^2 + 1536a -576

  14. anonymous
    • 5 years ago
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    okay i think i can take it from here. Thank you oh so much :D

  15. amistre64
    • 5 years ago
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    no problem; just be aware that any value you get for the new variable "a" needs to be equated to a^2 = x^4 and a = x^2... dont just think that "a" is the answer half way thru ;) youll do fine....

  16. anonymous
    • 5 years ago
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    One last question. If i use x=\[-b\pm\[\sqrt{?}\]b^2-4ac \div\]2a are the answers for x squared or can i even use that equation

  17. anonymous
    • 5 years ago
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    I meant the quadratic formula

  18. amistre64
    • 5 years ago
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    you can use that equation to get your answers for "a"...and then plug that value for "a" into the: x^4 = a^2 and x^2 = a equations. but I am not sure how well the quad formula works when you have common factors still in it, I think it might give you a false reading: try it on a simpler equation to find out like: 2x^2 +12x +18.... that is just 2(x^2 +6x +9) so if the answers are the same, your good to go, but if they are different, yo need to factor out the common number ;)

  19. anonymous
    • 5 years ago
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    We are good to go!

  20. amistre64
    • 5 years ago
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    yay!! ..simplicity at work lol

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