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anonymous

  • 5 years ago

critical point of 2x^4 - 4x^2+3

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  1. anonymous
    • 5 years ago
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    Take the derivative of 2x^4-4x^2+3, and set it equal to zero. You will get 2 values of x. x=0, x=1 These are the critical points of 2x^4-4x^2+3

  2. anonymous
    • 5 years ago
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    Don't forget to fan :$

  3. anonymous
    • 5 years ago
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    i get the derivative: 8x^3 - 8x is this correct?

  4. anonymous
    • 5 years ago
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    Yes, this is correct.

  5. anonymous
    • 5 years ago
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    then: 8x(x+1)(x-1) = 0 is the correct? How do you get to the point x=0, x=1

  6. anonymous
    • 5 years ago
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    8x=0 ;x=0 x+1=0 ;x=-1 x-1=0 ;x=1 So you actually get three values for x!

  7. anonymous
    • 5 years ago
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    thanks alot!

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