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Denebel

  • 4 years ago

What is the max value of the derivative of f9x) = 3x^2-x^3 ?

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  1. Rohangrr
    • 4 years ago
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    Given: y = x^3 - 3x^2 + 2 Therefore: dy/dx = 3x^2 - 3*2x = 3x(x - 2) Equating above expression for dy/dx to 0, we get two values of x. x 1 = 0 and x2 = 2 To get corresponding value of function, we substitute these values of y in equation for y. Thus: y1 = 0^3 - 3*0^2 + 2 = 2 y2 = 3^3 - 3*2^2 + 2 = 27 - 12 + 2 = 17 Answer: Maximum = 17 Minimum = 2

  2. Denebel
    • 4 years ago
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    Hmm.. the answer key says the minimum is 3, but I do not know how

  3. dpaInc
    • 4 years ago
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    |dw:1333626822884:dw|

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