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Math Help - local extrema problem?

  1. #1
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    local extrema problem?

    y=x square root of 8-x^2


    I did the derivative, -2x^2 square root of 8-x^2/2 square root 8-x^2
    but I don't know if it's right.

    Find the intervals on which the function is:
    a) increasing?
    b) decreasing?
    c) concave up?
    d) concave down?
    e) local extreme values?
    f) inflection points?
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  2. #2
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    Quote Originally Posted by Seigi View Post
    y=x square root of 8-x^2


    I did the derivative, -2x^2 square root of 8-x^2/2 square root 8-x^2
    but I don't know if it's right.

    Find the intervals on which the function is:
    a) increasing?
    b) decreasing?
    c) concave up?
    d) concave down?
    e) local extreme values?
    f) inflection points?
    I assume that your function is:

    f(x)=x \cdot \sqrt{8-x^2}

    Since the radicand must be positive or zero the domain is: [-\sqrt{8},\ \sqrt{8}]

    To calculate the first derivation you have to use the product rule:

    f'(x)=\sqrt{8-x^2} \cdot 1 + x\cdot \frac12 (8-x^2)^{-\frac12} \cdot (-2x) = \dfrac{8-2x^2}{\sqrt{8-x^2}} with D_{f'}=(-\sqrt{8},\ \sqrt{8})

    To calculate the concavity you need the second derivation. Use quotient rule:

    f''(x)=\dfrac{\sqrt{8-x^2} \cdot (-4x)-(8-2x^2)\cdot \frac12 (8-x^2)^{-\frac12}\cdot (-2x)}{8-x^2} = \dfrac{(8-x^2)\cdot (-4x)+x(8-2x^2)}{(\sqrt{8-x^2})^3}= \dfrac{2x^3-24x}{(\sqrt{8-x^2})^3}
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