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Math Help - Critical Points

  1. #1
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    Critical Points

    What are the critical point(s) of h(x)=(9-4x^2)^(1/2) ?
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  2. #2
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    Re: Critical Points

    The critical points are where \displaystyle \begin{align*} h'(x)  = 0 \end{align*}.
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  3. #3
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    Re: Critical Points

    after setting the derivative to 0 I got x=0 is that correct?
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    Re: Critical Points

    What did you get for the derivative?
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    Re: Critical Points

    I used my calculator for that I got ( -4x ) / ( -4 ( x^2 - 2.25) ^ (1/2) )
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  6. #6
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    Re: Critical Points

    Well if you were going to do it by hand, do it this way...

    \displaystyle \begin{align*} h &= \left( 9 - 4x^2 \right)^{\frac{1}{2}} \end{align*}

    Let \displaystyle \begin{align*} u = 9 - 4x^2 \end{align*} so that \displaystyle \begin{align*} h = u^{\frac{1}{2}} \end{align*}.

    Then \displaystyle \begin{align*} \frac{du}{dx} = -8x \end{align*} and \displaystyle \begin{align*} \frac{dh}{du} = \frac{1}{2}u^{-\frac{1}{2}} = \frac{1}{2\left( 9 - 4x^2 \right) ^{\frac{1}{2}}} \end{align*}

    So \displaystyle \begin{align*} \frac{dh}{dx} = \frac{-8x}{2 \left( 9 - 4x^2 \right)^{\frac{1}{2}} } = \frac{-4x}{\left( 9 - 4x^2 \right)^{\frac{1}{2}}} \end{align*}

    Yes, you are correct that when \displaystyle \begin{align*} \frac{dh}{dx} = 0 \end{align*} then \displaystyle \begin{align*} x = 0 \end{align*}. What is the value of h when x = 0?
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