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  1. #1
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    Post maximum value

     y = {x^{\frac{2}<br />
{3}}}(x - 4)

    What is the maximum value that this function takes on the interval [0,5]. (not plug and chug)
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  2. #2
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    Quote Originally Posted by genlovesmusic09 View Post
     y = {x^{\frac{2}<br />
{3}}}(x - 4)

    What is the maximum value that this function takes on the interval [0,5]. (not plug and chug)
    do a function analysis using the first derivative test to look for extrema on the given interval.
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    Quote Originally Posted by skeeter View Post
    do a function analysis using the first derivative test to look for extrema on the given interval.
    I found the derivative to be \frac{2}<br />
{3}{x^{\frac{{ - 1}}<br />
{3}}}(x - 4) + {x^{\frac{2}<br />
{3}}}(1)<br />
    i then set it equal to zero and found x=1.6
    and plugged 1.6 into the original equation and found the minimum of (about) -3.283
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    Quote Originally Posted by genlovesmusic09 View Post
    I found the derivative to be \frac{2}<br />
{3}{x^{\frac{{ - 1}}<br />
{3}}}(x - 4) + {x^{\frac{2}<br />
{3}}}(1)<br />
    i then set it equal to zero and found x=1.6
    and plugged 1.6 into the original equation and found the minimum of (about) -3.283
    however ...

    What is the maximum value that this function takes on the interval [0,5].
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    i know that is where i am lost
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  6. #6
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    Quote Originally Posted by genlovesmusic09 View Post
     y = {x^{\frac{2}<br />
{3}}}(x - 4)

    What is the maximum value that this function takes on the interval [0,5]. (not plug and chug)
    y = x^{\frac{5}{3}} - 4x^{\frac{2}{3}}

    y' = \frac{5}{3}x^{\frac{2}{3}} - \frac{8}{3}x^{-\frac{1}{3}}<br />

    note that x = 0 is a critical value

    set y' = 0

    \frac{5}{3}x^{\frac{2}{3}} - \frac{8}{3}x^{-\frac{1}{3}} =  0

    multiply every term by 3x^{\frac{1}{3}} ...

    5x - 8 = 0

    x = \frac{8}{5}

    for 0 < x < \frac{8}{5} , y'(x) < 0 ... y is decreasing

    for \frac{8}{5} < x < 5 , y'(x) > 0 ... y is increasing

    since the function decreases, then increases, that says the function has maxima at the endpoints. comparing the function values at the endpoints ...

    y(0) = 0

    y(5) = 5^{\frac{2}{3}} > 0

    y(5) = 5^{\frac{2}{3}} is the maximum over the interval.
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