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Math Help - [SOLVED] local min and max

  1. #1
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    [SOLVED] local min and max

    let f(x)= (x-1)^(4/5) x find the local minimum and maximum
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  2. #2
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    Quote Originally Posted by cole123 View Post
    let f(x)= (x-1)^(4/5) x find the local minimum and maximum
    f(x) = (x - 1)^{\frac{4}{5}}x


    To find local maxima and minima, you need to find the derivative, set it equal to 0 and solve for x.


    So f'(x) = (x - 1)^{\frac{4}{5}}\frac{d}{dx}(x) + x\frac{d}{dx}[(x - 1)^{\frac{4}{5}}

     = (x - 1)^{\frac{4}{5}} + \frac{4}{5}x(x - 1)^{-\frac{1}{5}}.


    Can you go from here?
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  3. #3
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    I am on the next step :

    I have (4/5)x + (x-1)= 0

    I know there is both a min and a max just dont know what to do from here
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  4. #4
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    Quote Originally Posted by cole123 View Post
    I am on the next step :

    I have (4/5)x + (x-1)= 0

    I know there is both a min and a max just dont know what to do from here
    No.

    It's (x - 1)^{\frac{4}{5}} + \frac{4}{5}x(x - 1)^{-\frac{1}{5}} = 0.
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  5. #5
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    I guess I dont then!!
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  6. #6
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    Can you help me?
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  7. #7
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    Prove It, cole123 was heading in the right direction. After you multiply (x-1)^{\frac{4}{5}}+ \frac{4}{5}x(x-1)^{-\frac{1}{5}}= 0 by (x-1)^{1}{5}, you are left with (x-1)+ \frac{4}{5}x= 0.

    cole123, solve that: x+ \frac{4}{5}x= 1.

    Remember that the local max and min must be at a point where the derivative is 0, or where there is NO derivative.
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  8. #8
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    Quote Originally Posted by cole123 View Post
    I am on the next step :

    I have (4/5)x + (x-1)= 0

    I know there is both a min and a max just dont know what to do from here


    I did that and someone said it was wrong
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  9. #9
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    so is x = 5/9 my only answer? Ive only answered max/min for parabolas. and I believe there could be another answer?!
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  10. #10
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    Quote Originally Posted by HallsofIvy View Post
    Prove It, cole123 was heading in the right direction. After you multiply (x-1)^{\frac{4}{5}}+ \frac{4}{5}x(x-1)^{-\frac{1}{5}}= 0 by (x-1)^{1}{5}, you are left with (x-1)+ \frac{4}{5}x= 0.

    cole123, solve that: x+ \frac{4}{5}x= 1.

    Remember that the local max and min must be at a point where the derivative is 0, or where there is NO derivative.
    OK point taken.

    Except when you multiply both sides by (x - 1)^{\frac{1}{5}} the equation actually becomes

    (x - 1)^4 + \frac{4}{5}x = 0

    x^4 - 4x^3 + 6x^2 - 4x + 1 + \frac{4}{5}x = 0.


    Now simplify and solve for x.
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  11. #11
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    Well, whoever told you that was wrong!

    I suspect they graphed that on a calculator and saw no graph for x< 1.

    Many perfectly good calculators (like my TI 89) will not handle odd roots of negative numbers correctly because they do them using a logarithm which is not defined for x\le 0.

    If you graph y1= (x-1)^(4/5)*x on a TI calculator, for example, you will see a graph that begins at (1, 0) and rises to the right.

    If you also graph y2= (1- x)^(4/5)*x, you will see the rest of the graph which has maximum at x= 5/9.
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  12. #12
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    Quote Originally Posted by Prove It View Post
    OK point taken.

    Except when you multiply both sides by (x - 1)^{\frac{1}{5}} the equation actually becomes

    (x - 1)^4 + \frac{4}{5}x = 0

    x^4 - 4x^3 + 6x^2 - 4x + 1 + \frac{4}{5}x = 0.


    Now simplify and solve for x.
    No! (x- 1)^{\frac{4}{5}}(x-1)^{\frac{1}{5}}= (x- 1)^1, not (x- 1)^4.
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