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Math Help - Minimum value for the function:

  1. #1
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    Minimum value for the function:

    y = 2^[(x^2) - 2x]

    I need solution!!

    Thanks in advance
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  2. #2
    Behold, the power of SARDINES!
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    When does the exponential part of the equation have its minimum value?

    x^2-2x is minimized when? Why will this solve your problem?
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  3. #3
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    The question is: What is the minimum value of the function f(x) = 2^[(x^2) - 2x] ?

    We have to complete the square and then find the minimum value for y.

    im stuck here:

    y = 2 ^ [ ((x-1)^2) -1 ] , now how can i get rid of the 2?

    the answer is 2^(-1), i want the solution

    sry about all the brackets
    Last edited by mr fantastic; November 5th 2010 at 07:34 PM.
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  4. #4
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    Like the previous question you asked in the geometry section, calculus is involved.

    Differentiation is required for this one, when dy/dx = 0, x will be your minimum point.
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  5. #5
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    Quote Originally Posted by Sorena View Post
    The question is: What is the minimum value of the function f(x) = 2^[(x^2) - 2x] ?

    We have to complete the square and then find the minimum value for y.

    im stuck here:

    y = 2 ^ [ ((x-1)^2) -1 ] , now how can i get rid of the 2?

    the answer is 2^(-1), i want the solution

    sry about all the brackets
    You should know what an exponent is right? if not we have a problem.

    So the question I asked you is what when we find the minimum value of

    x^2-2x why does this make

    2^{x^2-2x} a minimum?

    Completing the square gives

    x^2-2x=x^2-2x+1-1=(x-1)^2-1

    The minimum value occurs when the term (x-1)^2=0

    This gives the smallest value of the quadratic as -1

    So the minimum function value is 2^{-1}=\frac{1}{2}
    Last edited by mr fantastic; November 5th 2010 at 07:36 PM.
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  6. #6
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    Thank you soooo much.

    (sry i didnt mean to be rude earlier, I thought you're signature was your solution to my problem, lol (dumb I know!))
    Last edited by mr fantastic; November 5th 2010 at 07:36 PM.
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