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Math Help - Local and Absolute Maximums/Minimums

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    Senior Member sakonpure6's Avatar
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    Local and Absolute Maximums/Minimums

    If given a local or absolute max/min value, can I use them to come up with the equation of an even or odd function using vertex form y= a(x-b)^n + c ??
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    Re: Local and Absolute Maximums/Minimums

    I'm not sure what you mean by that. It is true that, if n is even, (x- b)^n is never negative so that a(x- b)^n is never negative if a is positive and never positive if a is negative. Then (x- b)^n+ c is
    1) equal to c at x= b and, if a>0, larger than c for x not equal to b.
    2) equal to c at x= b and, if a< 0, less than c for x not equal to b.

    In the first case, the graph goes down to a "vertex" at (a, c) and back up again.
    In the second case, the graph goes up to a "vertex" at (a, c) and back down again.
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    Re: Local and Absolute Maximums/Minimums

    Quote Originally Posted by sakonpure6 View Post
    If given a local or absolute max/min value, can I use them to come up with the equation of an even or odd function using vertex form y= a(x-b)^n + c ??
    The equation you wrote y= a(x-b)^n + c must have n zeros. if all of them are real and without multiplicity, then you will have n-1 local extrema. Also, if the domain is R, then it will have two absolute extrema if if n<0, one absolute extrema.
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    Re: Local and Absolute Maximums/Minimums

    Quote Originally Posted by votan View Post
    The equation you wrote y= a(x-b)^n + c must have n zeros. if all of them are real and without multiplicity, then you will have n-1 local extrema. Also, if the domain is R, then it will have two absolute extrema if if n<0, one absolute extrema.
    But "if all of them are real and without multiplicity" does not occure here. The zeros are obviously x= b+ (-c/a)^{1/n}. If n is odd, that has exactly one real root. If n is even and -c/a is positive, there are exactly two real roots. Most of what you say here is irrelevant to the question asked.

    (And, once again, I am not clear whether or not you are doing this intentionally!)
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