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Math Help - Movement Calc Problem, need help fast!!!!

  1. #1
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    Exclamation Movement Calc Problem, need help fast!!!!

    Consider the function f(x)=x^2-2ax-3x+a^2+3a, a-1 ≤ x ≤ a+3, a>0
    The absolute maximum value is __________
    and this occurs at equals _________
    The absolute minimum value is _____________
    and this occurs at equals ____________
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Ok here you go

    f(x)=x^2-2ax-3x+a^2+3a...then f'(x)=2x-2a-3...find the zeros... x=\frac{2a+3}{2}....it is undefined nowhere...and the original funtion is defined everywhere...off to the second derivative test.. f''(x)=2...so it is always positive so ... x=\frac{2a+3}{2} is a relative min...now we just need to check the endpoints.. f(a-1)=3a+1... f(a+3)=3a+9...and f(\frac{2a+3}{2})=a^2+9a+9...and even though its a relative min the max occurs at x=\frac{2a+3}{2} and the min at x=a-1
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  3. #3
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    Smile Thanks

    Thanks... I had nothing and was no where close to right!
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