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Thread: 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

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