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Math Help - Find the absolute maximum and absolute minimum

  1. #1
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    Find the absolute maximum and absolute minimum

    f(x)= 2x^4 - 16x^2 +3; on 0,2

    I got to 8x^3 -32x=0 what do I do now?

    absolute maximum=
    absolute minimum=
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  2. #2
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    Use the maximum-minimum principle:

    calculate f(x)'

    calculate a value c in the interval [0,2] such that f'(0)=c or doesnt exist

    list endpoints of interval: 0,c1,c2,c3....,2

    express f(x) for each value of c above

    The largest is absolute maximum and the smallest is absolute minimum

    OR you could just differentiate again OR just draw a graph to check your answer. Have you got mathcad or a similar graphing programme? There are online sites that calculate the graph based on an inputted function (but dont know if I can mention here - just google)
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  3. #3
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    Quote Originally Posted by moolimanj View Post
    Use the maximum-minimum principle:

    calculate f(x)'

    calculate a value c in the interval [0,2] such that f'(0)=c or doesnt exist

    list endpoints of interval: 0,c1,c2,c3....,2

    express f(x) for each value of c above

    The largest is absolute maximum and the smallest is absolute minimum

    OR you could just differentiate again OR just draw a graph to check your answer. Have you got mathcad or a similar graphing programme? There are online sites that calculate the graph based on an inputted function (but dont know if I can mention here - just google)
    What? I just need f'(x)
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  4. #4
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    Quote Originally Posted by bgwarhawk View Post
    f(x)= 2x^4 - 16x^2 +3; on 0,2

    I got to 8x^3 -32x=0 what do I do now?

    absolute maximum=
    absolute minimum=
    Quote Originally Posted by bgwarhawk View Post
    What? I just need f'(x)
    f'(x)=8x^3 -32x like you had it.

    Set f'(x)=0 and solve for x. That will give you the x value(s) where your local max/min occurs.

    But since f(x) is on an closed interval (I believe you meant to write [0,2]?), you also need to check the endpoints too.
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  5. #5
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    Quote Originally Posted by chabmgph View Post
    f'(x)=8x^3 -32x like you had it.

    Set f'(x)=0 and solve for x. That will give you the x value(s) where your local max/min occurs.

    But since f(x) is on an closed interval (I believe you meant to write [0,2]?), you also need to check the endpoints too.
    What do I do with the cube root of 4x?
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  6. #6
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    Quote Originally Posted by bgwarhawk View Post
    What do I do with the cube root of 4x?
    8x^3-32x=0
    8x(x^2-4)=0
    8x=0, x^2-4=0
    ...

    Check your work.
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