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Math Help - calculate smallest and highest value of the function

  1. #1
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    calculate smallest and highest value of the function

    calculate smallest and highest value of the function y=x^3-2x*|x-2| on segment [0,3]
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  2. #2
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    You check the values of the function at the endpoints, that is at the points x=0 & x=3 .
    Now, since your function is not differentiable at x=2, you cannot use the derivative at x=2.

    So if you limit yourself to (0,2) and (2,3) when performing you differentiate and look for max/min, and check the function value at x=2.

    Now you compare all the values you have. That is, you have your endpoint values, your value at x=2, and possibly some values at local max/min which you may have found during the differentiation.
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  3. #3
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    Quote Originally Posted by ecnanif View Post
    You check the values of the function at the endpoints, that is at the points x=0 & x=3 .
    Now, since your function is not differentiable at x=2, you cannot use the derivative at x=2.

    So if you limit yourself to (0,2) and (2,3) when performing you differentiate and look for max/min, and check the function value at x=2.

    Now you compare all the values you have. That is, you have your endpoint values, your value at x=2, and possibly some values at local max/min which you may have found during the differentiation.
    Differentiating is not an option, this is not a calculus problem.

    CB
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  4. #4
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    Quote Originally Posted by Garas View Post
    calculate smallest and highest value of the function y=x^3-2x*|x-2| on segment [0,3]
    If x< 2, x-2< 0 so |x-2|= -(x-2)= 2- x. Then y= x^3- 2x(2- x)= x^3+ 2x^2- 4x= x(x^2+ 2x- 4). You can complete the square to determine where x^2+ 2x- 4 has max or min.

    If x> 2, x-2> 0 so |x-2|= x- 2. Then y= x^3-2x(x-2)= x^3- 2x^2+ 4x= x(x^2- 2x+ 4). Again, try completing the square in x^2- 2x+ 4.
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