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Math Help - Extrema problem

  1. #1
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    Extrema problem

    Problem: Given f(x) = x^3 - ax^2 + 3x + b,


    (a) Determine conditions on a and b so that f(x) has exactly one critical point.

    (b) Could f(x) have no critical points? If yes, determine the necessary
    conditions on a and b. If no, explain why not.

    (c) Determine conditions on a and b so that f(x) has an inflection point.




    What I've Done:
    Could anybody help me with this? All I've done so far is try to take the derivative, and I got:

    f\prime(x) = 3x^2 -2ax + 3

    Then I set that to 0, and realized that this would give me more than one critical point, so I didn't bother to try and solve for it.


    Could somebody walk me through HOW to solve each of the parts of the question, and explain what you do?

    Thank you SO much!
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  2. #2
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    Quote Originally Posted by lysserloo View Post
    Problem: Given f(x) = x^3 - ax^2 + 3x + b,


    (a) Determine conditions on a and b so that f(x) has exactly one critical point.

    (b) Could f(x) have no critical points? If yes, determine the necessary
    conditions on a and b. If no, explain why not.

    (c) Determine conditions on a and b so that f(x) has an inflection point.




    What I've Done:
    Could anybody help me with this? All I've done so far is try to take the derivative, and I got:

    f\prime(x) = 3x^2 -2ax + 3

    Then I set that to 0, and realized that this would give me more than one critical point, so I didn't bother to try and solve for it.


    Could somebody walk me through HOW to solve each of the parts of the question, and explain what you do?

    Thank you SO much!
    You are on the right track, just find what a has to be so that f\prime(x) = 3x^2 -2ax + 3, has only one solution (hint: if the discriminant is 0 then there is only 1 solution)
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  3. #3
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    ...What is a discriminant?

    EDIT: I was kind of hoping somebody could walk me through how to do this problem. I really have no idea where to start or where to go from where I am.
    Last edited by lysserloo; November 4th 2009 at 01:00 AM.
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  4. #4
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    The solution to basic 2-nd degree polynomial given by

    ax^2+bx+c=0

    is

    x=\frac{-b\pm\sqrt{b2-4ac}}{2a}

    The discriminant is the thing under the square root

    D=b^2-4ac

    In your case you have

    3x^2-2ax+3=0


    so the discriminant is

    D=(-2a)^2-4\cdot(3\cdot3)

    What does a have to be so that

    4a^2-36=0

    ?
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  5. #5
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    I got a = +- 3

    Problem is...the question isn't asking for a value of a or b. It's asking for limitations. How is a= +-3 giving me a limitation? And if it's plus OR minus 3, isn't that more than on critical point still?
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  6. #6
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    Quote Originally Posted by lysserloo View Post
    I got a = +- 3

    Problem is...the question isn't asking for a value of a or b. It's asking for limitations. How is a= +-3 giving me a limitation? And if it's plus OR minus 3, isn't that more than on critical point still?
    Remember that where f(x) has critical points f'(x)=0
    Well, if a=+- 3 then what are the roots of f'(x) ??
    It will have only one root, so f(x) has only one critical point.
    I would say that a=+- 3 is plenty of limitations, notice though that the value of b plays no part.

    Think a little bit about it, you dont even recognise that you have the solution.
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