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Math Help - critical number?

  1. #1
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    critical number?

    determine the value of a for which the function f has no critical number:

    f(x)= (a2+a-6)cos2x+(a-2)x+cos1
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  2. #2
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    Re: critical number?

    What do you find when you equate the derivative to zero? Is the last term actually cos(1)?
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    Re: critical number?

    yes the last term is actually cos(1) and how do you take the derivative and set it equal to zero? is a just a constant?
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    MHF Contributor MarkFL's Avatar
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    Re: critical number?

    Yes, you would treat a as a constant.
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    Re: critical number?

    i got the derivative being

    -2a^2sin2x-2asin2x+12sin2x+a-2 and then i dont know how to take the derivative of cos1 and i dont know how to set this derivative equal to 0 with the constants in it?
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: critical number?

    cos(1) is a constant, so its derivative is zero. I would write the derivative as:

    f'(x)=-2(a^2+a-6)\sin(2x)+(a-2)=-2(a+3)(a-2)\sin(2x)+(a-2)=

    (a-2)(-2(a+3)\sin(2x)+1)=0

    What condition on a causes the second factor to have no root?
    Last edited by MarkFL; November 6th 2012 at 12:22 PM.
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  7. #7
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    Re: critical number?

    i dont see anything where you said i would write the derivative as:
    Quote Originally Posted by MarkFL2 View Post
    cos(1) is a constant, so its derivative is zero. I would write the derivative as:
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  8. #8
    MHF Contributor MarkFL's Avatar
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    Re: critical number?

    I accidentally hit Post Quick Reply rather than Go Advanced...I have edited my post.
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  9. #9
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    Re: critical number?

    i dont know what you mean about the root and i still dont understand how to set it equal to zero and solve for whatever like that?
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    MHF Contributor MarkFL's Avatar
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    Re: critical number?

    A root is a value that causes an expression to be zero. Anyway, consider:

    -1\le\sin(2x)\le1

    Now, equate the second factor to zero, solve for \sin(2x) and substitute into the above inequality. This will tell the the values of a for which there are critical numbers.
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  11. #11
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    Re: critical number?

    so i got a=2 and then the second part (a+3)sin(2x)=1/2 and i dont know how to simplify it any further
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  12. #12
    MHF Contributor MarkFL's Avatar
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    Re: critical number?

    When a=2, your function is constant, so there will be no critical values.

    As for the second part, solve for sin(2x), then as an alternate to what I suggested before, set the magnitude of this to greater than 1. You will have 2 cases to consider:

    i) \frac{1}{2(a+3)}>1

    ii) \frac{1}{2(a+3)}<-1
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  13. #13
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    Re: critical number?

    i dont understand
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  14. #14
    MHF Contributor MarkFL's Avatar
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    Re: critical number?

    We found:

    \sin(2x)=\frac{1}{2(a+3)}

    We therefore know that if \left|\frac{1}{2(a+3)} \right|>1 then there is no solution, that is, there is no real value of a which will give us a critical value.
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  15. #15
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    Re: critical number?

    well how do you solve for a?
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