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    anyone?
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    (Original post by jordanwu)
    Ok thanks guys I got to the answer.
    So the next part is: find the value of x at which the gradient of the curve is 1/12, giving your answer to 3sf.
    So 1/12=(1/2 + cos (x/2)/(2 + cos (x/2))^2 , but I keep getting stuck on rearranging. Any tips?
    Quadratic

    \dfrac{1}{12}=\dfrac{\frac{1}{2} + cos (\frac{x}{2})}{(2 + cos (\frac{x}{2}))^2}
    \left( 2 + cos (\frac{x}{2}) \right)^2=12(\frac{1}{2} + cos \left( \frac{x}{2}) \right)

    Expand, solve for cos(x/2) and then x.
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    (Original post by keromedic)
    Quadratic

    \dfrac{1}{12}=\dfrac{\frac{1}{2} + cos (\frac{x}{2})}{2 + cos (\frac{x}{2})^2}
    \left( 2 + cos (\frac{x}{2}) \right)^2=12(1/2 + )cos (\frac{1}{2}x)^2

    Expand, solve for cos(x/2) and then x.
    Hmm I got cos^2 (x/2) - 6cos (x/2) + 4 = 0
    No idea how xD...
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    (Original post by keromedic)
    Quadratic

    \dfrac{1}{12}=\dfrac{\frac{1}{2} + cos (\frac{x}{2})}{(2 + cos (\frac{x}{2}))^2}
    \left( 2 + cos (\frac{x}{2}) \right)^2=12(\frac{1}{2} + cos \left( \frac{x}{2}) \right)

    Expand, solve for cos(x/2) and then x.
    Hang on I've now got cos^2(x/2) - 8cos(x/2) - 2 = 0, how would you factorise that?
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    (Original post by jordanwu)
    Hang on I've now got cos^2(x/2) - 8cos(x/2) - 2 = 0, how would you factorise that?
    I'm not really following your working but assuming you've done everything right so far..

    How would you factorize t^2-8t-2?
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    (Original post by keromedic)
    I'm not really following your working but assuming you've done everything right so far..

    How would you factorize t^2-8t-2?
    By completing the square/quadratic formula because it can't be factorised?
    So by cts I got x=4plusorminus sqrt(18), so cos(x/2)=4plusorminus sqrt(18)??
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    (Original post by keromedic)
    I'm not really following your working but assuming you've done everything right so far..

    How would you factorize t^2-8t-2?
    Am I in the right direction or completely wrong with my answer?
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    (Original post by jordanwu)
    By completing the square/quadratic formula because it can't be factorised?
    So by cts I got x=4plusorminus sqrt(18), so cos(x/2)=4plusorminus sqrt(18)??
    Yes, that looks right

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    How would I then determine the range of values of x for which y is increasing, leaving my answer in terms of pi?
    dy/dx>0 when y is increasing?
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    (Original post by jordanwu)
    How would I then determine the range of values of x for which y is increasing, leaving my answer in terms of pi?
    dy/dx>0 when y is increasing?
    Yes.
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    (Original post by SeanFM)
    Yes.
    So do I set the expression for dy/dx > 0? Where would I get the pi from?
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    (Original post by jordanwu)
    So do I set the expression for dy/dx > 0? Where would I get the pi from?
    The pi bit will become clear when you set dy/dx > 0, then think about the denominator, then the numerator.
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    (Original post by SeanFM)
    The pi bit will become clear when you set dy/dx > 0, then think about the denominator, then the numerator.
    I'm sorry but I'm not sure what you mean, am I meant to rearrange or solve for x or?
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    (Original post by jordanwu)
    I'm sorry but I'm not sure what you mean, am I meant to rearrange or solve for x or?
    Remember that dy/dx > 0 means that dy/dx is always positive.

    You do require an inequality from the expression for dy/dx but it's not from the whole fraction. Then look back at my previous hint.
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    (Original post by SeanFM)
    Remember that dy/dx > 0 means that dy/dx is always positive.

    You do require an inequality from the expression for dy/dx but it's not from the whole fraction. Then look back at my previous hint.
    Well am I meant to find x to find the critical value?
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    (Original post by jordanwu)
    Well am I meant to find x to find the critical value?
    You don't need to do that, it's a bit more simple than that.

    When is dy/dx a positive number? In terms of the expression of dy/dx given in this question, I'm not looking for 'when y is increasing wrt x'.)
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    (Original post by SeanFM)
    You don't need to do that, it's a bit more simple than that.

    When is dy/dx a positive number? In terms of the expression of dy/dx given in this question, I'm not looking for 'when y is increasing wrt x'.)
    dy/dx is positive when the gradient is positive which is the same thing.. sry I really don't know
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    Not to worry. If I said that \frac{dy}{dx} = \frac{a}{b^2}, where a and b are variables that can be any positive or negative number when is dy/dx positive? (For what values of a and/or b?)

    (Original post by jordanwu)
    dy/dx is positive when the gradient is positive which is the same thing.. sry I really don't know
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    (Original post by SeanFM)
    Not to worry. If I said that \frac{dy}{dx} = \frac{a}{b^2}, where a and b are variables that can be any positive or negative number when is dy/dx positive? (For what values of a and/or b?)
    The numerator would have to be positive for dy/dx to be positive, the denominator can be either positive or negative as squaring either a +ve or -ve number is still +ve?
 
 
 
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