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    (Original post by jordanwu)
    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?
    Correct! :borat:

    So you would say that the numerator has to be >0, right?

    Now how does that example relate to the question at hand?

    I've just realised that it doesn't technically matter as the square root of the denominator will never be negative, but it can be in some questions and you can still apply the same logic regardless.
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    (Original post by SeanFM)
    Correct! :borat:

    So you would say that the numerator has to be >0, right?

    Now how does that example relate to the question at hand?

    I've just realised that it doesn't technically matter as the square root of the denominator will never be negative, but it can be in some questions and you can still apply the same logic regardless.
    Well so I have 1/2 + cos(x/2)>0, so x/2=cos^-1(-1/2), x=4.1887...
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    That's the critical value, so how do you find the range of values (remember the domain given in the question!)

    (Original post by jordanwu)
    Well so I have 1/2 + cos(x/2)>0, so x/2=cos^-1(-1/2), x=4.1887...
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    (Original post by SeanFM)
    That's the critical value, so how do you find the range of values (remember the domain given in the question!)
    Well when I've done these types of questions I usually get 2 critical values and then sketch a graph to see where the function is increasing/decreasing, but I only have one here
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    (Original post by jordanwu)
    Well when I've done these types of questions I usually get 2 critical values and then sketch a graph to see where the function is increasing/decreasing, but I only have one here
    Remember that 0<x<2pi.so 0<x/2<pi. So you're looking of values for cosx/2 > -1/2 between 'actual inputs' of 0 and pi. So if you sketch a graph of cos between 0 and pi and look for y = -1/2, what do you notice?
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    (Original post by SeanFM)
    Remember that 0<x<2pi.so 0<x/2<pi. So you're looking of values for cosx/2 > -1/2 between 'actual inputs' of 0 and pi. So if you sketch a graph of cos between 0 and pi and look for y = -1/2, what do you notice?
    I'm got a graph of y=cos x on my calculator and the x values for y= -1/2 are:

    x=2.094... and x=4.1887...
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    (Original post by jordanwu)
    I'm got a graph of y=cos x on my calculator and the x values for y= -1/2 are:

    x=2.094... and x=4.1887...
    You've drawn cosx between 0 and 2pi, but we're not interested in anything past pi because if x/2 > pi then x > 2pi, but in the question it states that 0 < x < 2pi.

    Which is what I was saying in my previous post.
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    (Original post by SeanFM)
    You've drawn cosx between 0 and 2pi, but we're not interested in anything past pi because if x/2 > pi then x > 2pi, but in the question it states that 0 < x < 2pi.

    Which is what I was saying in my previous post.
    So is this just another trig question? So because I have x/2 instead of x I have to divide the domain by 2, so the new domain is 0<x/2<pi, and so I only use x/2=2.094... then I have x=4.1887...???
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    (Original post by jordanwu)
    So is this just another trig question? So because I have x/2 instead of x I have to divide the domain by 2, so the new domain is 0<x/2<pi, and so I only use x/2=2.094... then I have x=4.1887...???
    Yes, but the question doesn't want the critical point, it wants a range of values of x for which dy/dx > 0. So what is the range of values?
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    (Original post by SeanFM)
    Yes, but the question doesn't want the critical point, it wants a range of values of x for which dy/dx > 0. So what is the range of values?
    That's the bit I'm confused about..
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    (Original post by jordanwu)
    That's the bit I'm confused about..
    I can't say much more without giving it away or going in circles, but I'll try.

    (Original post by SeanFM)
    Remember that 0<x<2pi.so 0<x/2<pi. So you're looking of values for cosx/2 > -1/2 between 'actual inputs' of 0 and pi. So if you sketch a graph of cos between 0 and pi and look for y = -1/2, what do you notice?
    My advice would be to sketch (or use your calculator if you wish) costheta between 0 and pi, look at theta = 4pi/3 which is when costheta = - 1/2 (and since x/2 = theta, x = 8pi/3) and look at all the theta values to the left of that point and their y values - whether they are all greater than -1/2 or less than -1/2, etc.
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    (Original post by SeanFM)
    I can't say much more without giving it away or going in circles, but I'll try.



    My advice would be to sketch (or use your calculator if you wish) costheta between 0 and pi, look at theta = 4pi/3 which is when costheta = - 1/2 (and since x/2 = theta, x = 8pi/3) and look at all the theta values to the left of that point and their y values - whether they are all greater than -1/2 or less than -1/2, etc.
    Hmm but what do I write down? This question is worth 3 marks.
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    (Original post by jordanwu)
    Hmm but what do I write down? This question is worth 3 marks.
    My guess is 1st mark : recognising that it's the numerator that you're concerned with and rearranging.

    2nd mark - finding the critical value.

    3rd mark - writing down the correct answer, but you haven't found that yet.
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    Please can someone help me with this question:

    The function is y=e^x sin 2x. I've found dy/dx to be e^x(2cos 2x + sin 2x). I am asked to show that the equation of the normal to the curve y=e^x sin 2x at the point where
    x=pi is 2e^(pi)y + x=pi.

    I've tried plugging in x=pi into dy/dx to get e^pi(2cos2pi + sin2pi) which is the gradient of the tangent so the gradient of the normal is -1/e^pi(2cos2pi + sin2pi). I have tried many times already to rearrange to get the above expression but it's just not working out unfortunately. Please could someone help? Thanks.
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    (Original post by jordanwu)
    Please can someone help me with this question:

    The function is y=e^x sin 2x. I've found dy/dx to be e^x(2cos 2x + sin 2x). I am asked to show that the equation of the normal to the curve y=e^x sin 2x at the point where
    x=pi is 2e^(pi)y + x=pi.

    I've tried plugging in x=pi into dy/dx to get e^pi(2cos2pi + sin2pi) which is the gradient of the tangent so the gradient of the normal is -1/e^pi(2cos2pi + sin2pi). I have tried many times already to rearrange to get the above expression but it's just not working out unfortunately. Please could someone help? Thanks.
    cos 2pi = 1, sin 2pi = 0.
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    (Original post by SeanFM)
    My guess is 1st mark : recognising that it's the numerator that you're concerned with and rearranging.

    2nd mark - finding the critical value.

    3rd mark - writing down the correct answer, but you haven't found that yet.
    But 4pi/3 isn't between 0 and pi right?
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    (Original post by jordanwu)
    But 4pi/3 isn't between 0 and pi right?
    Correct. But x is between 0 and 2pi, and x = 4pi/3. x/2 is between 0 and pi, and cos(2pi/3) = -1/2.

    So if x/2 is less than 2pi/3 then x is less than 4pi/3.
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    (Original post by Zacken)
    cos 2pi = 1, sin 2pi = 0.
    Thanks
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    (Original post by jordanwu)
    Thanks
    Welcome.
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    (Original post by SeanFM)
    Correct. But x is between 0 and 2pi, and x = 4pi/3. x/2 is between 0 and pi, and cos(2pi/3) = -1/2.

    So if x/2 is less than 2pi/3 then x is less than 4pi/3.
    Well the y values are increasing as x increases (after 4pi/3) until 2pi then the y values start to decrease from 1
 
 
 
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