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Math Help - For what value of x does the function y = x + 2 cos x have a horizontal tangent line?

  1. #1
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    For what value of x does the function y = x + 2 cos x have a horizontal tangent line?

    For what value of x does the function y = x + 2 cos x have a horizontal tangent line?

    I said f'(x) = 1 + 0(cosx) + 2(-sinx) = 1 - 2sinx

    So if the line is horizontal, f'(x) must = 0, so 0 = 1 - 2sinx --> sin x = 1/2

    Now I am stuck. What value of x should I put? 30 degrees? pi divided by six?

    ???
    Last edited by mr fantastic; April 5th 2010 at 03:33 AM. Reason: Copied title into main body of post.
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    Quote Originally Posted by satx View Post
    I said f'(x) = 1 + 0(cosx) + 2(-sinx) = 1 - 2sinx

    So if the line is horizontal, f'(x) must = 0, so 0 = 1 - 2sinx --> sin x = 1/2

    Now I am stuck. What value of x should I put? 30 degrees? pi divided by six?

    ???
    better be \frac{\pi}{6}
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  3. #3
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    Quote Originally Posted by satx View Post
    I said f'(x) = 1 + 0(cosx) + 2(-sinx) = 1 - 2sinx

    So if the line is horizontal, f'(x) must = 0, so 0 = 1 - 2sinx --> sin x = 1/2

    Now I am stuck. What value of x should I put? 30 degrees? pi divided by six?

    ???
    Read the question carefully as it is written. (I'm not sure if you transcribed it exactly to us or not.)

    It may be asking for any value of x that satisfies the condition. If so, then you can list whichever solution you want.

    It may ask for the value(s) that occurs in a particular range (i.e. between 0 and pi). In this case you should find them all in that range.

    Or, it may ask you to list EVERY value of x that would work. For example, recalling trig, there are two solutions between 0 and 2pi:

    \sin\left(\tfrac{\pi}{6}\right)=\sin\left(\tfrac{5  \pi}{6}\right)=\tfrac{1}{2}

    But it is periodic, so we could list the solutions like so:

     \cdots , -\tfrac{11\pi}{6} , -\tfrac{7\pi}{6} , \tfrac{\pi}{6} , \tfrac{5\pi}{6} , \tfrac{13\pi}{6} , \tfrac{17\pi}{6} ,\cdots

    or, probably better:

    x=\tfrac{\pi}{6} + 2\pi n
    and
    x=\tfrac{5\pi}{6} + 2\pi n
    where n is an integer
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    No, I copied it word for word...
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    Quote Originally Posted by satx View Post
    I said f'(x) = 1 + 0(cosx) + 2(-sinx) = 1 - 2sinx

    So if the line is horizontal, f'(x) must = 0, so 0 = 1 - 2sinx --> sin x = 1/2

    Now I am stuck. What value of x should I put? 30 degrees? pi divided by six?

    ???
    You don't need to use the product rule to differentiate 2\cos(x) since differentiation is a linear process so if a and b are constants:

    \frac{d}{dx} \left[af(x)+bg(x)\right]=a \frac{df}{dx}+b\frac{dg}{dx}

    and for 2\cos(x):

    \frac{d}{dx} \left[af(x)\right]=a \frac{df}{dx}

    where a=2 and f(x)=\cos(x)

    CB
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  6. #6
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    Quote Originally Posted by satx View Post
    No, I copied it word for word...
    Well then, since it is singular (i.e., it says "for what value" and not "values"), I'd probably just list one solution -- probably the lowest positive value which was \pi/6.

    However it is sort of ambiguous as written so you might want to ask your teacher what he/she expects you to answer when it is worded like this. We can only guess if the problem is not worded more explicitly.
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