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Math Help - Finding the maximum

  1. #1
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    Finding the maximum

    How would you find the maximum value and smallest +ve value the maximum occurs at for f(x) = -cosx + sqrt3sint ?

    I've got as far as

    f'(x) = sqrt3cosx + sinx

    but I'm not sure where to go now?
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  2. #2
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    Quote Originally Posted by JQ2009 View Post
    How would you find the maximum value and smallest +ve value the maximum occurs at for f(x) = -cosx + sqrt3sint ?

    I've got as far as

    f'(x) = sqrt3cosx + sinx

    but I'm not sure where to go now?
    Set f'=0, then combine the two trig terms into one. Solve for x to find test points for critical values.
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  3. #3
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    Quote Originally Posted by JQ2009 View Post
    How would you find the maximum value and smallest +ve value the maximum occurs at for f(x) = -cosx + sqrt3sint ?

    I've got as far as

    f'(x) = sqrt3cosx + sinx

    but I'm not sure where to go now?
    Here's a completely different way to do that problem. A standard trig identity is cos(a+b)= cos(a)cos(b)- sin(a)sin(b). If we take b= x (I am assuming that is "sqrt 3 sin x", not "sqrt 3 sin t".) we have f(x)= (-1)cos x+ sqt(3) sin(x)= cos(a)cos(x)- sin(a)sin(x) so we would have to have cos(a)= -1 and sin(a)= -sqrt(3).

    Well that's impossible because sqrt(3)> 1 and we would have to have cos^2(a)+ sin^2(a) which is not true: (-1)^2+ sqrt(3)^2= 4, not 1. So multiply and divide by the square root of that, 2:

    f(x)= -cos(x)+ \sqrt{3}sin(x)= 2(-(1/2)cos(x)+ \sqrt{3}/2 sin(x) and now we must have cos(a)= -1/2 and sin(a)= -sqrt(3)/2 which says that a= -\pi/3.

    That is, f(x)= -cos(x)+ \sqrt{3}sin(x)= 2 cos(x- \pi/3). And it is easy to find maximum and minimum values for that.
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