Lara P.

asked • 12/05/16

What are the apexes of this formula: f(x) = sin(x) x cos(x)

I don't know how to fix this problem. Can anyone help me?

1 Expert Answer

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Mark M. answered • 12/05/16

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sin(2x) = 2(sinx)(cosx), so, f(x) = (sinx)(cosx) 
 
                                             = (1/2)sin(2x)
 
f(x) has amplitude 1/2.  So, the maximum value of f(x) is 1/2 and the minimum value is -1/2.  
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Lara P.

I don't really understand how you did that. 
If i try to find out what f'(x) is, I get: f'(x) = -sin2(x) + cos2(x) (which is basically -1) 
And then I don't know how to calculate f'(x) = 0 
 
Also I need the exact coordinates of the maximum and minimums of the formula on (0, pi) 
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12/08/16

Mark M.

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sin(2x) = 2sinxcosx is the double angle formula for the sine function. Divide both sides by 2 to get sinxcosx = (1/2)sin(2x).  The number in front is the amplitude.  So, (1/2)sin(2x) is never larger than 1/2 and is never smaller than -1/2.
 
(1/2)sin(2x) = 1/2 when sin(2x) =1.  So, 2x = π/2 + 2kπ, k=0,1,-1,...
 
                                                          x = π/4+kπ
 
                                                             = π/4, 5π/4,...
 
(1/2)sin(2x) is minimized when sin(2x) = -1
 
                                                 2x = 3π/2 + 2kπ
 
                                                   x = 3π/4 +kπ
 
                                                      = 3π/4, 7π/4, ...
 
On [0, π], the maximum point is (π/4, 1/2) and the minimum point is   (3π/4, -1/2).  
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12/08/16

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