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cant get this Question... pls help   f(x)=2 sin x cos x ;  f'(x) = 2 sin x (-sin x) +...

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cant get this Question... pls help


f(x)=2 sin x cos x ;

 f'(x) = 2 sin x (-sin x) + 2 cos x cos x


-- using the tool of calculus, it can be shown that function f(x) given, the zeroes of f'(x) give the location of any maximum and/or minimum values. find the location of these values in the inteval [0,2`pi` ], using trig identities as needed to solve f'(x) = 0


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We are given f(x)=2sinxcosx and f'(x)=2sinx(-sinx)+2cosxcosx; we are asked to find the zeros of f'(x) in the interval `[0,2pi]` :

f'(x) is the first derivative of f(x); a theorem of calculus states that local extrema (maximums and minimums) can only occur where f'(x)=0 or f'(x) fails to exist. Thus by finding the zeros of the first derivative, we can locate maximums and minimums.

Set f'(x) equal to zero and solve for x:

2sinx(-sinx)+2cosxcosx=0  Factor out the common 2:


Replace `cos^2x-sin^2x` with cos2x using trig identity



We want to solve this on `[0,2pi]` .

cos2x=0 ==> `2x=pi/2,(3pi)/2,(5pi)/2,(7pi)/2`


`==> x=pi/4,(3pi)/4,(5pi)/4,(7pi)/4`


Thus the original function f(x)=2sinxcosx has maximums/minimums that can only occur at these four points.

The graph of f(x):

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