# 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)...

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:

`2(-sin^2x+cos^2x)=0`

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

2cos2x=0

cos2x=0

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

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

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

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

The graph of f(x):