# Determine where the function is increasing and decreasing and identify all relative extrema. g(x) = sin^2(x) cos^2(x) on the interval [0, 2pi]

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### 1 Answer

You need to perform first derivative test to identify the intervals where the function increases or decreases, hence you should find derivative of function such that:

`f'(x) = (sin^2(x))'*(cos^2(x)) - (sin^2(x))*(cos^2(x))` '

`f'(x) = 2sin x*cos x*(cos^2(x)) + 2sin x*cos x*(sin^2(x))`

You need to factor out `2sin x*cos x` such that:

`f'(x) = 2sin x*cos x*(cos^2(x) +sin^2(x))`

You need to remember that `(cos^2(x) + sin^2(x)) = 1` such that:

`f'(x) = 2sin x*cos x =gt f'(x) = sin 2x`

You need to solve the equation f'(x) = 0 such that:

`sin 2x = 0 =gt 2x = sin^(-1) 0`

The sine function is zero if `alpha = 0 ,alpha = pi ,alpha = 2pi ` such that:

`2x = 0 =gt x = 0`

`2x = pi =gt x = pi/2`

`2x = 2pi =gt x = pi`

You need to select a value for x in the quadrant 1 such that:

`x = pi/6 =gt sin 2*(pi/6) = sin pi/3 = sqrt3/2 gt 0`

**You need to notice that the function increases over `[0,pi]` and it decreases over `[pi,2pi], ` hence, the function reaches its maximum at `x = pi.` **