# What are the points of inflection, if any, of #f(x) = 5xcos^2x − 10sinx^2 # on #x in [0,2pi]#?

graph{-20cos(x)sin(x)+10x(sinx)^2-10x(cosx)^2-20cos(x^2)+40x^2sin(x^2) [-0, 6.283, -1800, 1800]}

They are:

graph{10sin(x^2)–5x(cosx)^2 [0, 6.283, -15, 40]}

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To find the points of inflection, first find the second derivative of the function. Then, solve for where the second derivative equals zero and check the concavity of the function around those points.

The second derivative of ( f(x) = 5x\cos^2(x) - 10\sin^2(x) ) is:

( f''(x) = -20\sin(x)\cos(x) - 20\sin(x)\cos(x) - 10\cos(2x) )

Next, find where the second derivative equals zero:

( -20\sin(x)\cos(x) - 10\cos(2x) = 0 )

( -20\sin(x)\cos(x) - 10(2\cos^2(x) - 1) = 0 )

( -20\sin(x)\cos(x) - 20\cos^2(x) + 10 = 0 )

( -20\sin(x)\cos(x) - 20(1 - \sin^2(x)) + 10 = 0 )

( -20\sin(x)\cos(x) - 20 + 20\sin^2(x) + 10 = 0 )

( -20\sin(x)\cos(x) + 20\sin^2(x) - 10 = 0 )

( -20\sin(x)(\cos(x) - \sin(x)) - 10 = 0 )

( -20\sin(x)\sin(x - \pi/4) - 10 = 0 )

There are no solutions for ( x ) in the interval ( [0,2\pi] ). Hence, there are no points of inflection within this interval.

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