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Question Number 199424 by cortano12 last updated on 03/Nov/23

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Answered by Frix last updated on 04/Nov/23

f (x) = \frac{\cos x}{3} ({6 \sin}^{3} x - {4 \sin}^{2} x + 1)

f^{'} (x) = - \sin x (1 - 2 \sin x) (3 - {4 \sin}^{2} x)

The rest is easy .

I solved it for 0 ⩽ x < 2 π

f (0) = \frac{1}{3} (\max)

f (\frac{π}{6}) = \frac{\sqrt{3}}{8} (\min)

f (\frac{π}{3}) = - \frac{8 - 9 \sqrt{3}}{24} (\max)

f (\frac{2 π}{3}) = \frac{8 - 9 \sqrt{3}}{24} (\min)

f (\frac{5 π}{6}) = - \frac{\sqrt{3}}{8} (\max)

f (π) = - \frac{1}{3} (\min)

f (\frac{4 π}{3}) = \frac{8 + 9 \sqrt{3}}{24} (\max)

f (\frac{5 π}{3}) = - \frac{8 + 9 \sqrt{3}}{24} (\min)

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