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2014-07-16T10:56:27+03:00
f'\left(x\right)=4\sin\left(4x-\frac{\pi}{3}\right)\cos\left(4x-\frac{\pi}{3}\right)
Atunci f'\left(\frac{\pi}{6}\right)=4\sin\frac{\pi}{3}\cos\frac{\pi}{3}=4\cdot\frac{\sqrt{3}}{2}\cdot\frac{1}{2}=\sqrt{3}.
iar f\left(x_0\right)=f\left(\frac{\pi}{6}\right)=\frac{1}{2}\sin^2\frac{\pi}{3}=\frac{3}{8}.
Inlocuieste in ecuatia tangentei si obtii rezultatul.
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