Question

Ajutor urgent!!!
Sa se scrie ecuatia tangentei la graficul functie $f:R-R$ , $f(x)= \frac{1}{2} sin^{2} (4x- \frac{ \pi }{3} )$ in punctul cu abcisa $x_{0}= \frac{ \pi }{6}$ 
ecuatia tangentei este: $y= f ^{'}( x_{0})(x- x_{0}) + f( x_{0} )$  si raspunsul trebuie sa fie $y= \sqrt{3} x - \frac{ \pi \sqrt{3} }{6} + \frac{3}{8}$

Answer 1

$f'\left(x\right)=4\sin\left(4x-\frac{\pi}{3}\right)\cos\left(4x-\frac{\pi}{3}\right)$
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.