How do you find the linearization at a=pi/6 of $f (x) = sin 2 x$ $f(x)=sin2x$ ?

Question

2416 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-17T12:19:56

$L (x) = \frac{\sqrt{3}}{2} + (x - \frac{π}{6})$ $L(x) = sqrt3/2+(x-pi/6)$

$L(x) = f(a)+f'(a)(x-a)$

$f(pi/6) = sin(pi/3) = sqrt3/2$

$f'(x) = 2cos2x$ , so $f'(pi/6) = 2cos(pi/3) = 1$

$L(x) = sqrt3/2+(x-pi/6)$

How do you find the linearization at a=pi/6 of f(x)=sin2xf(x)=sin2xf(x)=sin2x?