Question

Find the linearization, say L(x)=ax+b, of fx=sin^(2)x at the point x= pi/4

Answer 1

a=f'[x] = 2sin(x)*cos(x), at (pi/4,f[pi/4])
f[pi/4] = (sqrt[2]/2)² = 1/2
f'[pi/4] = 2sqrt[2]/2 * sqrt[2]/2 = 1
1/2-y=f'[pi/4](pi/4-x) => 1/2-pi/4+x=y
y = x+(4-pi)/4