g(Ø) = 16Ø - 4tanØ
g ' (Ø) = 16 - 4sec^2 Ø
= 0
4sec^2 Ø = 16
sec^2 Ø = 4
secØ = ±2
cosØ = ± 1/2
we know cos 60° or cos π/3 = 1/2 , and because of the ± , Ø could be in any of the four quadrants
Ø = 60° or 120° or 240° or 300°
or
Ø = π/3, 2π/3 , 4π/3 , 5π/3
Finding critical numbers
g(theta) = 16(theta) - 4tan(theta)
I can only get the derivative of g to be g'(theta) = 16 - 4sec^2(theta).
Am I supposed to move on to get sec^2(theta) = 4, then sec(theta) = 2?
1 answer
1 answer