`f(theta) = 2cos(theta) + sin^2(theta)` Find the critical numbers of the function
2 Answers
sciencesolve | Teacher | (Level 3) Educator Emeritus
You need to find the critical points of the function, hence, you need to evaluate the solutions to the equation `f'(theta) = 0` .
You need to evaluate the first derivative:
`f'(theta) = -2 sin theta + 2sin theta*cos theta`
You need to solve for theta `f'(theta) = 0` , such that:
`-2 sin theta + 2sin theta*cos theta= 0 `
Factoring out `2sin theta` yields:
`2sin theta*(-1 + cos theta) = 0`
Put `2sin theta = 0 => sin theta = 0`
`theta = (-1)^k*arcsin 0 + kpi`
`theta = kpi`
Put `-1 + cos theta= 0 =>cos theta = 1`
`theta = +-arccos 1 + 2kpi`
`theta = 2kpi`
Hence, evaluating the critical numbers of the function for `f'(theta) = 0` , yields `theta =2kpi, theta = kpi.`
scisser | (Level 3) Honors
`f(x) = 2*cos(x) + sin(x)^2 `
`f'(x) = 2sin(x) * (cos(x) - 1) `
To find the zeros:
`0 = -2sin(x) + 2sin(x)*cos(x) `
`0 = 2* sin(x) * (-1 + cos(x)) `
`0 = sin(x) * (cos(x) -1) `
Therefore:
`sin(x) = 0 `
`x = arcsin(0) = 0 or pi or 2pi `
OR
`cos(x) -1 = 0 `
`cos(x) = 1 `
`x = arccos(1) `
`x = 0 or 2pi `