What is the critical point of f(x)= -(sinx)/ (2+cosx) and f(x)= (sinx) / (cosx-2)?

Answer 1

To find the critical point take the derivative and set it equal to zero.

y = f(x) = (-sinx) / (2 + cosx)

dy/dx = [(2 + cosx)(-cosx) - (-sinx)(-sinx)] / (2 + cosx)²

dy/dx = [-2cosx - cos²x - sin²x] / (2 + cosx)²

dy/dx = [-2cosx - 1] / (2 + cosx)² = 0

-2cosx - 1 = 0
-2cosx = 1

cosx = -1/2
x = 2π/3, 4π/3
for x on the interval [0, 2π]
______________________

The second one works the same way.

y = f(x)= (sinx) / (cosx - 2)

dy/dx = [(cosx - 2)(cosx) - (sinx)(-sinx)] / (cosx - 2)²

dy/dx = [cos²x - 2cosx + sin²x] / (cosx - 2)²

dy/dx = [1 - 2cosx] / (cosx - 2)² = 0

1 - 2cosx = 0
2cosx = 1

cosx = 1/2
x = π/3, 5π/3
for x on the interval [0, 2π]

Answer 2

a million and a pair of SINX COSX = 2X, it truly is particularly common (sinx + cos x)^2 + (sinx - cosx)^2 = (sin^ x + cos^ x + 2 sin x cos x) + (sin^ x + cos^ x - 2 sin x cos x ) do you recognize the courting : sin^ x + cos^ x = a million = a million + 2 sin x cos x + a million - 2 sin x cos x = 2 voila :-)

Answer 3

you need to exercise thereom 7.2 which says the critical point of any problem with (sinx) in it equals 1 .... duhhh