What is the extreme f(θ)=-5cos(3/4 θ-6)-2 ?

The extreme f(θ)=-5cos(3/4 θ-6)-2 is Maximum(-(4pi)/3+8+(8pi)/3 n,3),Minimum(8+(8pi)/3 n,-7)

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extreme f(θ)=-5cos(3/4 θ-6)-2

extreme points

f (θ) = - 5 cos (\frac{3}{4} θ - 6) - 2

Maximum (- \frac{4 π}{3} + 8 + \frac{8 π}{3} n, 3), Minimum (8 + \frac{8 π}{3} n, - 7)

Solution steps

Find the critical points:

θ = - \frac{4 π}{3} + 8 + \frac{8 π}{3} n, θ = 8 + \frac{8 π}{3} n

Domain of

- 5 cos (\frac{3}{4} θ - 6) - 2 : - \infty < θ < \infty

Find intervals:

Increasing

: \frac{8 π}{3} n \leq θ < - \frac{4 π}{3} + 8 + \frac{8 π}{3} n,

Decreasing

: - \frac{4 π}{3} + 8 + \frac{8 π}{3} n < θ < 8 + \frac{8 π}{3} n,

Increasing

: 8 + \frac{8 π}{3} n < θ < \frac{8 π}{3} + \frac{8 π}{3} n

Plug the extreme point

x = - \frac{4 π}{3} + 8 + \frac{8 π}{3} n

into

- 5 cos (\frac{3}{4} θ - 6) - 2 \Rightarrow y = 3

Maximum (- \frac{4 π}{3} + 8 + \frac{8 π}{3} n, 3)

Plug the extreme point

x = 8 + \frac{8 π}{3} n

into

- 5 cos (\frac{3}{4} θ - 6) - 2 \Rightarrow y = - 7

Minimum (8 + \frac{8 π}{3} n, - 7)

Maximum (- \frac{4 π}{3} + 8 + \frac{8 π}{3} n, 3), Minimum (8 + \frac{8 π}{3} n, - 7)

f (x) = 5 x + 2 x^{- 1}

f (x, y) = - x^{2} - y^{3} + 4 x + 3 y

x^{2} + x y + 3 x + 2 y + 5

f (x) = x^{3} + 2 x^{2} - 5 x - 6

What is the extreme f(θ)=-5cos(3/4 θ-6)-2 ?
The extreme f(θ)=-5cos(3/4 θ-6)-2 is Maximum(-(4pi)/3+8+(8pi)/3 n,3),Minimum(8+(8pi)/3 n,-7)