What is the extreme f(x)=4θ-6sin(θ) ?

The extreme f(x)=4θ-6sin(θ) is

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extreme f(x)=4θ-6sin(θ)

extreme points

f (x) = 4 θ - 6 sin (θ)

Minimum (arccos (\frac{2}{3}) + 2 πn, 4 (arccos (\frac{2}{3}) + 2 πn) - 6 sin (arccos (\frac{2}{3}) + 2 πn)), Maximum (2 π - arccos (\frac{2}{3}) + 2 πn, 4 (2 π - arccos (\frac{2}{3}) + 2 πn) - 6 sin (2 π - arccos (\frac{2}{3}) + 2 πn))

Solution steps

Find the critical points:

θ = arccos (\frac{2}{3}) + 2 πn, θ = 2 π - arccos (\frac{2}{3}) + 2 πn

Domain of

4 θ - 6 sin (θ) : - \infty < θ < \infty

Find intervals:

Decreasing

: 2 πn \leq θ < arccos (\frac{2}{3}) + 2 πn,

Increasing

: arccos (\frac{2}{3}) + 2 πn < θ < 2 π - arccos (\frac{2}{3}) + 2 πn,

Decreasing

: 2 π - arccos (\frac{2}{3}) + 2 πn < θ < 2 π + 2 πn

Plug the extreme point

x = arccos (\frac{2}{3}) + 2 πn

into

4 θ - 6 sin (θ) \Rightarrow y = 4 (arccos (\frac{2}{3}) + 2 πn) - 6 sin (arccos (\frac{2}{3}) + 2 πn)

Minimum (arccos (\frac{2}{3}) + 2 πn, 4 (arccos (\frac{2}{3}) + 2 πn) - 6 sin (arccos (\frac{2}{3}) + 2 πn))

Plug the extreme point

x = 2 π - arccos (\frac{2}{3}) + 2 πn

into

4 θ - 6 sin (θ) \Rightarrow y = 4 (2 π - arccos (\frac{2}{3}) + 2 πn) - 6 sin (2 π - arccos (\frac{2}{3}) + 2 πn)

Maximum (2 π - arccos (\frac{2}{3}) + 2 πn, 4 (2 π - arccos (\frac{2}{3}) + 2 πn) - 6 sin (2 π - arccos (\frac{2}{3}) + 2 πn))

Minimum (arccos (\frac{2}{3}) + 2 πn, 4 (arccos (\frac{2}{3}) + 2 πn) - 6 sin (arccos (\frac{2}{3}) + 2 πn)), Maximum (2 π - arccos (\frac{2}{3}) + 2 πn, 4 (2 π - arccos (\frac{2}{3}) + 2 πn) - 6 sin (2 π - arccos (\frac{2}{3}) + 2 πn))

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

f (x) = 2 x^{3} + 9 x^{2} - 24 x - 10

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

f (x) = x^{2} + 3 x - 9

f (x) = \frac{25 x ^{2} + 81}{x}