What is the extreme f(x)=(4x-8sin(x)) ?

The extreme f(x)=(4x-8sin(x)) is

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extreme f(x)=(4x-8sin(x))

extreme points

f (x) = (4 x - 8 sin (x))

Minimum (\frac{π}{3} + 2 πn, 4 (\frac{π}{3} + 2 πn) - 8 sin (\frac{π}{3} + 2 πn)), Maximum (\frac{5 π}{3} + 2 πn, 4 (\frac{5 π}{3} + 2 πn) - 8 sin (\frac{5 π}{3} + 2 πn))

Solution steps

Find the critical points:

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Domain of

4 x - 8 sin (x) : - \infty < x < \infty

Find intervals:

Decreasing

: 2 πn \leq x < \frac{π}{3} + 2 πn,

Increasing

: \frac{π}{3} + 2 πn < x < \frac{5 π}{3} + 2 πn,

Decreasing

: \frac{5 π}{3} + 2 πn < x < 2 π + 2 πn

Plug the extreme point

x = \frac{π}{3} + 2 πn

into

4 x - 8 sin (x) \Rightarrow y = 4 (\frac{π}{3} + 2 πn) - 8 sin (\frac{π}{3} + 2 πn)

Minimum (\frac{π}{3} + 2 πn, 4 (\frac{π}{3} + 2 πn) - 8 sin (\frac{π}{3} + 2 πn))

Plug the extreme point

x = \frac{5 π}{3} + 2 πn

into

4 x - 8 sin (x) \Rightarrow y = 4 (\frac{5 π}{3} + 2 πn) - 8 sin (\frac{5 π}{3} + 2 πn)

Maximum (\frac{5 π}{3} + 2 πn, 4 (\frac{5 π}{3} + 2 πn) - 8 sin (\frac{5 π}{3} + 2 πn))

Minimum (\frac{π}{3} + 2 πn, 4 (\frac{π}{3} + 2 πn) - 8 sin (\frac{π}{3} + 2 πn)), Maximum (\frac{5 π}{3} + 2 πn, 4 (\frac{5 π}{3} + 2 πn) - 8 sin (\frac{5 π}{3} + 2 πn))

f (x) = 3 x

(8, 3), (9.5, - 3)

f (x) = \frac{x ^{3} + x ^{2} + x + 1}{∣ x + 1 ∣}

f (x) = \frac{x}{1 - x}

x^{2} - 7, x \geq 0