What is the extreme f(x)=9x-18cos(x) ?

The extreme f(x)=9x-18cos(x) is

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extreme f(x)=9x-18cos(x)

extreme points

f (x) = 9 x - 18 cos (x)

Maximum (\frac{7 π}{6} + 2 πn, 9 (\frac{7 π}{6} + 2 πn) - 18 cos (\frac{7 π}{6} + 2 πn)), Minimum (\frac{11 π}{6} + 2 πn, 9 (\frac{11 π}{6} + 2 πn) - 18 cos (\frac{11 π}{6} + 2 πn))

Solution steps

Find the critical points:

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Domain of

9 x - 18 cos (x) : - \infty < x < \infty

Find intervals:

Increasing

: 2 πn \leq x < \frac{7 π}{6} + 2 πn,

Decreasing

: \frac{7 π}{6} + 2 πn < x < \frac{11 π}{6} + 2 πn,

Increasing

: \frac{11 π}{6} + 2 πn < x < 2 π + 2 πn

Plug the extreme point

x = \frac{7 π}{6} + 2 πn

into

9 x - 18 cos (x) \Rightarrow y = 9 (\frac{7 π}{6} + 2 πn) - 18 cos (\frac{7 π}{6} + 2 πn)

Maximum (\frac{7 π}{6} + 2 πn, 9 (\frac{7 π}{6} + 2 πn) - 18 cos (\frac{7 π}{6} + 2 πn))

Plug the extreme point

x = \frac{11 π}{6} + 2 πn

into

9 x - 18 cos (x) \Rightarrow y = 9 (\frac{11 π}{6} + 2 πn) - 18 cos (\frac{11 π}{6} + 2 πn)

Minimum (\frac{11 π}{6} + 2 πn, 9 (\frac{11 π}{6} + 2 πn) - 18 cos (\frac{11 π}{6} + 2 πn))

Maximum (\frac{7 π}{6} + 2 πn, 9 (\frac{7 π}{6} + 2 πn) - 18 cos (\frac{7 π}{6} + 2 πn)), Minimum (\frac{11 π}{6} + 2 πn, 9 (\frac{11 π}{6} + 2 πn) - 18 cos (\frac{11 π}{6} + 2 πn))

x^{1} - x^{2} + 2 \cdot x 1^{2} + 2 \cdot x^{1} \cdot x^{2} + x 2^{2}

y = \frac{x - 1}{x ^{2} + 35 x}

f (x) = - 3 x^{2} + 12 x - 7

f (x y) = 8 x^{3} + 20 x^{2} y + 40 y^{3}

f (x) = - 6 x - \frac{864}{x}