What is the extreme f(x)=8x-16cos(x) ?

The extreme f(x)=8x-16cos(x) is

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extreme f(x)=8x-16cos(x)

extreme points

f (x) = 8 x - 16 cos (x)

Maximum (\frac{7 π}{6} + 2 πn, 8 (\frac{7 π}{6} + 2 πn) - 16 cos (\frac{7 π}{6} + 2 πn)), Minimum (\frac{11 π}{6} + 2 πn, 8 (\frac{11 π}{6} + 2 πn) - 16 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

8 x - 16 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

8 x - 16 cos (x) \Rightarrow y = 8 (\frac{7 π}{6} + 2 πn) - 16 cos (\frac{7 π}{6} + 2 πn)

Maximum (\frac{7 π}{6} + 2 πn, 8 (\frac{7 π}{6} + 2 πn) - 16 cos (\frac{7 π}{6} + 2 πn))

Plug the extreme point

x = \frac{11 π}{6} + 2 πn

into

8 x - 16 cos (x) \Rightarrow y = 8 (\frac{11 π}{6} + 2 πn) - 16 cos (\frac{11 π}{6} + 2 πn)

Minimum (\frac{11 π}{6} + 2 πn, 8 (\frac{11 π}{6} + 2 πn) - 16 cos (\frac{11 π}{6} + 2 πn))

Maximum (\frac{7 π}{6} + 2 πn, 8 (\frac{7 π}{6} + 2 πn) - 16 cos (\frac{7 π}{6} + 2 πn)), Minimum (\frac{11 π}{6} + 2 πn, 8 (\frac{11 π}{6} + 2 πn) - 16 cos (\frac{11 π}{6} + 2 πn))

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

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

f (x, y) = 23 - 2 x^{3} - y^{2} - 6 x y

f (x y) = x y + \frac{4}{x} + \frac{2}{y}

f (x) = x y + y - 11 x