What is the extreme f(x)= x/2-cos(x) ?

The extreme f(x)= x/2-cos(x) is

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extreme f(x)= x/2-cos(x)

extreme points

f (x) = \frac{x}{2} - cos (x)

Maximum (\frac{7 π}{6} + 2 πn, \frac{7 π + 12 πn}{12} - cos (\frac{7 π}{6} + 2 πn)), Minimum (\frac{11 π}{6} + 2 πn, \frac{11 π + 12 πn}{12} - 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

\frac{x}{2} - 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

\frac{x}{2} - cos (x) \Rightarrow y = \frac{7 π + 12 πn}{12} - cos (\frac{7 π}{6} + 2 πn)

Maximum (\frac{7 π}{6} + 2 πn, \frac{7 π + 12 πn}{12} - cos (\frac{7 π}{6} + 2 πn))

Plug the extreme point

x = \frac{11 π}{6} + 2 πn

into

\frac{x}{2} - cos (x) \Rightarrow y = \frac{11 π + 12 πn}{12} - cos (\frac{11 π}{6} + 2 πn)

Minimum (\frac{11 π}{6} + 2 πn, \frac{11 π + 12 πn}{12} - cos (\frac{11 π}{6} + 2 πn))

Maximum (\frac{7 π}{6} + 2 πn, \frac{7 π + 12 πn}{12} - cos (\frac{7 π}{6} + 2 πn)), Minimum (\frac{11 π}{6} + 2 πn, \frac{11 π + 12 πn}{12} - cos (\frac{11 π}{6} + 2 πn))

f (x) = - 2 (4 x - 5) e^{4 x}

f (x) = (2 cos (θ) + cos^{2} (θ))

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

f (x) = 2 sin (x) cos (x)

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