What is the extreme f(x)=x^1-(4cos(x)) ?

The extreme f(x)=x^1-(4cos(x)) is

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extreme f(x)=x^1-(4cos(x))

extreme points

f (x) = x^{1} - (4 cos (x))

Minimum (arcsin (- \frac{1}{4}) + 2 πn, - arcsin (\frac{1}{4}) + 2 πn - 4 cos (- arcsin (\frac{1}{4}) + 2 πn)), Maximum (π + arcsin (\frac{1}{4}) + 2 πn, π + arcsin (\frac{1}{4}) + 2 πn - 4 cos (π + arcsin (\frac{1}{4}) + 2 πn))

Solution steps

Find the critical points:

x = arcsin (- \frac{1}{4}) + 2 πn, x = π + arcsin (\frac{1}{4}) + 2 πn

Domain of

x^{1} - 4 cos (x) : - \infty < x < \infty

Find intervals:

Constant

: E m pt y I n t er v a l,

Increasing

: arcsin (- \frac{1}{4}) + 2 πn < x < π + arcsin (\frac{1}{4}) + 2 πn,

Decreasing

: π + arcsin (\frac{1}{4}) + 2 πn < x < 2 π + 2 πn

Plug the extreme point

x = arcsin (- \frac{1}{4}) + 2 πn

into

x^{1} - 4 cos (x) \Rightarrow y = - arcsin (\frac{1}{4}) + 2 πn - 4 cos (- arcsin (\frac{1}{4}) + 2 πn)

Minimum (arcsin (- \frac{1}{4}) + 2 πn, - arcsin (\frac{1}{4}) + 2 πn - 4 cos (- arcsin (\frac{1}{4}) + 2 πn))

Plug the extreme point

x = π + arcsin (\frac{1}{4}) + 2 πn

into

x^{1} - 4 cos (x) \Rightarrow y = π + arcsin (\frac{1}{4}) + 2 πn - 4 cos (π + arcsin (\frac{1}{4}) + 2 πn)

Maximum (π + arcsin (\frac{1}{4}) + 2 πn, π + arcsin (\frac{1}{4}) + 2 πn - 4 cos (π + arcsin (\frac{1}{4}) + 2 πn))

Minimum (arcsin (- \frac{1}{4}) + 2 πn, - arcsin (\frac{1}{4}) + 2 πn - 4 cos (- arcsin (\frac{1}{4}) + 2 πn)), Maximum (π + arcsin (\frac{1}{4}) + 2 πn, π + arcsin (\frac{1}{4}) + 2 πn - 4 cos (π + arcsin (\frac{1}{4}) + 2 πn))

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f (x) = 5 x ln (x)

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

f (x, y) = 9 e^{4 y - x^{2} - y^{2}}

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