What is the extreme f(x)=2cos(x)+sin^2(x) ?

The extreme f(x)=2cos(x)+sin^2(x) is Maximum(2pin,2),Minimum(pi+2pin,-2)

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extreme f(x)=2cos(x)+sin^2(x)

extreme points

f (x) = 2 cos (x) + sin^{2} (x)

Maximum (2 πn, 2), Minimum (π + 2 πn, - 2)

Solution steps

Find the critical points:

x = 2 πn, x = π + 2 πn

Domain of

2 cos (x) + sin^{2} (x) : - \infty < x < \infty

Find intervals:

Decreasing

: 2 πn < x < π + 2 πn,

Increasing

: π + 2 πn < x < 2 π + 2 πn

Plug the extreme point

x = 2 πn

into

2 cos (x) + sin^{2} (x) \Rightarrow y = 2

Maximum (2 πn, 2)

Plug the extreme point

x = π + 2 πn

into

2 cos (x) + sin^{2} (x) \Rightarrow y = - 2

Minimum (π + 2 πn, - 2)

Maximum (2 πn, 2), Minimum (π + 2 πn, - 2)

z (3 + x^{0.25})^{3} \cdot x^{- 0.75}

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

f (x) = P (x) = x^{3} + x^{2} - 5 x + 3

2 x^{3} - 3 x^{2} - 12 x + 12

f (x) = \frac{10 x}{1 + x ^{2}}

What is the extreme f(x)=2cos(x)+sin^2(x) ?
The extreme f(x)=2cos(x)+sin^2(x) is Maximum(2pin,2),Minimum(pi+2pin,-2)