What is the extreme f(x)=2cos(x)-1 ?

The extreme f(x)=2cos(x)-1 is Maximum(2pin,1),Minimum(pi+2pin,-3)

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extreme f(x)=2cos(x)-1

extreme points

f (x) = 2 cos (x) - 1

Maximum (2 πn, 1), Minimum (π + 2 πn, - 3)

Solution steps

Find the critical points:

x = 2 πn, x = π + 2 πn

Domain of

2 cos (x) - 1 : - \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) - 1 \Rightarrow y = 1

Maximum (2 πn, 1)

Plug the extreme point

x = π + 2 πn

into

2 cos (x) - 1 \Rightarrow y = - 3

Minimum (π + 2 πn, - 3)

Maximum (2 πn, 1), Minimum (π + 2 πn, - 3)

f (x, y) = y ln (5 x + 9 y)

f (x) = ∣ - 3 x - 2 ∣, - 4 < x < 2

f (x) = 2 + 4 x + 100 x^{- 1}

f (x, y) = x^{2} + y^{2} - 12 x + 16 y

x^{\frac{1}{7}} + 1

What is the extreme f(x)=2cos(x)-1 ?
The extreme f(x)=2cos(x)-1 is Maximum(2pin,1),Minimum(pi+2pin,-3)