What is the extreme f(x)=cos((2pi)/3 x) ?

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

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extreme f(x)=cos((2pi)/3 x)

extreme points

f (x) = cos (\frac{2 π}{3} x)

Maximum (3 n, 1), Minimum (\frac{3}{2} + 3 n, - 1)

Solution steps

Find the critical points:

x = 3 n, x = \frac{3}{2} + 3 n

Domain of

cos (\frac{2 π}{3} x) : - \infty < x < \infty

Find intervals:

Decreasing

: 3 n < x < \frac{3}{2} + 3 n,

Increasing

: \frac{3}{2} + 3 n < x < 3 + 3 n

Plug the extreme point

x = 3 n

into

cos (\frac{2 π}{3} x) \Rightarrow y = 1

Maximum (3 n, 1)

Plug the extreme point

x = \frac{3}{2} + 3 n

into

cos (\frac{2 π}{3} x) \Rightarrow y = - 1

Minimum (\frac{3}{2} + 3 n, - 1)

Maximum (3 n, 1), Minimum (\frac{3}{2} + 3 n, - 1)

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f (x, y) = 3 x^{4} + 8 x^{3} - 18 x^{2} + 6 y^{2} + 12 y - 4

f (x) = - x^{3} + 3 x^{2} + 45 x - 7

f (x) = \frac{3 x ^{2}}{x + 5}

f (x) = 4 x^{2} + y^{2} - 12 x - 6 y - 2 x y + 3

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