What is the extreme f(x)=4cos(3x) ?

The extreme f(x)=4cos(3x) is Maximum((2pi)/3 n,4),Minimum(pi/3+(2pi)/3 n,-4)

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extreme f(x)=4cos(3x)

extreme points

f (x) = 4 cos (3 x)

Maximum (\frac{2 π}{3} n, 4), Minimum (\frac{π}{3} + \frac{2 π}{3} n, - 4)

Solution steps

Find the critical points:

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

Domain of

4 cos (3 x) : - \infty < x < \infty

Find intervals:

Decreasing

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

Increasing

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

Plug the extreme point

x = \frac{2 π}{3} n

into

4 cos (3 x) \Rightarrow y = 4

Maximum (\frac{2 π}{3} n, 4)

Plug the extreme point

x = \frac{π}{3} + \frac{2 π}{3} n

into

4 cos (3 x) \Rightarrow y = - 4

Minimum (\frac{π}{3} + \frac{2 π}{3} n, - 4)

Maximum (\frac{2 π}{3} n, 4), Minimum (\frac{π}{3} + \frac{2 π}{3} n, - 4)

ln (e^{\frac{u + v}{2}}) - v

f (x) = 6 x^{2} - 6 x - 72

f (x) = 4 x^{5} - 3 x^{4} + x^{3} - 2 x^{2} + x - 1

f (x, y) = 5 x^{2} + 3 x y + 2 y^{2}

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

What is the extreme f(x)=4cos(3x) ?
The extreme f(x)=4cos(3x) is Maximum((2pi)/3 n,4),Minimum(pi/3+(2pi)/3 n,-4)