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

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

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

extreme points

f (x) = - cos (3 x)

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

Solution steps

Find the critical points:

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

Domain of

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

Find intervals:

Increasing

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

Decreasing

: \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

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

Minimum (\frac{2 π}{3} n, - 1)

Plug the extreme point

x = \frac{π}{3} + \frac{2 π}{3} n

into

- cos (3 x) \Rightarrow y = 1

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

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

f (x) = c

(1.2) (- 3.6)

- ln (x^{2} - 1)

5 x - 20

x^{2} - 4

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