What is the extreme y=cos(3x)-2 ?

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

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extreme y=cos(3x)-2

extreme points

y = cos (3 x) - 2

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

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) - 2 : - \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

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

Maximum (\frac{2 π}{3} n, - 1)

Plug the extreme point

x = \frac{π}{3} + \frac{2 π}{3} n

into

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

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

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

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

\frac{x ^{2} + 36}{x ^{2}}

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

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

f (x) = - \frac{1}{2} x^{4} + 16 x^{2} - 15

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