What is the extreme y=-5cos(3x)-2 ?

The extreme y=-5cos(3x)-2 is Minimum((2pi)/3 n,-7),Maximum(pi/3+(2pi)/3 n,3)

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extreme y=-5cos(3x)-2

extreme points

y = - 5 cos (3 x) - 2

Minimum (\frac{2 π}{3} n, - 7), Maximum (\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

- 5 cos (3 x) - 2 : - \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

- 5 cos (3 x) - 2 \Rightarrow y = - 7

Minimum (\frac{2 π}{3} n, - 7)

Plug the extreme point

x = \frac{π}{3} + \frac{2 π}{3} n

into

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

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

Minimum (\frac{2 π}{3} n, - 7), Maximum (\frac{π}{3} + \frac{2 π}{3} n, 3)

f (x, y) = 3 x^{\frac{1}{3}} y^{\frac{1}{3}}

f (x) = \frac{x ^{2}}{x - 1}

P (- 8.5) Q (- 4.11)

x^{3} + 8 x^{2} - 6 x - 4

f (x) = - \frac{13}{9} x^{2} + \frac{8}{3} x^{2} + \frac{37}{10}

What is the extreme y=-5cos(3x)-2 ?
The extreme y=-5cos(3x)-2 is Minimum((2pi)/3 n,-7),Maximum(pi/3+(2pi)/3 n,3)