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

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

extreme f(x)=-2cos(2x)

extreme points

f (x) = - 2 cos (2 x)

Minimum (πn, - 2), Maximum (\frac{π}{2} + πn, 2)

Solution steps

Find the critical points:

x = πn, x = \frac{π}{2} + πn

Domain of

- 2 cos (2 x) : - \infty < x < \infty

Find intervals:

Increasing

: πn < x < \frac{π}{2} + πn,

Decreasing

: \frac{π}{2} + πn < x < π + πn

Plug the extreme point

x = πn

into

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

Minimum (πn, - 2)

Plug the extreme point

x = \frac{π}{2} + πn

into

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

Maximum (\frac{π}{2} + πn, 2)

Minimum (πn, - 2), Maximum (\frac{π}{2} + πn, 2)

f (x) = x^{\frac{2}{3}} - 7

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

f (x) = 2 x - 11 ln (8 x), 2 \leq x \leq 9

f (x) = 3 x^{\frac{2}{3}}, - 1 \leq x \leq 1

(2 x - x^{2}) (2 y - y^{2})

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