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

The extreme f(x)=-(cos(2x))/2-2sin(x) is Minimum(pi/2+2pin,-3/2),Maximum((3pi)/2+2pin, 5/2)

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

extreme points

f (x) = - \frac{cos ( 2 x )}{2} - 2 sin (x)

Minimum (\frac{π}{2} + 2 πn, - \frac{3}{2}), Maximum (\frac{3 π}{2} + 2 πn, \frac{5}{2})

Solution steps

Find the critical points:

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

Domain of

- \frac{cos ( 2 x )}{2} - 2 sin (x) : - \infty < x < \infty

Find intervals:

Decreasing

: 2 πn \leq x < \frac{π}{2} + 2 πn,

Increasing

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

Decreasing

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

Plug the extreme point

x = \frac{π}{2} + 2 πn

into

- \frac{cos ( 2 x )}{2} - 2 sin (x) \Rightarrow y = - \frac{3}{2}

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

Plug the extreme point

x = \frac{3 π}{2} + 2 πn

into

- \frac{cos ( 2 x )}{2} - 2 sin (x) \Rightarrow y = \frac{5}{2}

Maximum (\frac{3 π}{2} + 2 πn, \frac{5}{2})

Minimum (\frac{π}{2} + 2 πn, - \frac{3}{2}), Maximum (\frac{3 π}{2} + 2 πn, \frac{5}{2})

f (x) = e^{x^{2} - 4 x - 1}

y = e^{x} + I n x - x^{2} + \frac{3}{x ^{5}} + 3

f (x) = ln (1 + x^{2})

- 3 x^{5} + 5 x^{3}

f (x, y) = x^{2} + y^{2} - x - y + 1

What is the extreme f(x)=-(cos(2x))/2-2sin(x) ?
The extreme f(x)=-(cos(2x))/2-2sin(x) is Minimum(pi/2+2pin,-3/2),Maximum((3pi)/2+2pin, 5/2)