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

The extreme f(x)=-2sin(x)cos(x) is Minimum(pi/4+pin,-1),Maximum((3pi)/4+pin,1)

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

extreme points

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

Minimum (\frac{π}{4} + πn, - 1), Maximum (\frac{3 π}{4} + πn, 1)

Solution steps

Find the critical points:

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Domain of

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

Find intervals:

Decreasing

: πn \leq x < \frac{π}{4} + πn,

Increasing

: \frac{π}{4} + πn < x < \frac{3 π}{4} + πn,

Decreasing

: \frac{3 π}{4} + πn < x < π + πn

Plug the extreme point

x = \frac{π}{4} + πn

into

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

Minimum (\frac{π}{4} + πn, - 1)

Plug the extreme point

x = \frac{3 π}{4} + πn

into

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

Maximum (\frac{3 π}{4} + πn, 1)

Minimum (\frac{π}{4} + πn, - 1), Maximum (\frac{3 π}{4} + πn, 1)

f (x, y) = - 2 x^{3} + 6 x y + 3 y^{3}

f (x) = 14 x^{2} - x^{3}

f (x) = x^{3} - 3 x^{2} + 6 x - 8

f (x) = \frac{4 ( x - 10 ) ^{2}}{x - 6}, (0, 6)

f (x) = 15 x^{\frac{2}{3}} - x

What is the extreme f(x)=-2sin(x)cos(x) ?
The extreme f(x)=-2sin(x)cos(x) is Minimum(pi/4+pin,-1),Maximum((3pi)/4+pin,1)