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

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

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

extreme points

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

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

Increasing

: πn \leq x < \frac{π}{4} + πn,

Decreasing

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

Increasing

: \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

Maximum (\frac{π}{4} + πn, 1)

Plug the extreme point

x = \frac{3 π}{4} + πn

into

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

Minimum (\frac{3 π}{4} + πn, - 1)

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

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

f (x) = x^{2} + x y + y^{2} - 7 y + 16

f (x, y) = x^{2} + x y + y^{2} + 7 y

f (x) = \frac{7 ( x - 1 ) ^{2}}{x + 9}, (- 21, - 9)

\frac{7 x - 2}{3}

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