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

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

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

extreme points

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

Maximum (\frac{π}{4} + πn, \frac{1}{2}), Minimum (\frac{3 π}{4} + πn, - \frac{1}{2})

Solution steps

Find the critical points:

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

Domain of

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

sin (x) cos (x) \Rightarrow y = \frac{1}{2}

Maximum (\frac{π}{4} + πn, \frac{1}{2})

Plug the extreme point

x = \frac{3 π}{4} + πn

into

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

Minimum (\frac{3 π}{4} + πn, - \frac{1}{2})

Maximum (\frac{π}{4} + πn, \frac{1}{2}), Minimum (\frac{3 π}{4} + πn, - \frac{1}{2})

\frac{x}{x ^{2} - 9}

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

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

f (x) = ∣ x ∣

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

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