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

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

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extreme sin(x)cos(x)

extreme points

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})

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

f (x) = x^{\frac{1}{5}} (x + 24)

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

\frac{6 ∣ x ∣}{∣ x - 1 ∣}

f (x) = \frac{1}{3} x^{3} - \frac{3}{2} x^{2} - 18 x + \frac{1}{3}

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