What is the extreme 1/2 sin(x+pi/4) ?

The extreme 1/2 sin(x+pi/4) is Maximum(pi/4+2pin, 1/2),Minimum((5pi)/4+2pin,-1/2)

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extreme 1/2 sin(x+pi/4)

extreme points

\frac{1}{2} sin (x + \frac{π}{4})

Maximum (\frac{π}{4} + 2 πn, \frac{1}{2}), Minimum (\frac{5 π}{4} + 2 πn, - \frac{1}{2})

Solution steps

Find the critical points:

x = \frac{π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

Domain of

\frac{1}{2} sin (x + \frac{π}{4}) : - \infty < x < \infty

Find intervals:

Increasing

: 2 πn \leq x < \frac{π}{4} + 2 πn,

Decreasing

: \frac{π}{4} + 2 πn < x < \frac{5 π}{4} + 2 πn,

Increasing

: \frac{5 π}{4} + 2 πn < x < 2 π + 2 πn

Plug the extreme point

x = \frac{π}{4} + 2 πn

into

\frac{1}{2} sin (x + \frac{π}{4}) \Rightarrow y = \frac{1}{2}

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

Plug the extreme point

x = \frac{5 π}{4} + 2 πn

into

\frac{1}{2} sin (x + \frac{π}{4}) \Rightarrow y = - \frac{1}{2}

Minimum (\frac{5 π}{4} + 2 πn, - \frac{1}{2})

Maximum (\frac{π}{4} + 2 πn, \frac{1}{2}), Minimum (\frac{5 π}{4} + 2 πn, - \frac{1}{2})

f (x) = x^{\frac{6}{7}} - 7

f (x) = x^{3} + y^{3} - 3 x^{2} - 3 y^{2} - 9 x

f (x) = \frac{( x ^{3} )}{3} - 2 x^{2} - 5 x

f (x, y) = - 4 x^{2} + y (8 - y)

f (x, y) = x + y - x y

What is the extreme 1/2 sin(x+pi/4) ?
The extreme 1/2 sin(x+pi/4) is Maximum(pi/4+2pin, 1/2),Minimum((5pi)/4+2pin,-1/2)