What is the extreme f(x)=sin(x+pi/4) ?

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

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extreme f(x)=sin(x+pi/4)

extreme points

f (x) = sin (x + \frac{π}{4})

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

Solution steps

Find the critical points:

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

Domain of

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

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

Maximum (\frac{π}{4} + 2 πn, 1)

Plug the extreme point

x = \frac{5 π}{4} + 2 πn

into

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

Minimum (\frac{5 π}{4} + 2 πn, - 1)

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

8 x^{2} + 3 ln (x + 1)

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

f (x) = x^{3} - 5 x^{2} + 3 x + 13

f (x, y) = x + 4 y + \frac{2}{x y}

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

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