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

The extreme f(x)=sin(x)-cos(x) is Maximum((3pi)/4+2pin,sqrt(2)),Minimum((7pi)/4+2pin,-sqrt(2))

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

extreme points

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

Maximum (\frac{3 π}{4} + 2 πn, 2), Minimum (\frac{7 π}{4} + 2 πn, - 2)

Solution steps

Find the critical points:

x = \frac{3 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Domain of

sin (x) - cos (x) : - \infty < x < \infty

Find intervals:

Increasing

: 2 πn \leq x < \frac{3 π}{4} + 2 πn,

Decreasing

: \frac{3 π}{4} + 2 πn < x < \frac{7 π}{4} + 2 πn,

Increasing

: \frac{7 π}{4} + 2 πn < x < 2 π + 2 πn

Plug the extreme point

x = \frac{3 π}{4} + 2 πn

into

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

Maximum (\frac{3 π}{4} + 2 πn, 2)

Plug the extreme point

x = \frac{7 π}{4} + 2 πn

into

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

Minimum (\frac{7 π}{4} + 2 πn, - 2)

Maximum (\frac{3 π}{4} + 2 πn, 2), Minimum (\frac{7 π}{4} + 2 πn, - 2)

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

f (x) = x^{2} + y^{2} + 5

x^{3} - 3 x^{2} + 1

f (x) = \frac{8}{135} x^{3} - \frac{1}{20} x^{2} - \frac{7}{16} x

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

What is the extreme f(x)=sin(x)-cos(x) ?
The extreme f(x)=sin(x)-cos(x) is Maximum((3pi)/4+2pin,sqrt(2)),Minimum((7pi)/4+2pin,-sqrt(2))