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

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

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

extreme points

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

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

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

Find intervals:

Decreasing

: 2 πn \leq x < \frac{3 π}{4} + 2 πn,

Increasing

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

Decreasing

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

Plug the extreme point

x = \frac{3 π}{4} + 2 πn

into

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

Minimum (\frac{3 π}{4} + 2 πn, - 2)

Plug the extreme point

x = \frac{7 π}{4} + 2 πn

into

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

Maximum (\frac{7 π}{4} + 2 πn, 2)

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

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

x^{3} - 12 x + 1

f (x) = \frac{1}{x}, - 2 \leq x \leq 3

f (x) = (x - 2)^{3} (x - 1)

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

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