What is the extreme f(x)=9sin(x)+9cos(x) ?

The extreme f(x)=9sin(x)+9cos(x) is Maximum(pi/4+2pin,9sqrt(2)),Minimum((5pi)/4+2pin,-9sqrt(2))

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extreme f(x)=9sin(x)+9cos(x)

extreme points

f (x) = 9 sin (x) + 9 cos (x)

Maximum (\frac{π}{4} + 2 πn, 92), Minimum (\frac{5 π}{4} + 2 πn, - 92)

Solution steps

Find the critical points:

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

Domain of

9 sin (x) + 9 cos (x) : - \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

9 sin (x) + 9 cos (x) \Rightarrow y = 92

Maximum (\frac{π}{4} + 2 πn, 92)

Plug the extreme point

x = \frac{5 π}{4} + 2 πn

into

9 sin (x) + 9 cos (x) \Rightarrow y = - 92

Minimum (\frac{5 π}{4} + 2 πn, - 92)

Maximum (\frac{π}{4} + 2 πn, 92), Minimum (\frac{5 π}{4} + 2 πn, - 92)

What is the extreme f(x)=9sin(x)+9cos(x) ?
The extreme f(x)=9sin(x)+9cos(x) is Maximum(pi/4+2pin,9sqrt(2)),Minimum((5pi)/4+2pin,-9sqrt(2))