What is the extreme f(x)=4sin(x)+3cos(x) ?

The extreme f(x)=4sin(x)+3cos(x) is Maximum(arctan(4/3)+2pin,5),Minimum(arctan(4/3)+pi+2pin,-5)

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extreme f(x)=4sin(x)+3cos(x)

extreme points

f (x) = 4 sin (x) + 3 cos (x)

Maximum (arctan (\frac{4}{3}) + 2 πn, 5), Minimum (arctan (\frac{4}{3}) + π + 2 πn, - 5)

Solution steps

Find the critical points:

x = arctan (\frac{4}{3}) + 2 πn, x = arctan (\frac{4}{3}) + π + 2 πn

Domain of

4 sin (x) + 3 cos (x) : - \infty < x < \infty

Find intervals:

Increasing

: 2 πn \leq x < arctan (\frac{4}{3}) + 2 πn,

Decreasing

: arctan (\frac{4}{3}) + 2 πn < x < arctan (\frac{4}{3}) + π + 2 πn,

Increasing

: arctan (\frac{4}{3}) + π + 2 πn < x < 2 π + 2 πn

Plug the extreme point

x = arctan (\frac{4}{3}) + 2 πn

into

4 sin (x) + 3 cos (x) \Rightarrow y = 5

Maximum (arctan (\frac{4}{3}) + 2 πn, 5)

Plug the extreme point

x = arctan (\frac{4}{3}) + π + 2 πn

into

4 sin (x) + 3 cos (x) \Rightarrow y = - 5

Minimum (arctan (\frac{4}{3}) + π + 2 πn, - 5)

Maximum (arctan (\frac{4}{3}) + 2 πn, 5), Minimum (arctan (\frac{4}{3}) + π + 2 πn, - 5)

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

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

f (x) = x^{2} + x y + 12 y^{2} - 4 x + y

f (x, y) = 6 x + 6 x y + y

f (x) = (x t)

What is the extreme f(x)=4sin(x)+3cos(x) ?
The extreme f(x)=4sin(x)+3cos(x) is Maximum(arctan(4/3)+2pin,5),Minimum(arctan(4/3)+pi+2pin,-5)