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

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)

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)