How do you find the maximum value of #f(x)=2sin(x)+cos(x)#?

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Answer 1 · 2016-09-01T01:34:41

#sqrt 5#.
Range is #[-sqrt5, sqrt 5]#-

f'=2 cos x - sin x = 0, when 2 cos x = sin x that gives

x = arc tan 2. The principal value is in Q1. Indeed, there are general

values in Q1 and Q3.

#f''=-2 sin x - cos x < 0#, for Q1 values and #> 0#

for Q3 values, as both sin x and cos x are negative in Q3.

The maximum is obtained when tan x = 2, with x in Q1. And this is

2sin x + cos x , with tan x = 2

#= 2(2/sqrt 5)+1/sqrt 5#

#=5/sqrt 5#

#=sqrt 5#.

Of course, the minimum is #-sqrt 5#.

Alternative method sans differentiation:

#f=sqrt 5((2/sqrt 5) sin x+(1/sqrt 5)cosx)#

#=sqrt 5 sin (x+alpha)#, where

#sin alpha = 2/sqrt 5 and cos alpha =1/sqrt 5#

#Max f = sqrt 5 max sin(x+alpha)#

#=sqrt 5 (1)#'