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

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

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

$=sqrt 5 (1)$ '

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