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

Question

William Abdalla · Accepted Answer

#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)#'

Samantha Adkison · Answer

undefined To find the maximum value of $ f(x) = 2\sin(x) + \cos(x) $, you can use calculus.

1. Take the derivative of $ f(x) $ with respect to $ x $ to find critical points.
2. Set the derivative equal to zero and solve for $ x $.
3. Once you have the critical points, evaluate $ f(x) $ at these points as well as at the endpoints of the domain to find the maximum value.

Here's the breakdown:

1. $ f'(x) = 2\cos(x) - \sin(x) $
2. Set $ f'(x) = 0 $:
   $ 2\cos(x) - \sin(x) = 0 $
3. Solve for $ x $:
   $ \sin(x) = 2\cos(x) $
   $ \tan(x) = 2 $
   $ x \approx 1.107 $ (using inverse tangent function)
4. Evaluate $ f(x) $ at the critical point and endpoints:
   $ f(1.107) $ and $ f(0) $ and $ f(2\pi) $
5. Compare the values obtained in step 4 to find the maximum value of $ f(x) $.