find the maximum value of

Question

f(x)= 3cos(x)-2sin(x)

I know you have to take the derivative but after the derivative I don't know how to solve it.

f'(x)=-3sin(x)-2cos(x)=0

I don't know what to do after this please help.

Answer 1

so far ok

-3sin(x)-2cos(x)=0
3sinx = -2cosx
sinx/cosx = -2/3
tanx = -2/3

construct your triangles in quad II and IV
for both the hypotenuse is √13

in quad II
y = 3(-3/√13) - 2(2/√13) = -13/√13

in quad IV
y = 3(3/√13) -2(-2/√13) = 13/√13

so the max value is 13/√13