How would you find the maximum value and smallest +ve value the maximum occurs at for f(x) = -cosx + sqrt3sint ?
I've got as far as
f'(x) = sqrt3cosx + sinx
but I'm not sure where to go now?
Here's a completely different way to do that problem. A standard trig identity is cos(a+b)= cos(a)cos(b)- sin(a)sin(b). If we take b= x (I am assuming that is "sqrt 3 sin x", not "sqrt 3 sin t".) we have f(x)= (-1)cos x+ sqt(3) sin(x)= cos(a)cos(x)- sin(a)sin(x) so we would have to have cos(a)= -1 and sin(a)= -sqrt(3).
Well that's impossible because sqrt(3)> 1 and we would have to have $\displaystyle cos^2(a)+ sin^2(a)$ which is not true: $\displaystyle (-1)^2+ sqrt(3)^2= 4$, not 1. So multiply and divide by the square root of that, 2:
$\displaystyle f(x)= -cos(x)+ \sqrt{3}sin(x)= 2(-(1/2)cos(x)+ \sqrt{3}/2 sin(x)$ and now we must have cos(a)= -1/2 and sin(a)= -sqrt(3)/2 which says that $\displaystyle a= -\pi/3$.
That is, $\displaystyle f(x)= -cos(x)+ \sqrt{3}sin(x)= 2 cos(x- \pi/3)$. And it is easy to find maximum and minimum values for that.