Find the critical values if `f'(x)=3/(2sqrt(3x))+2sinx` for `0<=x<=2pi`
The critical values are where `f'(x)=0` or `f'(x)` fails to exist (if the function is defined at that point.)
** I assume the original function was `y=sqrt(3x)-2cosx` **
(1) `f'(x)` fails to exist at x=0, but f(x) is defined at zero. So x=0 is a critical point.
(2) Setting `f'(x)=0` we get:
`3/(2sqrt(3x))+2sinx=0` Using a graphing utility we find 2 zeros on the interval:`x~~3.379` and `x~~6.107`
So the critical values are x=0,3.379,6.107
The graph of f(x) in black, f'(x) in red: