# can some one show me how to get the critical values for y'=(3/2)(3x)^(-1/2)+2sin(x) on 0<=x<=2pi I know I need to set it equal to 0 to find the cv's but I can't seem to get the right answers. 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...

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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:

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