Verify Rolle's theorem for function `f(x)=x/2-sin((pix)/(6))` on `[-1,\ 0]`

Explanation

Given, `f(x)=x/2-sin((pix)/(6))`


is continuous on `[-1,0]` and differentiable in `(-1,0)` as `x/2` is polynomial function and other is sine function which is continuous and differentiable everywhere


`f(-1)=-1/2-sin((-pi)/(6))=-1/2+1/2=0`


`f(0)=0/2-sin((0*pi)/(6))=0+0=0`


Rolle's theorem is verified.


there exist `c in (-1,0)` such that `f'(c)=0`


`f(x)=x/2-sin((pix)/(6))`


`f'(x)=1/2-(pi/6)cos((pix)/(6))`


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