Verify Lagranges mean value theorem for function `f(x)=2x-x^2` on `[0,\ 1]` and find a point `' c '` in the indicated interval:

Explanation

Lagrange's theorem is only applicable if function is continuous in `[0, 1]`, differentiable in `(0, 1)`

and
There exist `c in (0, 1)` such that `f'(c)=(f(b)-f(a))/(b-a)`


`f'(x)=2-2x`


`f'(c)=2-2c=(f(1)-f(0))/(1-0)`


`2-2c=(2-1-0)/(1-0)`


`2-2c=1`


`c=1/2 in (0,1)`


LMUT is verified