# If x is from 0 to 1 0 ≤ x ≤ 1 Find x, when the value of `(x^2-x)/((3-x^2)(2+2x-x^2))^(1/2)` is the biggest (max).

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You are looking for global maximum of function

`f(x)=(x^2-x)/((3-x^2)(2+2x-x^2))^(1/2)`

for `x in [0,1]`.

We search for the global maximum (or minimum) in endpoints of the segment (in this case they are 0 and 1) and critical points (points where function is not differentiable or its derivative is equal to 0). Since our function is differentiable on whole segment `[0,1]` we only need to find function values in endpoints and stationary points (where `f'(x)=0` ).

`f(0)=0`

`f(1)=0`

Now we find derivative

`f'(x)=(-6x^3+9x^2+9x-6)/(x^4-2x^3-5x^2+6x+6)^(3/2)`

Now to find solution of equation `f'(x)=0` we only need to find where numerator is equal to 0 i.e.

`-6x^3+9x^2+9x-6=0`

Solutions to that equation are `x_1=-1`, `x_2=1/2` and `x_3=2`. Since we are looking for maximum on segment [0,1] we are only interested in second solution `x_2=1/2`. Now we calculate function value at that point to see whether it is greater than function values at endpoints or not.

`f(1/2)=-1/11`

**Since `-1/11` is less than 0**, **global maximum is achieved at both endpoints that is for** `x=0` **and** `x=1.`

Stationary point `x=1/2` is actually global minimum.