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).
You are looking for global maximum of function
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` ).
Now we find derivative
Now to find solution of equation `f'(x)=0` we only need to find where numerator is equal to 0 i.e.
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.
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.