If x is a real number with 0 < x < 2,
then determine the minimum value of:
4 √6
------ + -----
2-x √x
This feels like it came out of a textbook.
So differentiate to get f'(x) = 4/(2-x)^2 + -sqrt(6)/(2*x^(3/2))
Set equal to zero and then isolate the radical:
4/(2-x)^2 = sqrt(6)/(2*x^(3/2))
Take reciprocals and square each side:
(x-2)^4/16 = 4x^3/6
Expand everything and simplify into a quartic:
3x^4-56x^3+72x^2-96x+48
Then rational root theorem lets us find the solution on the interval (0,2) is x=2/3.
As a final step we need the second derivative test.
f''(x) = 8/(2-x)^3 + 3*sqrt(6)/(4*x^(5/2))
It is easy to see that at x=2/3 we get f''(2/3) is a positive value so it must be the minimum.
Finally f(2/3) is 4/(2-2/3) + sqrt(6/(2/3)) = 3 + sqrt(9) = 6.
Solution
