Let k and K be the minimum and the maximum values of the function f(x) = (1+x)^0.6 + x^0.6 in [0,1] respectively, then the ordered pair (k,K) is equal to (2√4.4,1), (2√4.6,1), (2√4.4,20.6), (1,2√4.6)

The correct option is A (2√4.4,1)
f(x) = (1+x)^0.6 + x^0.6
f'(x) = 0.6(1+x)^-0.4(1+x) + 0.6x^-0.4 = 0
0.6(1+x)^-0.4(1+x) = -0.6x^-0.4
(1+x)^-0.4(1+x) = -x^-0.4
(1+x)^0.6 = -x^0.6
This equation has no solution in [0,1].
Let's check the boundary points:
f(0) = (1+0)^0.6 + 0^0.6 = 1
f(1) = (1+1)^0.6 + 1^0.6 = 2^0.6 + 1 ≈ 1.5157 + 1 = 2.5157 ≈ 2√4.4
Minimum occurs at x = 0, value is 1
Maximum occurs at x = 1, value is 2√4.4