Let a > 0 be a constant and f (x) = x ＾ (1 / 2) - ln (x + a) be the maximum of F (x) when a = 3 / 4,

Let a > 0 be a constant and f (x) = x ＾ (1 / 2) - ln (x + a) be the maximum of F (x) when a = 3 / 4,

f（x）=√x－ln（x+3/4）
If the root sign is meaningful and the true number is greater than 0, X ≥ 0, x + 3 / 4 > 0, the simultaneous solution is x ≥ 0
The derivation of F (x) leads to
f’（x）=(1/2)√x－1/（x+3/4）
Let f '(x) ≥ 0 be the increasing interval of the original function, and we get (1 / 2) √ X-1 / (x + 3 / 4) ≥ 0
(x+3/4-2√x)/[2(x+3/4)*2√x] ≥0
x+3/4-2√x≥0
(√x)²-2√x+3/4≥0
(√x)²-2√x+3/4≥0
(2√x-3)*(√x-1)≥0
0≤x3/2
Let f '(x) ≥ 0, take the increasing interval of the original function, and get (1 / 2) √ X-1 / (x + 3 / 4) ≥ 0
(x+3/4-2√x)/[2(x+3/4)*2√x] ≥0
x+3/4-2√x≥0
(√x)²-2√x+3/4≥0
(√x)²-2√x+3/4≥0
(2√x-3)*(√x-1)≥0
0 ≤ x ≤ 1 or X ≤ 3 / 2
Similarly, Let f '(x) ≤ 0, take the subtraction interval of the original function, and get (1 / 2) √ X-1 / (x + 3 / 4) ≤ 0
1≤x≤3/2
therefore
The maximum value of F (x) is f (1) = √ 1-LN (1 + 3 / 4) = 1-LN (7 / 4)
The minimum value of F (x) is f (3 / 2) = √ (3 / 2) - ln (3 / 2 + 3 / 4) = √ (3 / 2) - ln (9 / 4)
=√6/2－2ln（3/2）