Find the smallest value of the function f (x) = (4 / x + 1) + x. In the interval [0; 3]

Find the smallest value of the function f (x) = 4 / (x + 1) + x on the interval [0; 3].
1) f ‘(x) = (4 / (x + 1) + x)’ = – 4 / (x + 1) ^ 2 + 1;
2) To find the critical points, we equate the derivative of the function to zero. That is, we get:
– 4 / (x + 1) ^ 2 + 1 = 0;
– 4 / (x + 1) ^ 2 = – 1;
4 = (x + 1) ^ 2
x ^ 2 + 2 * x + 2 = 4;
x ^ 2 + 2 * x – 2 = 0;
x1 = -1 – √3 ≈ -2.73 does not belong to [0; 3];
x2 = -1 + √3 ≈ 0.73 belongs to [0; 3];
3) Find f (0), f (3) and f (-1 + √3).
f (0) = 4/1 + 0 = 4;
f (3) = 4 / (3 + 1) + 3 = 4/4 + 3 = 1 + 3 = 4;
f (-1 + √3) = 4 / (-1 + √3 + 1) -1 + √3 = 4 / √3 – 1 + √3 = (4 – √3 + 3) / √3 = (7 – √3) / √3 ≈ (7 – 1.7) / 1.7 ≈ 3.12;
Answer: (7 – √3) / √3 ≈ 3.12.