Find the minimum value of f(x) = 3 + |2x + 11|.

For a function of the form f(x) = c + |g(x)|, since the absolute value |g(x)| is always greater than or equal to 0 for every real x, the smallest value it can take is 0. Therefore f attains its minimum exactly when g(x) = 0, and that minimum value equals c.

1. Here f(x) = 3 + |2x + 11|, so c = 3 and g(x) = 2x + 11.

2. Set g(x) = 0: 2x + 11 = 0, which gives x = -11/2.

3. At x = -11/2, |2x + 11| = |0| = 0, so f(-11/2) = 3 + 0 = 3.

Checking nearby points confirms this is indeed the minimum: at x = -5, |2(-5) + 11| = |1| = 1, giving f(-5) = 4, which is larger than 3; at x = -6, |2(-6) + 11| = |-1| = 1, giving f(-6) = 4 as well — both exceed 3, consistent with 3 being the least possible value.

Hence the minimum value of f(x) is 3.