Answer is: option1

Solution:

From \( a < x < b \):

1. \( f'(x) < 0 \) → Function is decreasing.
2. \( f''(x) > 0 \) → Function is concave up.

At \( x = b \):

1. \( f'(x) = 0 \) → Critical point (possible min or max).
2. \( f''(x) > 0 \) → Since concave up, this is a local minimum.

From \( b < x < 0 \):

1. \( f'(x) > 0 \) → Function is increasing.
2. \( f''(x) > 0 \) → Still concave up.

At \( x = 0 \):

1. \( f'(x) = 3 \) → Still increasing.
2. \( f''(x) = 0 \) → Possible inflection point.

From \( 0 < x < c \):

1. \( f'(x) > 0 \) → Still increasing.
2. \( f''(x) < 0 \) → Now concave down.

A graph that decreases and is concave up from \( a \) to \( b \).

At \( b \), a local minimum.

Then increases and is concave up from \( b \) to 0.

At 0, possible inflection point.

Still increasing, but now concave down from 0 to \( c \).

Option A:

1. Function is decreasing until point \( b \), and then increasing afterward.
2. Looks concave up until the origin, then concave down.
3. Minimum at \( b \), inflection at 0.
4. This matches all criteria.

Option B:

1. Function is increasing until \( b \), which would contradict \( f'(x) < 0 \) for \( x < b \).
2. ✘ Doesn't match.

Option C:

1. Looks like an inflection point is at \( b \), but we want a minimum there.
2. ✘ Doesn't match.

Option D:

1. Function is decreasing throughout from \( a \) to \( c \), not increasing after \( b \).
2. ✘ Doesn't match.

Correct Answer: Option A.