Answer is: option4

Solution:

1. Behavior around \( x = a \):

• \( f' \) is negative to the left of \( a \) and negative to the right of \( a \), but it reaches zero at \( x = a \). This suggests \( x = a \) is a critical point where the derivative touches zero but does not change sign. Thus, \( x = a \) is not a relative extremum (maximum or minimum) but rather a point where the graph of \( f \) has a horizontal tangent (sometimes called a "saddle point" or "stationary point of inflection").

2. Behavior on the interval \( b < x < c \):

• The graph of \( f' \) is increasing on this interval. Since \( f' \) is the derivative of \( f \), an increasing \( f' \) implies that the second derivative \( f'' \) is positive. Therefore, \( f \) is concave up on \( b < x < c \).

3. Behavior at \( x = c \):

• \( f' \) changes from negative to positive at \( x = c \). This means \( f \) changes from decreasing to increasing at \( x = c \), which implies \( f \) has a relative minimum at \( x = c \).

4. Comparison of \( f(c) \) and \( f(a) \):

• Since \( f' \) is negative on \( (-\infty, a) \) and \( (a, c) \), the function \( f \) is decreasing on these intervals. Therefore, \( f(c) < f(a) \). This contradicts statement I, which claims \( f(c) > f(a) \).

I. \( f(c) > f(a) \):

False. As explained above, \( f \) is decreasing on \( (-\infty, c) \), so \( f(c) < f(a) \).

II. The graph of \( f \) is concave up on the interval \( b < x < c \):

True. The graph of \( f' \) is increasing on \( b < x < c \), so \( f'' > 0 \), meaning \( f \) is concave up.

III. \( f \) has a relative minimum at \( x = c \):

True. \( f' \) changes from negative to positive at \( x = c \), indicating a relative minimum.