The first and second order derivatives of a function f(x) exit at all point in (a,b) with f'( c) =0 , where `altcltb` , of c and `f'(x)gt0` for all points on the immediate right of c, and `f'(x)lt0` for all points on the immediate left of c then at x=c , , f(x) has a

Options

A. local maximum
B. local minimum
C. point of inflexion
D. none of these

Answer

To solve the problem, we need to analyze the behavior of the function \( f(x) \) at the point \( x = c \) based on the given conditions regarding its first derivative \( f'(x) \).

### Step-by-Step Solution:

1. **Understanding the Given Conditions**:
We are given that:
- \( f'(c) = 0 \)
- \( f'(x) > 0 \) for \( x \) immediately to the right of \( c \)
- \( f'(x) < 0 \) for \( x \) immediately to the left of \( c \)

2. **Interpreting the Derivative Conditions**:
- Since \( f'(c) = 0 \), this indicates that the slope of the tangent to the curve at \( x = c \) is horizontal.
- The condition \( f'(x) > 0 \) for \( x > c \) implies that the function is increasing in the interval \( (c, b) \).
- The condition \( f'(x) < 0 \) for \( x < c \) implies that the function is decreasing in the interval \( (a, c) \).

3. **Graphical Representation**:
- We can visualize the function \( f(x) \) as follows:
- From \( a \) to \( c \), the function is decreasing.
- At \( x = c \), the function reaches a minimum point (since the slope is zero).
- From \( c \) to \( b \), the function is increasing.

4. **Conclusion about the Point \( c \)**:
- Given that the function decreases before \( c \) and increases after \( c \), we conclude that \( c \) is a local minimum of the function \( f(x) \).

5. **Final Answer**:
Therefore, at \( x = c \), the function \( f(x) \) has a **local minimum**.