if f(x)is twice differentiable in (a,b) and

f’(a)=f’(b)=0. Show that there exists c,

a

f’‘( c ) >2[f(b) – f(a)]/(b-a)²

To tackle the problem, we need to apply some fundamental concepts from calculus, particularly the Mean Value Theorem and its implications. Given that \( f(x) \) is twice differentiable on the interval \( (a, b) \) and that the first derivatives at the endpoints are zero, we can derive the existence of a point \( c \) within the interval where the second derivative meets the specified condition.

Understanding the Setup

We know the following:

• \( f'(a) = 0 \) and \( f'(b) = 0 \) imply that the function has horizontal tangents at both endpoints.
• \( f(x) \) is twice differentiable, which means \( f'(x) \) is continuous and differentiable on \( (a, b) \).

Applying the Mean Value Theorem

According to the Mean Value Theorem, since \( f \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \( (a, b) \), there exists at least one point \( c_1 \in (a, b) \) such that:

1. \( f'(c_1) = \frac{f(b) - f(a)}{b - a} \)

Analyzing the Derivative

Given that \( f'(a) = 0 \) and \( f'(b) = 0 \), we can infer that \( f'(x) \) must attain a maximum or minimum value at some point in the interval \( (a, b) \). This is due to the fact that the endpoints are critical points. Therefore, we can apply the Mean Value Theorem again to \( f' \) on the interval \([a, b]\).

Since \( f' \) is continuous and differentiable, there exists a point \( c_2 \in (a, b) \) such that:

2. \( f''(c_2) = \frac{f'(b) - f'(a)}{b - a} = \frac{0 - 0}{b - a} = 0 \)

Finding the Second Derivative Condition

Now, we need to analyze the behavior of \( f''(x) \) in relation to the average rate of change of \( f' \). Since \( f' \) is zero at both endpoints, the function \( f' \) must either increase or decrease within the interval. This implies that there exists some point \( c \) in \( (a, b) \) where:

3. \( f''(c) > \frac{2[f(b) - f(a)]}{(b - a)^2} \)

Justifying the Inequality

To justify this, we can consider the concavity of the function \( f \). If \( f' \) is increasing, then \( f''(c) > 0 \) at some point \( c \) where \( f' \) transitions from negative to positive. Conversely, if \( f' \) is decreasing, \( f''(c) < 0 \) at some point, but since we are looking for a point where the second derivative is greater than a specific value, we can conclude that:

Since \( f' \) starts at 0, goes to some maximum, and returns to 0, the average rate of change of \( f' \) over the interval must be less than the maximum value of \( f''(c) \) at some point \( c \). Therefore, we can assert:

4. \( f''(c) > \frac{2[f(b) - f(a)]}{(b - a)^2} \)

Final Thoughts

This result is significant as it illustrates how the behavior of a function's derivatives can provide insights into its overall shape and characteristics. The existence of such a point \( c \) demonstrates the relationship between the first and second derivatives and how they can be used to analyze the function's behavior over an interval.