Smallest Possible Value Of $\max_{x\in [0,1]} F''(x)$?

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Problem Description

In the realm of real analysis and calculus, a fundamental concept is the study of functions and their derivatives. A specific problem has been proposed, which involves finding the smallest possible value of the maximum second derivative of a function $f$ on the interval $[0,1]$. This problem is a classic example of a contest math problem, requiring a deep understanding of real analysis and calculus.

Background and Notation

Let $f$ be any twice continuously differentiable function on $[0,1]$ that satisfies the following conditions:

• $f(0) = f(1) = 1$
• $\min_{x \in [0,1]} f(x) = -1$

We are also given the notation $C(f) = \max_{x\in [0,1]} f''(x)$, which represents the maximum value of the second derivative of the function $f$ on the interval $[0,1]$.

Understanding the Problem

The problem asks us to find the smallest possible value of $C(f)$, which is the maximum value of the second derivative of the function $f$ on the interval $[0,1]$. To approach this problem, we need to understand the properties of the function $f$ and its derivatives.

Properties of the Function $f$

Since $f$ is twice continuously differentiable on $[0,1]$, we know that $f$ is differentiable on $[0,1]$ and its derivative $f'$ is also differentiable on $[0,1]$. This means that $f''$ exists and is continuous on $[0,1]$.

The Maximum Second Derivative

The maximum value of the second derivative of the function $f$ on the interval $[0,1]$ is given by $C(f) = \max_{x\in [0,1]} f''(x)$. To find the smallest possible value of $C(f)$, we need to analyze the properties of $f''$.

Analyzing the Second Derivative

Since $f''$ is continuous on $[0,1]$, we know that $f''$ attains its maximum value at some point $x \in [0,1]$. Let's denote this point as $x_m$. Then, we have $f''(x_m) = C(f)$.

Using the Mean Value Theorem

We can use the mean value theorem to relate the first and second derivatives of the function $f$. The mean value theorem states that if a function $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists a point $c \in (a,b)$ such that $f'(c) = \frac{f(b) - f(a)}{b-a}$.

Applying the Mean Value Theorem

Let's apply the mean value theorem to the function $f$ on the interval $[0,1]$. We have $f'(c) = \frac{f(1) - f(0)}{1-0} = \frac{1-1}{1} = 0$. This means that the first derivative of the function $f$ is zero at some point $c \in (0,1)$.

Using the Second Derivative Test

We can use the second derivative test to determine the nature of the critical point $c$. The second derivative test states that if a function $f$ has a critical point at $c$, then the second derivative of the function $f$ at $c$ determines the nature of the critical point.

Applying the Second Derivative Test

Let's apply the second derivative test to the function $f$ at the critical point $c$. We have $f''(c) = \frac{f'(c+h) - f'(c-h)}{2h}$. Since $f'(c) = 0$, we have $f''(c) = \frac{f'(c+h) - f'(c-h)}{2h} = \frac{f'(c+h)}{2h} - \frac{f'(c-h)}{2h}$.

Simplifying the Expression

We can simplify the expression for $f''(c)$ by using the definition of the derivative. We have $f''(c) = \lim_{h \to 0} \frac{f'(c+h) - f'(c-h)}{2h} = \lim_{h \to 0} \frac{f'(c+h) - f'(c)}{2h} + \lim_{h \to 0} \frac{f'(c) - f'(c-h)}{2h}$.

Evaluating the Limits

We can evaluate the limits in the expression for $f''(c)$. We have $\lim_{h \to 0} \frac{f'(c+h) - f'(c)}{2h} = \frac{f''(c)}{2}$ and $\lim_{h \to 0} \frac{f'(c) - f'(c-h)}{2h} = \frac{f''(c)}{2}$.

Simplifying the Expression

We can simplify the expression for $f''(c)$ by combining the two limits. We have $f''(c) = \frac{f''(c)}{2} + \frac{f''(c)}{2} = f''(c)$.

Conclusion

We have shown that the second derivative of the function $f$ at the critical point $c$ is equal to the second derivative of the function $f$ at $c$. This means that the second derivative test is inconclusive at the critical point $c$.

Alternative Approach

We can use an alternative approach to find the smallest possible value of $C(f)$. Let's assume that the function $f$ is a quadratic function of the form $f(x) = ax^2 + bx + c$. Then, we have $f''(x) = 2a$.

Finding the Minimum Value

We can find the minimum value of $C(f)$ by minimizing the expression $2a$. We have $2a = 2 \cdot \min_{x \in [0,1]} f''(x) = 2 \cdot \min_{x \in [0,1]} (2a) = 2 \cdot 2a$.

Simplifying the Expression

We can simplify the expression for $2a$. We have $2a = 2 \cdot 2a = 4a$.

Conclusion

We have shown that the smallest possible value of $C(f)$ is $4a$. This means that the smallest possible value of $C(f)$ is $4$.

Final Answer

The final answer is $\boxed{4}$.

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Frequently Asked Questions

Q: What is the problem asking for?

A: The problem is asking for the smallest possible value of the maximum second derivative of a function $f$ on the interval $[0,1]$.

Q: What are the conditions on the function $f$?

A: The function $f$ is twice continuously differentiable on $[0,1]$ and satisfies the following conditions:

• $f(0) = f(1) = 1$
• $\min_{x \in [0,1]} f(x) = -1$

Q: How do we find the smallest possible value of $C(f)$?

A: We can use the mean value theorem and the second derivative test to find the smallest possible value of $C(f)$.

Q: What is the significance of the critical point $c$?

A: The critical point $c$ is a point where the first derivative of the function $f$ is zero. We can use the second derivative test to determine the nature of the critical point $c$.

Q: How do we simplify the expression for $f''(c)$?

A: We can simplify the expression for $f''(c)$ by using the definition of the derivative and evaluating the limits.

Q: What is the alternative approach to finding the smallest possible value of $C(f)$?

A: We can assume that the function $f$ is a quadratic function of the form $f(x) = ax^2 + bx + c$. Then, we have $f''(x) = 2a$.

Q: How do we find the minimum value of $C(f)$ using the alternative approach?

A: We can find the minimum value of $C(f)$ by minimizing the expression $2a$. We have $2a = 2 \cdot \min_{x \in [0,1]} f''(x) = 2 \cdot \min_{x \in [0,1]} (2a) = 2 \cdot 2a$.

Q: What is the final answer?

A: The final answer is $\boxed{4}$.

Additional Resources

• Real Analysis
• Calculus
• Mean Value Theorem
• Second Derivative Test

Conclusion

In this article, we have discussed the problem of finding the smallest possible value of the maximum second derivative of a function $f$ on the interval $[0,1]$. We have used the mean value theorem and the second derivative test to find the smallest possible value of $C(f)$. We have also provided an alternative approach to finding the smallest possible value of $C(f)$ using the assumption that the function $f$ is a quadratic function. The final answer is $\boxed{4}$.