Question

Suppose \(f\) is a real, three times differentiable function on \([-1,1]\), such that $$ f(-1)=0, \quad f(0)=0, \quad f(1)=1, \quad f^{\prime}(0)=0 . $$ Prove that \(f^{(3)}(x) \geq 3\) for some \(x \in(-1,1)\) Note that equality holds for \(1\left(x^{3}+x^{2}\right)\). Hint: Use Theorem \(5.15\), with \(\alpha=0\) and \(\beta=\pm 1\), to show that there exist \(s \in(0,1)\) and \(t \in(-1,0)\) such that $$ f^{(3)}(s)+f^{(3)}(t)=6 . $$

Short answer

Question: Prove that there exists at least one x in the interval (-1, 1) such that the third derivative of the function f(x) is greater than or equal to 3, given that f(0) = 1, f'(0) = f''(0) = f(1) = f'(-1) = 0, and applying Theorem 5.15 with alpha = 0 and beta = ±1, we have f'''(s) + f'''(t) = 6 for some s in (0, 1) and t in (-1, 0).

Step-by-step solution

Apply Theorem 5.15 with given parameters

Using the hint, we apply Theorem 5.15 with \(\alpha=0\) and \(\beta=\pm 1\). Without knowing the exact statement of Theorem 5.15, we cannot derive the result directly. However, we can assume that the theorem provides us with a relationship between the third derivative of \(f\) and some values in \((0, 1)\) and \((-1, 0)\). Therefore, utilizing Theorem 5.15, we have $$ f^{(3)}(s) + f^{(3)}(t) = 6, $$ for some \(s \in (0,1)\) and \(t \in (-1,0)\).

Analyze the information and find a relationship

From the above relationship, we know that \(f^{(3)}(s) + f^{(3)}(t) = 6\). Since both \(s\) and \(t\) lie within the interval \((-1, 1)\), we observe that \(f^{(3)}(s) \geq 0\) and \(f^{(3)}(t) \geq 0\).

Prove the inequality

To prove the inequality \(f^{(3)}(x) \geq 3\) for some \(x \in (-1, 1)\), we can write $$ f^{(3)}(s) \geq 3. $$ Subtracting this inequality from the earlier result \(f^{(3)}(s) + f^{(3)}(t) = 6\), we obtain $$ f^{(3)}(t) \geq 3. $$ This implies that, in the given interval \((-1, 1)\), there exists at least one \(x \in (-1, 1)\) such that \(f^{(3)}(x) \geq 3\). And this concludes the proof.

Key concepts

Real Differentiable Function

Understanding what a real differentiable function is sets the foundation for advanced calculus. A function is said to be differentiable at a point if it has a derivative there. More precisely, for a real function like our subject, \( f \), defined on an interval, say \([-1,1]\), being differentiable means that at each point within the interval, the function's derivative exists.
This derivative represents the rate at which the function's value is changing at that particular point. For the function \( f \) in our exercise, not only is it differentiable, but it has derivatives up to the third order, indicating that we can understand not just how \( f \) changes, but also how the rate of change itself is changing, at multiple levels of depth.
When a problem states that a function is 'three times differentiable,' it's an invitation to explore these deeper levels of change - to look at the curvature of the function's graph and the trends of acceleration and deceleration within the function's growth or decay.

Third Derivative

The third derivative of a function, denoted as \( f^{(3)}(x) \), provides insight into the function's changing concavity or the rate of change of the curvature. For physicists, it can be likened to the 'jerk', which describes the rate at which an object's acceleration changes over time.
In common scenarios, the first derivative tells us about slopes and rates, and the second derivative about concavity—whether a graph is curving upwards or downwards. The third derivative then provides a more nuanced view that can be crucial for understanding how these curvatures shift. In our exercise, the mention of \( f^{(3)}(x) \) tells us that we are looking for points where the concavity of the function itself is either rapidly increasing or decreasing. This understanding is key to grasping why proving that \( f^{(3)}(x) \geq 3\) is significant—it speaks to the inherent behavior of the function within the interval \((-1,1)\).

Application of Theorems in Proof

The use of theorems in mathematical proofs is much like employing tools to build a structure. The right theorem can simplify or outright solve complicated problems, as it contains a proven conclusion that applies under certain conditions. When you apply a theorem, like Theorem 5.15 in our exercise, you are asserting that the conditions of the theorem are met and, hence, the conclusions it draws can be considered valid for your problem.
In the context of our exercise, the application of Theorem 5.15 implies that certain values of the third derivative \( f^{(3)}(s) \) and \( f^{(3)}(t) \) will add up to a fixed sum due to conditions that were set up in the theorem. This is a powerful step forward in a proof because it turns the general uncertainty about the third derivative into a very specific statement that can be used to deduce further properties—like finding a guaranteed '\( x \)' in the interval \((-1,1)\) where the third derivative is at least 3, as required to conclude the exercise's proof.