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Chapter 11: Problem 42

(a) Prove that if \(f\) is a twice differentiable function with \(f(0)=0\) and \(f(1)=1\) and \(f^{\prime}(0)=f^{\prime}(1)=0,\) then \(\left|f^{\prime \prime}(x)\right| \geq 4\) for some \(x\) in \((0,1).\) In more picturesque terms: A particle which travels a unit distance in a unit time, and starts and ends with velocity \(0,\) has at some time an acceleration \(\geq 4 .\) Hint: Prove that cither \(f^{\prime \prime}(x) \geq 4\) for some \(x\) in \(\left(0, \frac{1}{2}\right),\) or else \(f^{\prime \prime}(x) \leq-4\) for some \(x\) in \(\left(\frac{1}{2}, 1\right).\) (b) Show that in fact we must have \(\left|f^{\prime \prime}(x)\right|>4\) for some \(x\) in \((0,1).\)

Short Answer

Expert verified
\(|f''(x)| \geq 4\) for some \(x\) in \((0,1)\). \(|f''(x)| > 4\) for some \(x\) in \((0,1)\).

Step by step solution

01

Apply Mean Value Theorem to the First Derivative

Since the function is twice differentiable (and therefore differentiable and continuous), we can apply the Mean Value Theorem (MVT) to the interval \([0,1]\). By MVT, there exists a point \(c\) in \((0,1)\) such that \[f'(c) = \frac{f(1) - f(0)}{1 - 0} = 1\].
02

Analyze the First Derivative in Smaller Intervals

Consider the intervals \([0, \frac{1}{2}]\) and \([\frac{1}{2}, 1]\). Apply the MVT to \(f'\) in these intervals. There exist points \(c_1\) in \((0, \frac{1}{2})\) and \(c_2\) in \((\frac{1}{2}, 1)\) such that: \[ f''(c_1) = \frac{f'(\frac{1}{2}) - f'(0)}{\frac{1}{2} - 0} = \frac{f'(\frac{1}{2})}{\frac{1}{2}}, \] \[ f''(c2) = \frac{f'(1) - f'(\frac{1}{2})}{1 - \frac{1}{2}} = \frac{-f'(\frac{1}{2})}{\frac{1}{2}}. \]
03

Relate the Second Derivative to the Acceleration

Combining the equations from Step 2, we get that either \(f''(c_1) > 4\) or \(f''(c_2) < -4\). This is equivalent to stating that \[|f''(x)| \geq 4\] for some \(x\) in \((0,1)\).
04

Part (b): Strengthen the Inequality

To show that \(|f''(x)| > 4\), note that any palindrome-like function that satisfy \(f(0) = 0\), \(f(1) = 1\) with zero derivative at the endpoints and having non-negative values in \(f'\) on \([0,1]\) will necessarily imply a 'steeper' slope somewhere in between. As a result, we have \[|f''(x)| > 4\] for some \(x\) in \((0,1)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean Value Theorem
The Mean Value Theorem (MVT) is a fundamental concept in calculus. It states that for a function that is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), there exists at least one point \(c\) in \((a, b)\) such that the derivative at that point \(c\) is equal to the average rate of change of the function over the interval \([a, b]\).
Mathematically, MVT can be stated as follows: if \(f\) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists a number \(c\) in \((a, b)\) such that \[f'(c) = \frac{f(b) - f(a)}{b - a} \] In simple terms, there is at least one point where the instantaneous rate of change (the derivative) is equal to the average rate of change over the entire interval. This concept helps us to bridge the gap between the average behavior and the instantaneous behavior of functions.
Second Derivative
The second derivative, denoted as \(f''(x)\), is the derivative of the first derivative \(f'(x)\). It provides information about the curvature and the concavity of the function.
In this context, the second derivative is related to the acceleration of a particle. While the first derivative \(f'(x)\) represents the velocity, the second derivative \(f''(x)\) measures how the velocity is changing over time (i.e., the acceleration).
To solve the problem, we apply the Mean Value Theorem to the first derivative \(f'(x)\) over smaller intervals to find the possible values of \(f''(x)\). This reveals critical points where the function's concavity changes, thus providing insights into the changes in acceleration.
Unit Distance Travel
In the given problem, a particle travels a unit distance within a unit time while starting and ending with zero velocity. Mathematically, this means that \(f(0) = 0\) and \(f(1) = 1\) with \(f'(0) = 0\) and \(f'(1) = 0\).
To understand this scenario better, imagine a particle moving from point A to point B (which are 1 unit apart) in exactly 1 unit of time. The particle starts from rest at A and comes to rest at B.
The challenging insight here is that for a particle to cover this distance given the constraints, there must be a substantial change in velocity at some point, which is captured by the second derivative (acceleration).
By analyzing the function鈥檚 behavior, we can deduce that there must be at least one point where the particle's acceleration is significant enough, in this case, \( \left|f''(x)\right| \geq 4\).
Acceleration
Acceleration is the rate at which the velocity of a particle changes with respect to time. In the context of this problem, acceleration is given by the second derivative \(f''(x)\).
Here's why acceleration matters: Since the particle's initial and final velocities are both zero, and it travels a unit distance in unit time, it must experience some form of acceleration and deceleration during its journey.
By applying the Mean Value Theorem (MVT) to the first derivative \(f'(x)\), we can deduce that the second derivative, or the acceleration, must have significant magnitude at some point within the interval \((0, 1)\). This means that \( \left|f''(x)\right| \textgreater 4\) for some \(x\) in \((0, 1)\).
This results from the fact that a particle must accelerate quickly enough to cover the unit distance in the given constraints, leading to a high value of the second derivative at some point.

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Most popular questions from this chapter

Suppose that \(f(0)=0\) and \(f^{\prime}\) is increasing. Prove that the function \(g(x)=\) \(f(x) / x\) is increasing on \((0 . \infty) .\) Hint: Obviously you should look at \(g^{\prime}(x)\) Prove that it is positive by applying the Mean Value Theorem to \(f\) on the right interval (it will help to remember that the hypothesis \(f(0)=0\) is essential, as shown by the function \(f(x)=1+x^{2}\) ).

A function \(f\) is Lipsclitz of order \(\alpha\) at \(x\) if there is a constant \(C\) such that $$|f(x)-f(y)| \leq C|x-y|^{\alpha}$$ for all \(y\) in an interval around \(x .\) The function \(f\) is Lipschitz of order \(\alpha\) on an intereal if \((*)\) holds for all \(x\) and \(y\) in the interval. (a) If \(f\) is Lipschitz of order \(\alpha>0\) at \(x,\) then \(f\) is continuous at \(x.\) (b) If \(f\) is Lipschitz of order \(\alpha>0\) on an interval, then \(f\) is uniformly continuous on this interval (see Chapter \(8,\) Appendix). (c) If \(f\) is differentiable at \(x\), then \(f\) is lipschitz of order 1 at \(x\). Is the converse true? (d) If \(f \text { is differentiable on } | a . b],\) is \(f\) lipschitz of order 1 on \([a, b] ?\) (e) If \(f\) is Lipschitz of order \(\alpha>1\) on \([a, b],\) then \(f\) is constant on \([a, b].\)

Although it is true that a weight dropped from rest will fall \(s(t)=16 t^{2}\) feet after \(t\) seconds, this experimental fact does not mention the behavior of weights which are thrown upwards or downwards. On the other hand, the law \(s^{\prime \prime}(t)=32\) is always true and has just enough ambiguity to account for the behavior of a weight released from any height, with an initial velocity. For simplicity let us agree to measure heights upwards from ground level; in this case, velocities are positive for rising bodies and negative for falling bodies, and all bodies fall according to the law \(s^{\prime \prime}(t)=-32.\) (a) Show that \(s\) is of the form \(s(t)=-16 t^{2}+\alpha t+\beta.\) (b) By setting \(t=0\) in the formula for \(s\), and then in the formula for \(s^{\prime},\) show that \(s(t)=-16 t^{2}+v_{0} t+s_{0},\) where \(s_{0}\) is the height from which the body is released at time \(0,\) and \(v_{0}\) is the velocity with which it is released. (c) A weight is thrown upwards with velocity \(v\) feet per second, at ground level. How high will it go? ("How high" means "what is the maximum height for all times".) What is its velocity at the moment it achieves its greatest height? What is its acceleration at that moment? When will it hit the ground again? What will its velocity be when it hits the ground again?

A right triangle with hypotenuse of length \(a\) is rotated about one of its legs to generate a right circular cone. Find the greatest possible volume of such a cone.

(a) Suppose that the polynomial function \(f(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}\) has exactly \(k\) critical points and \(f^{\prime \prime}(x) \neq 0\) for all critical points \(x\). Show that \(n-k\) is odd. (b) For each \(n,\) show that if \(n-k\) is odd, then there is a polynomial function \(f\) of degree \(n\) with \(k\) critical points, at each of which \(f^{\prime \prime}\) is non-zero. (c) Suppose that the polynomial function \(f(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}\) has \(k_{1}\) local maximum points and \(k_{2}\) local minimum points. Show that \(k_{2}=k_{1}+1\) if \(n\) is even, and \(k_{2}=k_{1}\) if \(n\) is odd. (d) Let \(n, k_{1}, k_{2}\) be three integers with \(k_{2}=k_{1}+1\) if \(n\) is even, and \(k_{2}=k_{1}\) if \(n\) is odd, and \(k_{1}+k_{2} < n .\) Show that there is a polynomial function \(f\) of degree \(n,\) with \(k_{1}\) local maximum points and \(k_{2}\) local minimum points. Hint: Pick \(a_{1} < a_{2} < \cdots< a_{k_{1}+k_{2}}\) and try \(f^{\prime}(x)=\prod_{i=1}^{k_{1}+k_{2}}\left(x-a_{i}\right) \cdot\left(1+x^{2}\right)^{l}\) for an appropriate number \(l.\)
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