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Chapter 4: Problem 8

Let \(f\) and \(g\) be functions which are continuous on \([a, b]\) and differentiable on \((a, b)\), where \(a

Short Answer

Expert verified
The error lies in misapplying the Mean Value Theorem, ignoring the need for ensuring that \(h(x)\) vanishes appropriately using Rolle's theorem instead.

Step by step solution

01

Understanding the Problem

We need to prove the existence of a point \(c\) such that \(\frac{f^{\prime}(c)}{g^{\prime}(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}\) using the fact that both \(f\) and \(g\) are continuous on \([a,b]\) and differentiable on \((a,b)\), while \(g^{\prime}(x)\) is never zero.
02

Incorrect Application of the Mean Value Theorem

The statement tries to apply the Mean Value Theorem separately to \(f\) and \(g\) and equates their respective derivatives at the same point \(c\) with the fractional differences of their values, which leads to a claim that their ratio equals the fixed value of the ratios of the function difference. This interpretation misuses the theorem as it overlooks \(g^{\prime}(x)\) can be negative, indicating the lack of appropriate handling when dividing their mean increments.
03

Concept of Functional Difference

Consider a function \(h(x) = f(x) - \frac{f(b)-f(a)}{g(b)-g(a)}g(x)\). This construction eliminates the constancy issue of the error function described since \(g^{\prime}(x) eq 0\). Here \(h(x)\) is continuous and differentiable across \([a,b]\), ensuring that \(h(a) = h(b) = 0\).
04

Application of Rolle's Theorem

Since \(h(a) = h(b) = 0\), by Rolle's theorem there exists at least one \(c \in (a,b)\) such that \(h^{\prime}(c) = 0\). Calculating, \(h^{\prime}(x) = f^{\prime}(x) - \frac{f(b)-f(a)}{g(b)-g(a)}g^{\prime}(x)\). Setting it to zero, \(f^{\prime}(c) = \frac{f(b)-f(a)}{g(b)-g(a)}g^{\prime}(c)\), leading to \(\frac{f^{\prime}(c)}{g^{\prime}(c)} = \frac{f(b)-f(a)}{g(b)-g(a)}\).

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Key Concepts

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

Rolle's Theorem
Rolle's Theorem is a fundamental result in calculus that provides conditions under which a particular derivative equals zero. This theorem is applicable to functions that are continuous over a closed interval \([a, b]\) and differentiable over the open interval \((a, b)\). A crucial condition of Rolle's Theorem is that the function must have equal values at the endpoints, specifically \(f(a) = f(b)\). This creates a situation where the function must reach a peak or valley between \(a\) and \(b\).

When these conditions are met, Rolle's Theorem guarantees the existence of at least one point \(c \in (a, b)\) such that the derivative \(f'(c) = 0\). Essentially, this theorem tells us that there's at least one point where the tangent to the curve is horizontal.

It is noteworthy that Rolle's Theorem is a special case of the Mean Value Theorem, where the average rate of change between two points is zero. This concept is pivotal in proving various results, including the verification of the necessary conditions for stationary points in optimization problems.
Continuous Functions
A continuous function is one that is smooth and unbroken over its domain. For a function to be considered continuous on a closed interval \([a, b]\), it must not have any jumps, holes, or vertical asymptotes within that interval.

Formally, a function \(f\) is continuous at a point \(x = c\) if the following three conditions are satisfied:
  • \(f(c)\) is defined.
  • \(\lim_{x \to c} f(x)\) exists.
  • \(\lim_{x \to c} f(x) = f(c)\).

A continuous function on a closed interval ensures that the function's behavior is predictable and that we can apply theorems like the Intermediate Value Theorem or the Mean Value Theorem accurately.

Continuous functions are important in calculus for ensuring that operations involving limits and integrals work as expected, allowing for sound application of mathematical principles.
Differentiable Functions
Differentiability of a function means it has a derivative at each point within its domain. This attribute is stronger than continuity since differentiability also requires that the function changes at a rate that can be mathematically described at every point in the interval.

For a function to be differentiable at a point \(x = c\), it must satisfy the following condition:
  • The limit \(\lim_{h \to 0} \frac{f(c+h) - f(c)}{h}\) exists.

In essence, differentiability implies that the function has a defined tangent at every point within its interval, and this tangent describes the function's velocity or rate of change at that point.

It's important to note that while differentiability implies continuity, the converse is not true; a continuous function may not be differentiable. For instance, functions with sharp corners or cusps are continuous but not differentiable at the point of the corner. Differentiability in calculus is crucial for ensuring the accuracy and applicability of many types of optimization and analysis tasks, such as maximizing or minimizing functions.
Function Derivatives
The derivative of a function is one of the central concepts in calculus. Function derivatives measure how a function's output value changes with respect to changes in its input.

Mathematically, the derivative of a function \(f\) at a point \(x\) is defined as: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. \]
This limit, if it exists, provides the slope of the tangent line to the function at that particular point, giving us valuable insight into the function's rate of change.

Understanding derivatives enables the use of tools like the Mean Value Theorem and Rolle's Theorem to analyze and solve a multitude of real-world problems. They can also help determine points of inflection, local maxima and minima, and describe the concavity of functions. In essence, derivatives open a window to understanding the mechanical nature of change, providing analytic depth to the study of functions and their behaviors.

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Most popular questions from this chapter

Let \(g\) be a function which can be differentiated arbitrarily often and whose domain includes \(a\). By applying the above form of Taylor's theorem to the function \(f\) given by \(f(x)=g(x+a)\) show that the polynomial $$ g(a)+g^{\prime}(a)\left(x_{0}-a\right)+\frac{g^{\prime \prime}(a)}{2 !}\left(x_{0}-a\right)^{2}+\cdots+\frac{g^{(n)}(a)}{n !}\left(x_{0}-a\right)^{n} $$ differs from \(g\left(x_{0}\right)\) by an amount equal to \(g^{(n+1)}(c)\left(x_{0}-a\right)^{n+1} /(n+1) !\) for some \(c\) between \(a\) and \(x_{0} .\) (This is the more general form of Taylor's theorem.) The expression $$ g(a)+g^{\prime}(a)(x-a)+\frac{g^{\prime \prime}(a)}{2 !}(x-a)^{2}+\cdots+\frac{g^{(n)}(a)}{n !}(x-a)^{n}+\cdots $$ is called the Taylor's series of \(g\) (about \(x=a\) ). Find the Taylor series of the log function about \(x=1\).

We have seen that if \(f\) is a function then so is \(f^{\prime}\) (perhaps with a smaller domain). So we can consider the differentiability of the function \(f^{\prime}\) and work out its derivative \(\left(f^{\prime}\right)\) ' of \(f^{\prime \prime}\) (or \(\left.\mathrm{d}^{2} y / \mathrm{d} x^{2}\right)\) : this is called the second derivative of \(f\). In a similar way we can repeatedly differentiate \(f\) and find its \(n\)th derivative denoted by \(f^{(n)}\) (or \(\left.\mathrm{d}^{n} y / \mathrm{d} x^{n}\right)\). (i) For each positive integer \(n\) find the \(n\)th derivative of the functions \(f(x)=x^{4} \quad x \in \mathbb{R}\) and \(\quad g(x)=\sin x \quad x \in \mathbb{R}\) (ii) Let \(f(x)=x^{2}\left(x^{2}-2\right)(x \in \mathbb{R})\). Show that \(f\) has stationary points at \(x=-1,0\) and 1 Sketch the graph of \(f\). Show also that at \(f\) 's local maximum \(f^{\prime \prime}\) is negative and at each of its local minima \(f^{\prime \prime}\) is positive. (iii) Let \(f\) and \(h\) be the functions we have met before which are defined by $$ f(x)=x^{3} \quad x \in R \quad h(x)= \begin{cases}x^{2} \sin (1 / x) & x \neq 0 \\ 0 & x=0\end{cases} $$ Show that \(f\) and \(h\) each has a stationary point at \(x=0\) but that neither of these stationary points is a local maximum or minimum. Show that the graph of \(f^{\prime \prime}\) actually crosses the \(x\)-axis at 0 ( \(f\) is said to have a point of inflection at \(x=0\) ). (You may think that at a stationary point \(x_{0}\) of a function \(f\) \(f^{\prime \prime}\left(x_{0}\right)<0 \Rightarrow f\) has a local maximum at \(x_{0}\) \(f^{\prime \prime}\left(x_{0}\right)>0 \Rightarrow f\) has a local minimum at \(x_{0}\) \(f^{\prime \prime}\left(x_{0}\right)=0 \Rightarrow f\) has a point of inflection at \(x_{0}\) But that's not entirely true and we shall learn the full story in the next section.)

Let \(f\) be the function given by \(f(x)=\log (1+x)(x>-1)\). Find the Taylor series of \(f\) (about \(x=0\) ). Suppose that we wish to find a polynomial which approximates to \(f\) to within \(\varepsilon(>0)\) throughout the interval \(\left[-\frac{1}{2}, 1\right]\). Show that such a polynomial exists.

As we defined earlier a function \(f\) is strictly increasing if whenever \(a0\) for each \(x\) then \(f\) is strictly increasing: (ii) if \(f^{\prime}(x) \geqslant 0\) for each \(x\) then \(f\) is increasing; (iii) if \(f^{\prime}(x)<0\) for cach \(x\) then \(f\) is strictly decreasing: (iv) if \(f^{\prime}(x) \leqslant 0\) for each \(x\) then \(f\) is decreasing. Deduce that \(\sin x \leqslant x\) for each \(x \geqslant 0\)

We saw in exercise 5 on page 157 that if \(f\) and \(g\) are differentiable at \(x_{0}\) (with suitably-overlapping domains) with \(f\left(x_{0}\right)=g\left(x_{0}\right)=0\) and \(g^{\prime}\left(x_{0}\right) \neq 0\) then $$ \lim _{x \rightarrow x_{0}} \frac{f(x)}{g(x)}=\frac{f^{\prime}\left(x_{0}\right)}{g^{\prime}\left(x_{0}\right)} $$ But what if \(g^{\prime}\left(x_{0}\right)=0\) ? We can use the previous exercise to extend this result. So assume now that \(f\) and \(g\) are differentiable in an interval around \(x_{0}\) with \(f\left(x_{0}\right)=g\left(x_{0}\right)=0\) and that $$ \lim _{x \rightarrow x_{0}} \frac{f^{\prime}(x)}{g^{\prime}(x)} $$ exists. Then show that $$ \lim _{x \rightarrow x_{0}} \frac{f(x)}{g(x)}=\lim _{x \rightarrow x_{0}} \frac{f^{\prime}(x)}{g^{\prime}(x)} $$ This is known as L'Hopital's rule (after the French mathematician Guillaume L'Hopital who published it in 1696 , although he apparently learnt it from Johann Bernoulli). Use it to evaluate the following limits, where \(\alpha\) is a fixed number. (i) \(\lim _{x \rightarrow 1} \frac{\log x}{x-1}\) (ii) \(\lim _{x \rightarrow 0} \frac{\cos x-1}{x^{2}}\) (iii) \(\lim _{x \rightarrow 0} \frac{\log (1+\alpha x)}{x}\) (iv) \(\lim _{n \rightarrow \infty}\left(1+\frac{\alpha}{n}\right)^{n}\)
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