/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 As we defined earlier a function... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 10

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 \(x \geqslant 0\)

Short Answer

Expert verified
The function \( f \) shows different behaviors based on \( f'(x) \), proving \( \sin x \leq x \) for \( x \geq 0 \).

Step by step solution

01

Understand what the derivative tells us about the function

The derivative, \( f'(x) \), represents the rate at which the function \( f \) is changing at the point \( x \). If \( f'(x) > 0 \), the function is increasing at point \( x \). Conversely, if \( f'(x) < 0 \), the function is decreasing at \( x \). When \( f'(x) = 0 \), the function has a horizontal tangent and might be at a local maximum, minimum or an inflection point.
02

Prove part (i) - f is strictly increasing if f'(x) > 0

Assume \( a < b \). Since \( f \) is differentiable and \( f'(x) > 0 \) for all \( x \) in the interval, by Mean Value Theorem, \( f'(c) = \frac{f(b) - f(a)}{b - a} > 0 \) for some \( c \) in \( (a, b) \). Therefore, \( f(b) - f(a) > 0 \), implying \( f(b) > f(a) \), so \( f \) is strictly increasing.
03

Prove part (ii) - f is increasing if f'(x) 鈮� 0

Again by Mean Value Theorem, \( f'(c) = \frac{f(b) - f(a)}{b - a} \geq 0 \) when \( a < b \). Therefore, \( f(b) - f(a) \geq 0 \), implying \( f(b) \geq f(a) \), so \( f \) is increasing.
04

Prove part (iii) - f is strictly decreasing if f'(x) < 0

Using similar logic, if \( f'(x) < 0 \) for each \( x \), then by Mean Value Theorem, \( f'(c) = \frac{f(b) - f(a)}{b - a} < 0 \). Therefore, \( f(b) - f(a) < 0 \), implying \( f(b) < f(a) \), so \( f \) is strictly decreasing.
05

Prove part (iv) - f is decreasing if f'(x) 鈮� 0

Likewise, if \( f'(x) \leq 0 \) for each \( x \), then by Mean Value Theorem, \( f'(c) = \frac{f(b) - f(a)}{b - a} \leq 0 \). Therefore, \( f(b) - f(a) \leq 0 \), implying \( f(b) \leq f(a) \), so \( f \) is decreasing.
06

Deduce sin(x) 鈮� x for each x 鈮� 0

Consider the function \( g(x) = x - \sin(x) \). Its derivative is \( g'(x) = 1 - \cos(x) \geq 0 \) for all \( x \geq 0 \), since \( 0 \leq \cos(x) \leq 1 \). Thus, \( g(x) \) is increasing for \( x \geq 0 \). Since \( g(0) = 0 \), it follows that \( g(x) \geq 0 \) for \( x \geq 0 \), meaning \( x - \sin(x) \geq 0 \). Hence, \( \sin(x) \leq x \) for all \( x \geq 0 \).

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Key Concepts

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

Strictly Increasing Functions
A function is called strictly increasing if, whenever you have two points, say \(a\) and \(b\), with \(a < b\), it follows that the function's value at \(a\) is less than its value at \(b\). This concept is critical because it implies that no two points will have the same function value as long as you move in a forward direction along the domain.

In mathematical terms, if \(f\) is strictly increasing, then for every pair of numbers \(a\) and \(b\) in the domain, \(a
Understanding strictly increasing functions helps us discern the behavior of a function just by looking at its derivative.
  • If the derivative, \(f'(x)\), is greater than zero for all \(x\) in the interval, the function is uniformly moving upwards as \(x\) increases, which confirms it's strictly increasing.
  • This knowledge is invaluable when analyzing continuous functions, as it provides insight into their overall trend and direction without plotting every single point.
Mean Value Theorem
The Mean Value Theorem (MVT) is a fundamental principle in calculus. It essentially bridges the behavior of a function over an interval with its derivative. The theorem states that for any differentiable function \(f\) over a closed interval \([a, b]\), there exists at least one point \(c\) in this interval where the instantaneous rate of change (or derivative) \(f'(c)\) equals the average rate of change of the function over the interval.

Mathematically, it can be expressed as: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \] This principle is pivotal in proving whether functions are increasing or decreasing by using derivatives.
  • The MVT implies that if \(f'(x) > 0\) throughout an interval \([a, b]\), then \(f\) will exhibit a positive change over that interval, making \(f\) strictly increasing.
  • The theorem facilitates a deeper understanding of function behavior by linking local derivative behavior with global function trends.
Derivative Analysis
Analyzing the derivative of a function provides a wealth of information about the function鈥檚 behavior. The derivative tells us the rate of change of the function's output with respect to changes in its input.

There are several insights we can obtain:
  • If the derivative \( f'(x) > 0 \), the function is increasing at that point as it suggests the output is growing as input increases.
  • If \( f'(x) < 0 \), the function is decreasing, signaling that the output declines as the input grows.
  • A derivative of zero, \( f'(x) = 0 \), suggests a potential local maximum, minimum, or inflection point depending on the surrounding values.
Understanding and performing derivative analysis is crucial for predicting future behavior without needing graph plotting.

Derivatives simplify the complex task of function trend analysis into simple algebraic calculations.
Decreasing Functions
A function is considered decreasing if as you move from left to right across its domain, the function values never increase. More formally, a function \(f\) is said to be decreasing on an interval if for any two points \(a\) and \(b\) within it where \(a < b\), it holds that \(f(a) \geq f(b)\).

There are two scenarios for decreasing functions:
  • Strictly Decreasing: This occurs when \(f(a) > f(b)\). This means every step forward in the domain results in a lower function value, i.e., \(f'(x) < 0\).
  • Decreasing: This occurs when \(f(a) \geq f(b)\), allowing for constant values across \(a\) and \(b\), indicated by \(f'(x) \leq 0\).
Understanding decreasing functions through their derivative gives a clear picture of their downward behavior across any interval. Thus, through analyzing derivatives, we can rightly predict whether a function continuously falls across a domain.

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Most popular questions from this chapter

Find the Taylor series (about \(x=0\) ) of the function cos and of the function \(\cosh\). In each case suppose that we wish to find a polynomial which approximates to the function to within \(\varepsilon(>0)\) throughout the interval \([-k, k]\). Show that such a polynomial exists.

(i) The function $$ y=F(x)=\sqrt{\left(1-x^{2}\right)} \quad x \in(-1,1) $$ can be expressed in the alternative 'parametric' form $$ x=f(t)=\cos t \quad y=g(t)=\sin t \quad t \in(0, \pi) $$ Find \(F^{\prime}(x)\) and verify that it equals \(g^{\prime}(t) / f^{\prime}(t)\). (ii) In general suppose that the function \(y=F(x)\) is expressed in parametric form \(x=f(t), y=g(t)\) where \(f\) and \(g\) are differentiable and \(f^{\prime}(t) \neq 0\). Show that $$ F^{\prime}(x)=\frac{g^{\prime}(t)}{f^{\prime}(t)} $$ Express this result in terms of \(\mathrm{d} y / \mathrm{d} x\) etc. and note that once again this appears to behave as a sensible fraction.

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 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) \(\lim _{n \rightarrow \infty}\left(1+\frac{\alpha}{n}\right)^{n}\)

Let \(f\) and \(g\) be functions with suitably-overlapping domains and assume that each of the functions can be differentiated \(n\) times. Use the principle of mathematical induction to show that the product \(f g\) is differentiable \(n\) times and that $$ \begin{aligned} (f g)^{(n)}=&\left(\begin{array}{l} n \\ 0 \end{array}\right) f g^{(n)}+\left(\begin{array}{l} n \\ 1 \end{array}\right) f^{\prime} g^{(n-1)}+\left(\begin{array}{l} n \\ 2 \end{array}\right) f^{\prime \prime} g^{(n-2)}+\cdots \\ &+\left(\begin{array}{l} n \\ r \end{array}\right) f^{(0)} g^{(n-r)}+\cdots+\left(\begin{array}{l} n \\ n \end{array}\right) f^{(n)} g \end{aligned} $$ where \(\left(\begin{array}{l}n \\ r\end{array}\right)\) is the "binomial coefficient' \(n ! /(r !(n-r) !)\). This is known as Leibniz' rule. Find \((f g)^{(7)}(x)\) where \(f(x)=x^{2}\) and \(g(x)=\mathrm{e}^{2 x}(x \in \mathbb{R})\).
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