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## 数学代写|微积分代写Calculus代考|Average and Instantaneous Speed

In the late sixteenth century, Galileo discovered that a solid object dropped from rest (initially not moving) near the surface of the earth and allowed to fall freely will fall a distance proportional to the square of the time it has been falling. This type of motion is called free fall. It assumes negligible air resistance to slow the object down, and that gravity is the only force acting on the falling object. If $y$ denotes the distance fallen in feet after $t$ seconds, then Galileo’s law is
$$y=16 t^2 \mathrm{ft}$$
where 16 is the (approximate) constant of proportionality. (If $y$ is measured in meters instead, then the constant is close to 4.9.)

More generally, suppose that a moving object has traveled distance $f(t)$ at time $t$. The object’s average speed during an interval of time $\left[t_1, t_2\right]$ is found by dividing the distance traveled $f\left(t_2\right)-f\left(t_1\right)$ by the time elapsed $t_2-t_1$. The unit of measure is length per unit time: kilometers per hour, feet (or meters) per second, or whatever is appropriate to the problem at hand.
Average Speed
When $f(t)$ measures the distance traveled at time $t$,
$$\text { Average speed over }\left[t_1, t_2\right]=\frac{\text { distance traveled }}{\text { elapsed time }}=\frac{f\left(t_2\right)-f\left(t_1\right)}{t_2-t_1}$$

## 数学代写|微积分代写Calculus代考|Average Rates of Change and Secant Lines

Given any function $y=f(x)$, we calculate the average rate of change of $y$ with respect to $x$ over the interval $\left[x_1, x_2\right]$ by dividing the change in the value of $y, \Delta y=f\left(x_2\right)-f\left(x_1\right)$, by the length $\Delta x=x_2-x_1=h$ of the interval over which the change occurs. (We use the symbol $h$ for $\Delta x$ to simplify the notation here and later on.)
DEFINITION The average rate of change of $y=f(x)$ with respect to $x$ over the interval $\left[x_1, x_2\right]$ is
$$\frac{\Delta y}{\Delta x}=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}=\frac{f\left(x_1+h\right)-f\left(x_1\right)}{h}, \quad h \neq 0 .$$
Geometrically, the rate of change of $f$ over $\left[x_1, x_2\right]$ is the slope of the line through the points $P\left(x_1, f\left(x_1\right)\right)$ and $Q\left(x_2, f\left(x_2\right)\right)$ (Figure 2.1). In geometry, a line joining two points of a curve is called a secant line. Thus, the average rate of change of $f$ from $x_1$ to $x_2$ is identical with the slope of secant line $P Q$. As the point $Q$ approaches the point $P$ along the curve, the length $h$ of the interval over which the change occurs approaches zero. We will see that this procedure leads to the definition of the slope of a curve at a point.

Defining the Slope of a Curve
We know what is meant by the slope of a straight line, which tells us the rate at which it rises or falls – its rate of change as a linear function. But what is meant by the slope of a curve at a point $P$ on the curve? If there is a tangent line to the curve at $P$-a line that grazes the curve like the tangent line to a circle-it would be reasonable to identify the slope of the tangent line as the slope of the curve at $P$. We will see that, among all the lines that pass through the point $P$, the tangent line is the one that gives the best approximation to the curve at $P$. We need a precise way to specify the tangent line at a point on a curve.
Specifying a tangent line to a circle is straightforward. A line $L$ is tangent to a circle at a point $P$ if $L$ passes through $P$ and is perpendicular to the radius at $P$ (Figure 2.2). But what does it mean to say that a line $L$ is tangent to a more general curve at a point $P$ ?

To define tangency for general curves, we use an approach that analyzes the behavior of the secant lines that pass through $P$ and nearby points $Q$ as $Q$ moves toward $P$ along the curve (Figure 2.3). We start with what we can calculate, namely the slope of the secant line $P Q$. We then compute the limiting value of the secant line’s slope as $Q$ approaches $P$ along the curve. (We clarify the limit idea in the next section.) If the limit exists, we take it to be the slope of the curve at $P$ and define the tangent line to the curve at $P$ to be the line through $P$ with this slope.

The next example illustrates the geometric idea for finding the tangent line to a curve.

## 数学代写|微积分代写Calculus代考|Average and Instantaneous Speed

$$y=16 t^2 \mathrm{ft}$$

$$\text { Average speed over }\left[t_1, t_2\right]=\frac{\text { distance traveled }}{\text { elapsed time }}=\frac{f\left(t_2\right)-f\left(t_1\right)}{t_2-t_1}$$

## 数学代写|微积分代写Calculus代考|Average Rates of Change and Secant Lines

$$\frac{\Delta y}{\Delta x}=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}=\frac{f\left(x_1+h\right)-f\left(x_1\right)}{h}, \quad h \neq 0 .$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 数学代写|微积分代写Calculus代考|Even Functions and Odd Functions: Symmetry

The graphs of even and odd functions have special symmetry properties.
DEFINITIONS A function $y=f(x)$ is an
\begin{aligned} & \text { even function of } x \text { if } f(-x)=f(x), \ & \text { odd function of } x \quad \text { if } f(-x)=-f(x), \end{aligned}
for every $x$ in the function’s domain.
The names even and odd come from powers of $x$. If $y$ is an even power of $x$, as in $y=x^2$ or $y=x^4$, it is an even function of $x$ because $(-x)^2=x^2$ and $(-x)^4=x^4$. If $y$ is an odd power of $x$, as in $y=x$ or $y=x^3$, it is an odd function of $x$ because $(-x)^1=-x$ and $(-x)^3=-x^3$

The graph of an even function is symmetric about the $\boldsymbol{y}$-axis. Since $f(-x)=f(x)$, a point $(x, y)$ lies on the graph if and only if the point $(-x, y)$ lies on the graph (Figure 1.12a). A reflection across the $y$-axis leaves the graph unchanged.

The graph of an odd function is symmetric about the origin. Since $f(-x)=-f(x)$, a point $(x, y)$ lies on the graph if and only if the point $(-x,-y)$ lies on the graph (Figure 1.12b). Equivalently, a graph is symmetric about the origin if a rotation of $180^{\circ}$ about the origin leaves the graph unchanged. Notice that the definitions imply that both $x$ and $-x$ must be in the domain of $f$.

## 数学代写|微积分代写Calculus代考|Common Functions

A variety of important types of functions are frequently encountered in calculus.
Linear Functions A function of the form $f(x)=m x+b$, where $m$ and $b$ are fixed constants, is called a linear function. Figure 1.14a shows an array of lines $f(x)=m x$. Each of these has $b=0$, so these lines pass through the origin. The function $f(x)=x$ where $m=1$ and $b=0$ is called the identity function. Constant functions result when the slope is $m=0$ (Figure 1.14b).

DEFINITION Two variables $y$ and $x$ are proportional (to one another) if one is always a constant multiple of the other-that is, if $y=k x$ for some nonzero constant $k$.
If the variable $y$ is proportional to the reciprocal $1 / x$, then sometimes it is said that $y$ is inversely proportional to $x$ (because $1 / x$ is the multiplicative inverse of $x$ ).

Power Functions A function $f(x)=x^a$, where $a$ is a constant, is called a power function. There are several important cases to consider.

(a) $f(x)=x^a$ with $a=n$, a positive integer.
The graphs of $f(x)=x^n$, for $n=1,2,3,4,5$, are displayed in Figure 1.15. These functions are defined for all real values of $x$. Notice that as the power $n$ gets larger, the curves tend to flatten toward the $x$-axis on the interval $(-1,1)$ and to rise more steeply for $|x|>1$. Each curve passes through the point $(1,1)$ and through the origin. The graphs of functions with even powers are symmetric about the $y$-axis; those with odd powers are symmetric about the origin. The even-powered functions are decreasing on the interval $(-\infty, 0]$ and increasing on $[0, \infty)$; the odd-powered functions are increasing over the entire real line $(-\infty, \infty)$.

(b) $f(x)=x^a$ with $a=-1$ or $a=-2$.
The graphs of the functions $f(x)=x^{-1}=1 / x$ and $g(x)=x^{-2}=1 / x^2$ are shown in Figure 1.16. Both functions are defined for all $x \neq 0$ (you can never divide by zero). The graph of $y=1 / x$ is the hyperbola $x y=1$, which approaches the coordinate axes far from the origin. The graph of $y=1 / x^2$ also approaches the coordinate axes. The graph of the function $f$ is symmetric about the origin; $f$ is decreasing on the intervals $(-\infty, 0)$ and $(0, \infty)$. The graph of the function $g$ is symmetric about the $y$-axis; $g$ is increasing on $(-\infty, 0)$ and decreasing on $(0, \infty)$.

(c) $a=\frac{1}{2}, \frac{1}{3}, \frac{3}{2}$, and $\frac{2}{3}$.
The functions $f(x)=x^{1 / 2}=\sqrt{x}$ and $g(x)=x^{1 / 3}=\sqrt[3]{x}$ are the square root and cube root functions, respectively. The domain of the square root function is $[0, \infty)$, but the cube root function is defined for all real $x$. Their graphs are displayed in Figure 1.17, along with the graphs of $y=x^{3 / 2}$ and $y=x^{2 / 3}$. (Recall that $x^{3 / 2}=\left(x^{1 / 2}\right)^3$ and $x^{2 / 3}=\left(x^{1 / 3}\right)^2$.)
Polynomials A function $p$ is a polynomial if
$$p(x)=a_n x^n+a_{n-1} x^{n-1}+\cdots+a_1 x+a_0$$
where $n$ is a nonnegative integer and the numbers $a_0, a_1, a_2, \ldots, a_n$ are real constants (called the coefficients of the polynomial). All polynomials have domain $(-\infty, \infty)$. If the leading coefficient $a_n \neq 0$, then $n$ is called the degree of the polynomial. Linear functions with $m \neq 0$ are polynomials of degree 1 . Polynomials of degree 2 , usually written as $p(x)=a x^2+b x+c$, are called quadratic functions. Likewise, cubic functions are polynomials $p(x)=a x^3+b x^2+c x+d$ of degree 3. Figure 1.18 shows the graphs of three polynomials. Techniques to graph polynomials are studied in Chapter 4 .

## 数学代写|微积分代写Calculus代考|Even Functions and Odd Functions: Symmetry

\begin{aligned} & \text { even function of } x \text { if } f(-x)=f(x), \ & \text { odd function of } x \quad \text { if } f(-x)=-f(x), \end{aligned}

## 数学代写|微积分代写Calculus代考|Common Functions

(a) $f(x)=x^a$与$a=n$为正整数。

(b) $f(x)=x^a$与$a=-1$或$a=-2$。

(c) $a=\frac{1}{2}, \frac{1}{3}, \frac{3}{2}$和$\frac{2}{3}$。

$$p(x)=a_n x^n+a_{n-1} x^{n-1}+\cdots+a_1 x+a_0$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

Posted on Categories:Calculus Assignment, 微积分, 数学代写

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## 数学代写|微积分代写Calculus代考|Summation Notation

For adding up long series of numbers like the rectangle areas in a left, right, or midpoint sum, summation or sigma notation is handy.
Summing up the basics
Say you wanted to add up the first 100 multiples of 5 – that’s from 5 to 500 . You could write out the sum like this:
$$5+10+15+20+25+\ldots \ldots \ldots+490+495+500$$
But with sigma notation, the sum is much more condensed.
$$\sum_{i=1}^{100} 5 i$$

This notation just tells you to plug 1 in for the $i$ in $5 i$, then plug 2 into the $i$ in $5 i$, then 3 , then 4 , all the way up to 100 . Then you add up the results. So that’s $5 \times 1$ plus $5 \times 2$ plus $5 \times 3$, and so on, up to $5 \times 100$. It’s the same thing as writing out the sum the long way. By the way, the letter $i$ has no significance. You can write the sum with a $j, \sum_{j=1}^{100} 5 j$, or any other letter you like.

Here’s one more. If you want to add up $10^2+11^2+12^2+\ldots \ldots \ldots \ldots$. $+29^2+30^2$, you can write the sum with sigma notation as follows:
$$\sum_{k=10}^{30} k^2$$

## 数学代写|微积分代写Calculus代考|Writing Riemann sums with sigma notation

You can use sigma notation to write out the right-rectangle sum for the curve $x^2+1$ from the “Approximating Area” sections. Recall the formula for a right sum from the “Approximating area with right sums” section:
$$R_n=\frac{b-a}{n}\left[f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)+\ldots \ldots \ldots . . .+f\left(x_n\right)\right]$$
Here’s the same formula written with sigma notation:
$$R_n=\sum\left[f\left(x_i\right) \cdot\left(\frac{b-a}{n}\right)\right]$$
Now work this out for the six right rectangles in Figure 8-8. You’re figuring the area under $x^2 1$ between 0 and 3 with six rectangles, so the width of each, $\frac{b-a}{n}$, is $\frac{3-0}{6}$, or $\frac{3}{6}$, or $\frac{1}{2}$. So now you’ve got
$$R_6=\sum_{i=1}^6\left[f\left(x_i\right) \cdot \frac{1}{2}\right]$$
Now, because the width of each rectangle is $\frac{1}{2}$, the right edges of the six rectangles fall on the first six multiples of $\frac{1}{2}: 0.5,1,1.5,2$, 2.5, and 3. These numbers are the $x$-coordinates of the six points $x_1$ through $x_6$; they can be generated by the expression $\frac{1}{2} i$, where $i$

equals 1 through 6 . You can check that this works by plugging 1 in for $i$ in $\frac{1}{2} i$, then 2 , then 3 , up to 6 . So now you can replace the $x_i$ in the formula with $\frac{1}{2} i$, giving you
$$R_6=\sum_{i=1}^6\left[f\left(\frac{1}{2} i\right) \cdot \frac{1}{2}\right]$$

## 数学代写|微积分代写Calculus代考|Summation Notation

$$5+10+15+20+25+\ldots \ldots \ldots+490+495+500$$

$$\sum_{i=1}^{100} 5 i$$

$$\sum_{k=10}^{30} k^2$$

## 数学代写|微积分代写Calculus代考|Writing Riemann sums with sigma notation

$$R_n=\frac{b-a}{n}\left[f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)+\ldots \ldots \ldots . . .+f\left(x_n\right)\right]$$

$$R_n=\sum\left[f\left(x_i\right) \cdot\left(\frac{b-a}{n}\right)\right]$$

$$R_6=\sum_{i=1}^6\left[f\left(x_i\right) \cdot \frac{1}{2}\right]$$

$$R_6=\sum_{i=1}^6\left[f\left(\frac{1}{2} i\right) \cdot \frac{1}{2}\right]$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

Posted on Categories:Calculus Assignment, 微积分, 数学代写

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## 数学代写|微积分代写Calculus代考|Approximating area with midpoint sums

A third way to approximate areas with rectangles is to make each rectangle cross the curve at the midpoint of its top side. A midpoint sum is a much better estimate of area than either a left or a right sum. Figure 8-7 shows why.

You can see in Figure 8-7 that the part of each rectangle that’s above the curve looks about the same size as the gap between the rectangle and the curve. A midpoint sum produces such a good estimate because these two errors roughly cancel out each other.
For the three rectangles in Figure 8-8, the widths are 1 and the heights are $f(0.5)=1.25 f(1.5)=3.25$, and $f(2.5)=7.25$. The total area comes to 11.75. Table $8-3$ lists the midpoint sums for the same number of rectangles used in Tables 8-1 and 8-2.

Estimates of the Area under $\boldsymbol{f}(\boldsymbol{x})=\boldsymbol{x}^2+1$ Given by Increasing Numbers of “Midpoint” Rectangles
\begin{tabular}{l|l|}
\hline Number of Rectangles & Area Estimate \
\hline 3 & 11.75 \
\hline 6 & 11.9375 \
12 & $\sim 11.9844$ \
24 & $\sim 11.9961$ \
48 & $\sim 11.9990$ \
96 & $\sim 11.9998$ \
192 & $\sim 11.9999$ \
384 & $\sim 11.99998$ \
\hline
\end{tabular}

Sorry to give away the ending, but you’ll soon see that the exact area is 12. And to see how much faster the midpoint approximations approach the exact answer of 12 than the left or right approximations, compare the three tables. The error with 6 midpoint rectangles is about the same as the error with 192 left or right rectangles!

## 数学代写|微积分代写Calculus代考|The Midpoint Rule

The Midpoint Rule: You can approximate the exact area under a curve between $a$ and $b, \int_a^b f(x) d x$, with a sum of midpoint rectangles given by the following formula. In general, the more rectangles, the better the estimate.
$$\begin{gathered} M_n=\frac{b-a}{n}\left[f\left(\frac{x_0+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+\ldots \ldots \ldots . . .\right. \ \left.+f\left(\frac{x_{n-1}+x_n}{2}\right)\right] \end{gathered}$$
where $n$ is the number of rectangles, $\frac{b-a}{n}$ is the width of each, and the function values are the heights of the rectangles.

The left, right, and midpoint sums are called Riemann sums after the great German mathematician G. F. B. Riemann (1826-66).

## 数学代写|微积分代写Calculus代考|Approximating area with midpoint sums

\begin{tabular}{l|l|}
\hline Number of Rectangles & Area Estimate \hline 3 & 11.75 \hline 6 & 11.9375 \12 &$\sim 11.9844$ \24 &$\sim 11.9961$ \48 &$\sim 11.9990$ \96 &$\sim 11.9998$ \192 &$\sim 11.9999$ \384 &$\sim 11.99998$ \hline
\end{tabular}

## 数学代写|微积分代写Calculus代考|The Midpoint Rule

$$\begin{gathered} M_n=\frac{b-a}{n}\left[f\left(\frac{x_0+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+\ldots \ldots \ldots . . .\right. \ \left.+f\left(\frac{x_{n-1}+x_n}{2}\right)\right] \end{gathered}$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 数学代写|微积分代写Calculus代考|Approximating area with right sums

Now let’s estimate the same area under $f(x)=x^2+1$ from 0 to 3 with right rectangles. This method works just like the left sum method except that each rectangle is drawn so that its right upper corner touches the curve. See Figure 8-6.

The heights of the three rectangles in Figure $8-6$ are given by the function values at their right edges: $f(1)=2, f(2)=5$, and $f(3)=10$. Each rectangle has a width of 1 , so the areas are 2,5 , and 10 , which total 17. Table 8-2 shows the improving estimates you get with more and more right rectangles.

Estimates of the Area under $\boldsymbol{f}(\boldsymbol{x})=\boldsymbol{x}^2+\boldsymbol{1}$ Given by Increasing Numbers of “Right” Rectangles
\begin{tabular}{ll}
Number of Rectangles & Area Estimate \
\hline 3 & 17 \
6 & 14.375 \
12 & $\sim 13.156$ \
24 & $\sim 12.570$ \
48 & $\sim 12.283$ \
96 & $\sim 12.141$ \
192 & $\sim 12.070$ \
384 & $\sim 12.035$ \
\hline
\end{tabular}

## 数学代写|微积分代写Calculus代考|The Right Rectangle Rule

The Right Rectangle Rule: You can approximate the exact area under a curve between $a$ and $b, \int^b f(x) d x$, with a sum of right rectangles given by the following formula. In general, the more rectangles, the better the estimate.
$$R_n=\frac{b-a}{n}\left[f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)+\ldots \ldots \ldots . .+f\left(x_n\right)\right]$$
where $n$ is the number of rectangles, $\frac{b-a}{n}$ is the width of each, and the function values are the heights of the rectangles.

If you compare this formula to the one for a left rectangle sum, you get the complete picture about those subscripts. The two formulas are the same except for one thing. Look at the sums of the function values in both formulas. The right sum formula has one value, $f\left(x_n\right)$, that the left sum formula doesn’t have, and the left sum formula has one value, $f\left(x_0\right)$, that the right sum formula doesn’t have. All the function values between those two appear in both formulas. You can get a handle on this by comparing the three left rectangles from Figure 8-4 to the three right rectangles from Figure 8-6. Their areas and totals, which we calculated earlier, are
Three left rectangles: $\quad 1+2+5=8$
Three right rectangles: $2+5+10=17$

The sums of the areas are the same except for the left-most left rectangle and the right-most right rectangle. Both sums include the rectangles with areas 2 and 5 . Look at how the rectangles are constructed – the second and third rectangles in Figure 8-4 are the same as the first and second rectangles in Figure 8-6.

## 数学代写|微积分代写Calculus代考|Approximating area with right sums

\begin{tabular}{ll}
Number of Rectangles & Area Estimate \hline 3 & 17 \6 & 14.375\12 &$\sim 13.156$ \24 &$\sim 12.570$ \48 &$\sim 12.283$ \96 &$\sim 12.141$ \192 &$\sim 12.070$ \384 &$\sim 12.035$ \hline
\end{tabular}

## 数学代写|微积分代写Calculus代考|The Right Rectangle Rule

$$R_n=\frac{b-a}{n}\left[f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)+\ldots \ldots \ldots . .+f\left(x_n\right)\right]$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 数学代写|微积分代写Calculus代考|Finding Absolute Extrema on a Closed Interval

Every function that’s continuous on a closed interval has an absolute maximum and an absolute minimum value in that interval a highest and lowest point – though, as you see in the following example, there can be a tie for highest or lowest value.

A closed interval like $[2,5]$ includes the endpoints 2 and 5. An open interval like $(2,5)$ excludes the endpoints.

Finding the absolute max and min is a snap. All you do is compute the critical numbers of the function in the given interval, determine the height of the function at each critical number, and then figure the height of the function at the two endpoints of the interval. The greatest of this set of heights is the absolute max; and the least is the absolute min. Example: Find the absolute max and min of $h(x)=\cos (2 x)-2 \sin x$ in the closed interval $\left[\frac{\pi}{2}, 2 \pi\right]$.

1. Find the critical numbers of $h$ in the open interval $\left(\frac{\pi}{2}, 2 \pi\right)$.
\begin{aligned} h(x) & =\cos (2 x)-2 \sin x \ h^{\prime}(x) & =-\sin (2 x) \cdot 2-2 \cos x \ 0 & =-2 \sin (2 x)-2 \cos x \ 0 & =\sin (2 x)+\cos x \ 0 & =2 \sin x \cos x+\cos x \ 0 & =\cos x(2 \sin x+1) \end{aligned}
(by the chain rule)
(now divide both sides by -2 )
(now use a trig identity)
(factor out $\cos x$ )
$$\begin{array}{rlrlrl} \cos x & =0 & \text { or } & & 2 \sin x+1 & =0 \ x=\frac{3 \pi}{2} & & \sin x & =-\frac{1}{2} \ & & x & =\frac{7 \pi}{6}, \frac{11 \pi}{6} \end{array}$$
Thus, the zeros of $h^{\prime}$ are $\frac{7 \pi}{6}, \frac{3 \pi}{2}$, and $\frac{11 \pi}{6}$, and because $h^{\prime}$ is defined for all input numbers, this is the complete list of critical numbers.
2. Compute the function values (the heights) at each critical number.
\begin{aligned} & h\left(\frac{7 \pi}{6}\right)=\cos \left(2 \cdot \frac{7 \pi}{6}\right)-2 \sin \left(\frac{7 \pi}{6}\right)=0.5-2 \cdot(-0.5)=1.5 \ & h\left(\frac{3 \pi}{2}\right)=\cos \left(2 \cdot \frac{3 \pi}{2}\right)-2 \sin \left(\frac{3 \pi}{2}\right)=-1-2 \cdot(-1)=1 \ & h\left(\frac{11 \pi}{6}\right)=\cos \left(2 \cdot \frac{11 \pi}{6}\right)-2 \sin \left(\frac{11 \pi}{6}\right)=0.5-2 \cdot(-0.5)=1.5 \end{aligned}
1. Determine the function values at the endpoints of the interval.
\begin{aligned} & h\left(\frac{\pi}{2}\right)=\cos \left(2 \cdot \frac{\pi}{2}\right)-2 \sin \left(\frac{\pi}{2}\right)=-1-2 \cdot 1=-3 \ & h(2 \pi)=\cos (2 \cdot 2 \pi)-2 \sin (2 \pi)=1-2 \cdot 0=1 \end{aligned}
So, from Steps 2 and 3, you’ve found five heights: 1.5, 1, 1.5, -3 , and 1 . The largest number in this list, 1.5 , is the absolute max; the smallest, -3 , is the absolute min.

The absolute max occurs at two points: $\left(\frac{7 \pi}{6}, 1.5\right)$ and $\left(\frac{11 \pi}{6}, 1.5\right)$. The absolute min occurs at one of the endpoints, $\left(\frac{\pi}{2},-3\right)$. Figure 6-7 shows the graph of $h$.

## 数学代写|微积分代写Calculus代考|Finding Absolute Extrema over a Function’s Entire Domain

A function’s absolute max and absolute min over its entire domain are the highest and lowest values of the function anywhere it’s defined. A function can have an absolute max or min or both or neither. For example, the parabola $y=x^2$ has an absolute min at the point $(0,0)$ – the bottom of its cup shape – but no absolute max because it goes up forever to the left and the right. You could say that its absolute max is infinity if it weren’t for the fact that infinity is not a number and thus it doesn’t qualify as a maximum (ditto, of course, for negative infinity as a minimum).

The basic idea is this: Either a function will max out somewhere or it will go up forever to infinity. And the same idea applies to a min and going down to negative infinity.

To locate a function’s absolute max and min over its domain, find the height of the function at each of its critical numbers – just like in the previous section, except that here you consider all the critical numbers, not just those in a given interval. The highest of these values is the absolute max unless the function rises to positive infinity somewhere, in which case you say that it has no absolute max. The lowest of these values is the absolute min, unless the function goes down to negative infinity, in which case it has no absolute min.

If a function goes up or down infinitely, it does so at its extreme right or left or at a vertical asymptote. So, your last step (after evaluating all the critical points) is to evaluate $\lim {x \rightarrow x} f(x)$ and $\lim {x \rightarrow-\infty} f(x)$ – the so-called end behavior of the function – and the limit of the function as $x$ approaches each vertical asymptote from the left and from the right. If any of these limits equals positive infinity, then the function has no absolute max; if none equals positive infinity, then the absolute max is the function value at the highest of the critical points. And if any of these limits is negative infinity, then the function has no absolute min; if none of them equals negative infinity, then the absolute min is the function value at the lowest of the critical points.

Figure 6-8 shows a couple functions where the above method won’t work. The function $f(x)$ has no absolute max despite the fact that it doesn’t go up to infinity. Its max isn’t 4 because it never gets to 4, and its max can’t be anything less than 4, like 3.999, because it gets higher than that, say 3.9999. The function $g(x)$ has no absolute min despite the fact that it doesn’t go down to negative infinity. Going left, $g(x)$ crawls along the horizontal asymptote at $y=0$. $g$ gets lower and lower, but it never gets as low as zero, so neither zero nor any other number can be the absolute min.

## 数学代写|微积分代写Calculus代考|Finding Absolute Extrema on a Closed Interval

\begin{aligned} h(x) & =\cos (2 x)-2 \sin x \ h^{\prime}(x) & =-\sin (2 x) \cdot 2-2 \cos x \ 0 & =-2 \sin (2 x)-2 \cos x \ 0 & =\sin (2 x)+\cos x \ 0 & =2 \sin x \cos x+\cos x \ 0 & =\cos x(2 \sin x+1) \end{aligned}
(根据链式法则)
(现在两边同时除以-2)
(现在用三角恒等式)
(提出$\cos x$)
$$\begin{array}{rlrlrl} \cos x & =0 & \text { or } & & 2 \sin x+1 & =0 \ x=\frac{3 \pi}{2} & & \sin x & =-\frac{1}{2} \ & & x & =\frac{7 \pi}{6}, \frac{11 \pi}{6} \end{array}$$

\begin{aligned} & h\left(\frac{7 \pi}{6}\right)=\cos \left(2 \cdot \frac{7 \pi}{6}\right)-2 \sin \left(\frac{7 \pi}{6}\right)=0.5-2 \cdot(-0.5)=1.5 \ & h\left(\frac{3 \pi}{2}\right)=\cos \left(2 \cdot \frac{3 \pi}{2}\right)-2 \sin \left(\frac{3 \pi}{2}\right)=-1-2 \cdot(-1)=1 \ & h\left(\frac{11 \pi}{6}\right)=\cos \left(2 \cdot \frac{11 \pi}{6}\right)-2 \sin \left(\frac{11 \pi}{6}\right)=0.5-2 \cdot(-0.5)=1.5 \end{aligned}

\begin{aligned} & h\left(\frac{\pi}{2}\right)=\cos \left(2 \cdot \frac{\pi}{2}\right)-2 \sin \left(\frac{\pi}{2}\right)=-1-2 \cdot 1=-3 \ & h(2 \pi)=\cos (2 \cdot 2 \pi)-2 \sin (2 \pi)=1-2 \cdot 0=1 \end{aligned}

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 数学代写|微积分代写Calculus代考|The First Derivative Test

The First Derivative Test is based on the Nobel-prize-caliber ideas that as you go over the top of a hill, first you go up and then you go down, and that when you drive into and out of a valley, you go down and then up. This calculus stuff is pretty amazing, isn’t it?
Take a number line and put down the critical numbers you found above: 0, -2, and 2. See Figure 6-3.

This number line is now divided into four regions: to the left of -2 , from -2 to 0 , from 0 to 2 , and to the right of 2 . Pick a value from each region, plug it into the derivative, and note whether your result is positive or negative. Let’s use the numbers $-3,-1,1$, and 3 to test the regions.
\begin{aligned} f^{\prime}(x) & =15 x^4-60 x^2 \ f^{\prime}(-3) & =15(-3)^4-60(-3)^2=15 \cdot 81-60 \cdot 9=675 \ f^{\prime}(-1) & =15(-1)^4-60(-1)^2=15-60=-45 \ f^{\prime}(1) & =15(1)^4-60(1)^2=15-60=-45 \ f^{\prime}(3) & =15(3)^4-60(3)^2=15 \cdot 81-60 \cdot 9=675 \end{aligned}
These four results are, respectively, positive, negative, negative, or positive. Now, take your number line, mark each region with the appropriate positive or negative sign, and indicate whether the function is increasing (derivative is positive) or decreasing (derivative is negative). The result is a sign graph. See Figure 6-4.

Figure 6-4 simply tells you what you already know from the graph of $f-$ that the function goes up until -2 , down from -2 to 0 , further down from 0 to 2 , and up again from 2 on.

The function switches from increasing to decreasing at -2 ; you go up to -2 and then down. So at -2 you have a hill or a local maximum. Conversely, because the function switches from decreasing to increasing at 2, you have a valley there or a local minimum. And because the signs of the first derivative don’t switch at zero, there’s neither a min nor a max at that $x$-value (you get an inflection point when this happens).

The last step is to obtain the function values (heights) of these two local extrema by plugging the $x$-values into the original function:
$$\begin{gathered} f(-2)=3(-2)^5-20(-2)^3=64 \ f(2)=3(2)^5-20(2)^3=-64 \end{gathered}$$
The local $\max$ is at $(-2,64)$ and the local $\min$ is at $(-2,64)$.

## 数学代写|微积分代写Calculus代考|The Second Derivative Test

The Second Derivative Test is based on two more prize-winning ideas: That at the crest of a hill, a road has a hump shape, so it’s curving down or concave down; and that at the bottom of a valley, a road is cup-shaped, so it’s curving up or concave up.

The concavity of a function at a point is given by its second derivative: A positive second derivative means the function is concave up, a negative second derivative means the function is concave down, and a second derivative of zero is inconclusive (the function could be concave up, concave down, or there could be an inflection point there). So, for our function $f$, all you have to do is find its second derivative and then plug in the critical numbers you found $(-2,0$, and 2$)$ and note whether your results are positive, negative, or zero. To wit –

\begin{aligned} & f(x)=3 x^5-20 x^3 \ & f^{\prime}(x)=15 x^4-60 x^2 \quad \text { (power rule) } \ & f^{\prime \prime}(x)=60 x^3-120 x \quad \text { (power rule) } \ & f^{\prime \prime}(-2)=60(-2)^3-120(-2)=-240 \ & f^{\prime \prime}(0)=60(0)^3-120(0)=0 \ & f^{\prime \prime}(2)=60(2)^3-120(2)=240 \end{aligned}
At -2 , the second derivative is negative $(-240)$. This tells you that $f$ is concave down where $x$ equals -2 , and therefore that there’s a local max at $x=-2$. The second derivative is positive (240) where $x$ is 2 , so $f$ is concave up and thus there’s a local min at $x=2$. Because the second derivative equals zero at $x=0$, the Second Derivative Test fails – it tells you nothing about the concavity at $x=0$ or whether there’s a local min or max there. When this happens, you have to use the First Derivative Test.

## 数学代写|微积分代写Calculus代考|The First Derivative Test

\begin{aligned} f^{\prime}(x) & =15 x^4-60 x^2 \ f^{\prime}(-3) & =15(-3)^4-60(-3)^2=15 \cdot 81-60 \cdot 9=675 \ f^{\prime}(-1) & =15(-1)^4-60(-1)^2=15-60=-45 \ f^{\prime}(1) & =15(1)^4-60(1)^2=15-60=-45 \ f^{\prime}(3) & =15(3)^4-60(3)^2=15 \cdot 81-60 \cdot 9=675 \end{aligned}

$$\begin{gathered} f(-2)=3(-2)^5-20(-2)^3=64 \ f(2)=3(2)^5-20(2)^3=-64 \end{gathered}$$

## 数学代写|微积分代写Calculus代考|The Second Derivative Test

\begin{aligned} & f(x)=3 x^5-20 x^3 \ & f^{\prime}(x)=15 x^4-60 x^2 \quad \text { (power rule) } \ & f^{\prime \prime}(x)=60 x^3-120 x \quad \text { (power rule) } \ & f^{\prime \prime}(-2)=60(-2)^3-120(-2)=-240 \ & f^{\prime \prime}(0)=60(0)^3-120(0)=0 \ & f^{\prime \prime}(2)=60(2)^3-120(2)=240 \end{aligned}

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 数学代写|微积分代写Calculus代考|Exponential and logarithmic functions

If you can’t memorize the next rule, hang up your calculator.
$$\frac{d}{d x} e^x=e^x$$
That’s right, the derivative of $e^x$ is itself! This is a special function. $e^x$ and its multiples, like $5 e^x$, are the only functions that are their own derivatives.

If the base is a number other than $e$, you have to tweak the derivative by multiplying it by the natural log of the base:
If $y=2^x$, then $y^{\prime}=2^x \ln 2$.
If $y=10^x$, then $y^{\prime}=10^x \ln 10$.
Here’s the derivative of the natural log:
$$\frac{d}{d x} \ln x=\frac{1}{x}$$
If the log base is a number other than $e$, you tweak this derivative – as with exponential functions – except you divide by the natural $\log$ of the base instead of multiplying:
\begin{aligned} & \frac{d}{d x} \log 2 x=\frac{\frac{1}{x}}{\ln 2}=\frac{1}{x \ln 2}, \text { and } \ & \frac{d}{d x} \log x=\frac{1}{x \ln 10}\left(\text { Recall that } \log {10} x \text { is written without the } 10 .\right) \end{aligned}

## 数学代写|微积分代写Calculus代考|Derivative Rules for Experts

These rules, especially the chain rule, can be a bit tough. The product and quotient rules

The Product Rule (for the product (duh) of two functions):
If $y=$ this $\cdot$ that,
then $y^{\prime}$ this’ $\cdot$ that + this $\cdot$ that’
So, for $y=x^3 \cdot \sin x$,
\begin{aligned} y^{\prime} & =\left(x^3\right)^{\prime} \cdot \sin x+x^3 \cdot(\sin x)^{\prime} \ & =3 x^2 \sin x+x^3 \cos x \end{aligned}
The Quotient Rule (bet you can guess what this is for):
If $y=\frac{\text { top }}{\text { bottom’ }}$

Most calculus books give this rule in a slightly different form that’s harder to remember. And some give a “mnemonic” involving the words lodeehi and hideelo or hodeehi and hideeho, which is easy to get mixed up – great, thanks a lot.

Memorize the quotient rule as I’ve written it. You’ll remember what goes in the denominator – no one ever forgets it. The trick is knowing the order of the terms in the numerator. Think of it like this: You’re doing a derivative, so the first thing you do is to take a derivative. The natural place to begin is at the top of the fraction. So the quotient rule begins with the derivative of the top. Remember that, and the rest of the numerator is almost automatic.
Here’s the derivative of $y=\frac{\sin x}{x^4}$ :
\begin{aligned} y^{\prime} & =\frac{(\sin x)^{\prime} \cdot x^4-\sin x \cdot\left(x^4\right)^{\prime}}{\left(x^4\right)^2} \ & =\frac{x^4 \cos x-4 x^3 \sin x}{x^8} \ & =\frac{x^3(x \cos x-4 \sin x)}{x^8} \ & =\frac{x \cos x-4 \sin x}{x^5} \end{aligned}

## 数学代写|微积分代写Calculus代考|Exponential and logarithmic functions

$$\frac{d}{d x} e^x=e^x$$

$$\frac{d}{d x} \ln x=\frac{1}{x}$$

\begin{aligned} & \frac{d}{d x} \log 2 x=\frac{\frac{1}{x}}{\ln 2}=\frac{1}{x \ln 2}, \text { and } \ & \frac{d}{d x} \log x=\frac{1}{x \ln 10}\left(\text { Recall that } \log {10} x \text { is written without the } 10 .\right) \end{aligned}

## 数学代写|微积分代写Calculus代考|Derivative Rules for Experts

\begin{aligned} y^{\prime} & =\left(x^3\right)^{\prime} \cdot \sin x+x^3 \cdot(\sin x)^{\prime} \ & =3 x^2 \sin x+x^3 \cos x \end{aligned}

\begin{aligned} y^{\prime} & =\frac{(\sin x)^{\prime} \cdot x^4-\sin x \cdot\left(x^4\right)^{\prime}}{\left(x^4\right)^2} \ & =\frac{x^4 \cos x-4 x^3 \sin x}{x^8} \ & =\frac{x^3(x \cos x-4 \sin x)}{x^8} \ & =\frac{x \cos x-4 \sin x}{x^5} \end{aligned}

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 数学代写|微积分代写Calculus代考|Miscellaneous algebra

When factoring and conjugate multiplication don’t work, try other basic algebra like adding or subtracting fractions, multiplying or dividing fractions, canceling, or some other form of simplification.
Evaluate $\lim _{x \rightarrow 0} \frac{\frac{1}{x+4}-\frac{1}{4}}{x}$.

Try substitution.
Plug in 0: That gives you $\frac{0}{0}$ – no good.

Simplify the complex fraction (that’s a big fraction that contains little fractions) by multiplying the numerator and denominator by the least common denominator of the little fractions, namely $4(x+4)$.

Note: Adding or subtracting the little fractions in the numerator or denominator also works in this type of problem, but it’s a bit longer than the method here.
\begin{aligned} & \lim {x \rightarrow 0} \frac{\left(\frac{1}{x+4}-\frac{1}{4}\right)}{x} \cdot \frac{4(x+4)}{4(x+4)} \ = & \lim {x \rightarrow 0} \frac{4-(x+4)}{4 x(x+4)} \ = & \lim {x \rightarrow 0} \frac{-x}{4 x(x+4)} \ = & \lim {x \rightarrow 0} \frac{-1}{4(x+4)} \end{aligned}

1. Now substitution works.
$$=\frac{-1}{4(0+4)}=-\frac{1}{16}$$

## 数学代写|微积分代写Calculus代考|Limits at Infinity

In the limits in the last section, $x$ approaches a finite number, but there are also limits where $x$ approaches infinity or negative infinity. Consider the function $f(x)=\frac{1}{x}$. See Figure 3-1.

You can see on the graph (in the first quadrant) that as $x$ gets bigger and bigger – in other words, as $x$ approaches infinity – the height of the function gets lower and lower but never gets to zero. This is confirmed by considering what happens when you plug bigger and bigger numbers into $\frac{1}{x}$. The outputs get smaller and smaller. This graph thus has a horizontal asymptote of $y=0$ (the $x$-axis), and we say that $\lim {x \rightarrow \infty} \frac{1}{x}=0$. The fact that $x$ never actually reaches infinity and that $f$ never gets to zero has no relevance. When we say that $\lim {x \rightarrow \infty} \frac{1}{x}=0$, we mean that as $x$ gets bigger and bigger without end, $f$ gets closer and closer to zero. The function $f$ also approaches zero as $x$ approaches negative infinity, written as $\lim _{x \rightarrow-\infty} \frac{1}{x}=0$.

Horizontal asymptotes
Horizontal asymptotes and limits at infinity go hand in hand you can’t have one without the other. For a rational function like $f(x)=\frac{3 x-7}{2 x+8}$, determining the limit at infinity or negative infinity is the same as finding the location of the horizontal asymptote.
Here’s what you do. First, note the degree of the numerator (that’s the highest power of $x$ in the numerator) and the degree of the denominator. You’ve got three cases:
If the degree of the numerator is greater than the degree of the denominator, for example $f(x)=\frac{6 x^4+x^3-7}{2 x^2+8}$, there’s no horizontal asymptote and the limit of the function as $x$ approaches infinity (or negative infinity) does not exist.
If the degree of the denominator is greater than the degree of the numerator, for example $g(x)=\frac{4 x^2-9}{x^3+12}$, the $x$-axis (the line $y=0$ ) is the horizontal asymptote and $\lim {x \rightarrow \infty} g(x)=\lim {x \rightarrow-\infty} g(x)=0$.) If the degrees of the numerator and denominator are equal, take the coefficient of the highest power of $x$ in the numerator and divide it by the coefficient of the highest power of $x$ in the denominator. That quotient gives you the answer to the limit problem and the height of the asymptote. For example, if $h(x)=\frac{4 x^3-10 x+1}{5 x^3+3 x^2-x} \lim {x \rightarrow \infty} h(x)=\lim {x \rightarrow-\infty} h(x)=\frac{4}{5}$, and $h$ has a horizontal asymptote at $y=\frac{4}{5}$.

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 数学代写|微积分代写Calculus代考|Limits and horizontal asymptotes

Up till now, I’ve been looking at limits where $x$ approaches a regular, finite number. But $x$ can also approach infinity or negative infinity. Limits at infinity exist when a function has a horizontal asymptote. For example, the function in Figure 2-3 has a horizontal asymptote at $y=1$, which the function crawls along as it goes toward infinity to the right and negative infinity to the left. (Going left, the function crosses the horizontal asymptote at $x=-7$ and then gradually comes down toward the asymptote.) The limits equal the height of the horizontal asymptote and are written as
$$\lim {x \rightarrow \infty} f(x)=1 \text { and } \lim {x \rightarrow-\infty} f(x)=1$$

Instantaneous speed
Say you decide to drop a ball out of your second-story window. Here’s the formula that tells you how far the ball has dropped after a given number of seconds (ignoring air resistance):
$$h(t)=16 t^2$$
(where $h$ is the height the ball has fallen, in feet, and $t$ is the amount of time since the ball was dropped, in seconds)
If you plug 1 into $t, h$ is 16 ; so the ball falls 16 feet during the first second. During the first 2 seconds, it falls a total of $16 \cdot 2^2$, or 64 feet, and so on. Now, what if you wanted to determine the ball’s speed exactly 1 second after you dropped it? You can start by whipping out this trusty ol’ formula:
Distance $=$ rate $\cdot$ time, so Rate $=$ distance $/$ time

## 数学代写|微积分代写Calculus代考|Limits and Continuity

A continuous function is simply a function with no gaps – a function that you can draw without taking your pencil off the paper. Consider the four functions in Figure 2-5.

Whether or not a function is continuous is almost always obvious. The first two functions in Figure 2-5 $-f$ and $g$ – have no gaps, so they’re continuous. The next two $-p$ and $q-$ have gaps at $x=3$, so they’re not continuous. That’s all there is to it. Well, not quite. The two functions with gaps are not continuous everywhere, but because you can draw sections of them without taking your pencil off the paper, you can say that parts of them are continuous. And sometimes a function is continuous everywhere it’s defined. Such a function is described as being continuous over its entire domain, which means that its gap or gaps occur at $x-$ values where the function is undefined. The function $p$ is continuous over its entire domain; $q$, on the other hand, is not continuous over its entire domain because it’s not continuous at $x=3$, which is in the function’s domain.

Continuity and limits usually go hand in hand. Look at $x=3$ on the four functions in Figure 2-5. Consider whether each function is continuous there and whether a limit exists at that $x$-value. The first two, $f$ and $g$, have no gaps at $x=3$, so they’re continuous there. Both functions also have limits at $x=3$, and in both cases, the limit equals the height of the function at $x=3$, because as $x$ gets closer and closer to 3 from the left and the right, $y$ gets closer and closer to $f(3)$ and $g(3)$, respectively.

## 数学代写|微积分代写Calculus代考|Limits and horizontal asymptotes

$$\lim {x \rightarrow \infty} f(x)=1 \text { and } \lim {x \rightarrow-\infty} f(x)=1$$

$$h(t)=16 t^2$$
($h$是球下落的高度，单位是英尺，$t$是球下落的时间，单位是秒)

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。