### 数学代写|微积分代写Calculus代写|MATH1051

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|微积分代写Calculus代写|Complete curve-sketching examples

Example 1 provided the information needed to sketch the curve. What if we need to gather this information? The complete process is lengthy but highly informative. The next example illustrates the entire process for a relatively simple function.
Example 2 Sketch the graph of $f(x)=\sqrt[3]{x^2}$.
Solution For any function, a good place to start is to determine its domain. Because $f$ contains only an odd root and no denominator, it is defined on all real numbers.

Our earlier list of information that calculus can provide gives us a list of what to explore. We start with discontinuities. Because root functions are continuous where defined, the function is continuous everywhere; there are no discontinuities.

Next on the list is corners or vertical tangents, which may be identified as part of the process of finding intervals of increase/decrease and extreme points. We defer this item momentarily.

For intervals of increase/decrease, we need to find the derivative and identify critical numbers. Differentiating, we have
$$f^{\prime}(x)=\frac{2}{3} x^{-1 / 3}=\frac{2}{3 \sqrt[3]{x}} .$$
To find critical numbers, we set both the numerator and the denominator equal to zero and solve; $2=0$ has no solutions and $3 \sqrt[3]{x}=0$ has solution $x=0$. The only critical number is $x=0$, and our chart is completed easily:
\begin{tabular}{c|c|c|c}
interval & sign of $f^{\prime}$ & inc/dec & local extrema \
\hline$(-\infty, 0)$ & $-$ & decreasing & $\leftarrow$ local min at $x=0$ \
$(0, \infty)$ & $+$ & increasing &
\end{tabular}
For the purpose of graphing, we need the extreme points and not just their locations, so we find the $y$-coordinate as well:
$$f(0)=\sqrt[3]{0^2}=0 .$$
The local minimum point is $(0,0)$. Notice that because $f^{\prime}$ is undefined at $x=0$, there is no horizontal tangent line at this local min. In other words, the local min might be at a corner in the graph.

We continue by finding intervals of concavity and inflection points. The second derivative is
$$f^{\prime \prime}(x)=-\frac{2}{9} x^{-4 / 3}=-\frac{2}{9 x^{4 / 3}}=-\frac{2}{9 \sqrt[3]{x^4}}$$

## 数学代写|微积分代写Calculus代写|Optimization example: maximum enclosed area

The general solution strategy for an optimization problem is to determine the quantity to be optimized, make that quantity the value of a function, and then find the extreme values of that function. Let’s examine the details of this strategy using an example.

Example 1 A farmer’s child has purchased a piglet. The farmer has given the child 60 feet of fencing left over from another project. Using the side of the barn as one side of a rectangular pig pen, the child wishes to enclose the largest area possible. What dimensions should be used?

Solution Perhaps the first step in solving an optimization problem is to recognize that it is an optimization problem. This is accomplished by noticing that the stated task involves the largest, the smallest, the greatest, the best, the maximum, the minimum, or [insert optimum word here]. In this case, largest is the word used to indicate an optimum value.

As when working any word problem, we draw a picture, if possible. The pig pen is described as a rectangle, so we draw a rectangle. We are lonking at the ground from above (a top view, or aerial view). We depict a barn along one side of the rectangle. See figure 1.

Although nothing in this example is changing (this is not a related rates exercise), so our diagram is static (diagrams for related rates exercises are dynamic), it is still helpful to visualize various possibilities. We are told that there is 60 feet of fencing to make the rectangular pig pen.

We could make the pig pen very wide but not very long (figure 2 , left), very long but not very wide (figure 2, right), or something in between.

We cannot draw all possible configurations and check their areas, for there are infinitely many possibilities. For this reason, we introduce one or more variables to help. Let’s use $\ell$ for length and $w$ for width, as in figure 3 . Then area, which is the quantity we wish to maximize, is given by
$$A=\ell \cdot w .$$

# 微积分代考

## 数学代写|微积分代写Calculus代写|Complete curve-sketching examples

$$f^{\prime}(x)=\frac{2}{3} x^{-1 / 3}=\frac{2}{3 \sqrt[3]{x}} .$$

$$f(0)=\sqrt[3]{0^2}=0$$

$$f^{\prime \prime}(x)=-\frac{2}{9} x^{-4 / 3}=-\frac{2}{9 x^{4 / 3}}=-\frac{2}{9 \sqrt[3]{x^4}}$$

## 数学代写|微积分代写Calculus代写|Optimization example: maximum enclosed area

$$A=\ell \cdot w$$

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## MATLAB代写

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