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By admin 数学代写 2022年2月22日

This happens if $2 x-1>2$ or if $2 x-1<-2$. Thus the solution is $x>3 / 2$ or $x<-1 / 2 .$ Written in terms of intervals this is $\left(\frac{3}{2}, \infty\right) \cup\left(-\infty,-\frac{1}{2}\right)$. Example 2.4.17 Solve $|x+1|=|2 x-2|$ There are two ways this can happen. It could be the case that $x+1=2 x-2$ in which case $x=3$ or alternatively, $x+1=2-2 x$ in which case $x=1 / 3$. Example 2.4.18 Solve $|x+1| \leq|2 x-2|$ In order to keep track of what is happening, it is a very good idea to graph the two relations, $y=|x+1|$ and $y=|2 x-2|$ on the same set of coordinate axes. This is not a hard job. $|x+1|=x+1$ when $x>-1$ and $|x+1|=-1-x$ when $x \leq-1$. Therefore, it is not hard to draw its graph. Similar considerations apply to the other relation. Functions and their graphs are discussed formally later but I assume the reader has seen these things. The result is
Equality holds exactly when $x=3$ or $x=\frac{1}{3}$ as in the preceding example. Consider $x$ between $\frac{1}{3}$ and 3 . You can see these values of $x$ do not solve the inequality. For example $x=1$ does not work. Therefore, $\left(\frac{1}{3}, 3\right)$ must be excluded. The values of $x$ larger than 3 do not produce equality so either $|x+1|<|2 x-2|$ for these points or $|2 x-2|<|x+1|$ for these points. Checking examples, you see the first of the two cases is the one which holds. Therefore, $[3, \infty)$ is included. Similar reasoning obtains $\left(-\infty, \frac{1}{3}\right]$. It follows the solution set to this inequality is $\left(-\infty, \frac{1}{3}\right] \cup[3, \infty)$.
Example 2.4.19 Suppose $\varepsilon>0$ is a given positive number. Obtasn a number, $\delta>0$, such that if $|x-1|<\delta$, then $\left|x^{2}-1\right|<\varepsilon .$
First of all, note $\left|x^{2}-1\right|=|x-1||x+1| \leq(|x|+1)|x-1| .$ Now if $|x-1|<1$, it follows $|x|<2$ and so for $|x-1|<1,\left|x^{2}-1\right|<3|x-1|$. Now let $\delta=\min \left(1, \frac{\pi}{3}\right)$. This notation means to take the minimum of the two numbers, 1 and $\frac{2}{3}$. Then if $|x-1|<\delta,\left|x^{2}-1\right|<3|x-1|<3 \frac{\varepsilon}{3}=\varepsilon$.
2.5 Exercises
Solve $(3 x+2)(x-3) \leq 0$.
Solve $(x-1)(2 x+1)>2$.
Solve $(3 x+2)(x-3)>0$.
Solve $x^{2}-2 x \leq 0$.
Solve $\frac{x+2}{3 x-2}<0$.
Solve $(x+2)(x-2)^{2} \leq 0$.
Solve $\frac{x+1}{x+3}<1$.
Solve $\frac{3 x-4}{x^{2}+2 x+2} \geq 0$.
Solve $(x-1)(2 x+1) \leq 2$.
Solve $\frac{3 x+9}{x^{2}+2 x+1} \geq 1$.
Solve $\frac{z^{2}+2 x+1}{3 z+7}<1$.
Describe the set of numbers, $a$ such
Solve $|x+1|=|2 x-3|$. that there is no solution to $|x+1|=$ $4-|x+a|$
Solve $|x+1|=|2 x-3|$.
Solve $|3 x+1|<8$. Give your answer
Suppose $0b^{-1}$. in terms of intervals on the real line.
Sketch on the number line the solu- 25 . Show that if $|x-6|<1$, then $|x|<$ tion to the inequality $|x-3|>2$. $7 .$
Sketch on the number line the solu-
Suppose $|x-8|<2$. How large can tion to the inequality $|x-3|<2$. $|x-5|$ be?
Show $|x|=\sqrt{x^{2}}$.
Obtain a number, $\delta>0$, such that if
Solve $|x+2|<|3 x-3|$. $|x-1|<\delta$, then $\left|x^{2}-1\right|<1 / 10 .$
Tell when equality holds in the trian-
Obtain a number, $\delta>0$, such that if gle inequality. $|x-4|<\delta$, then $|\sqrt{x}-2|<1 / 10$
Solve $|x+2| \leq 8+|2 x-4|$.
Suppose $\varepsilon>0$ is a given positive
Solve $(x+1)(2 x-2) x \geq 0$. number. Obtain a number, $\delta>$ 0 , such that if $|x-1|<\delta$, then
Solve $\frac{x+3}{2 x+1}>1$. $|\sqrt{x}-1|<\varepsilon .$ Hint: This $\delta$ will de-
Solve $\frac{x+2}{3 x+1}>2$. pend in some way on $\varepsilon$. You need to tell how.
$2.6$ The Binomial Theorem
Consider the following problem: You have the integers $S_{n}={1,2, \cdots, n}$ and $k$ is an integer no larger than $n$. How many ways are there to fill $k$ slots with these integers starting from left to right if whenever an integer from $S_{n}$ has been used, it cannot be re used in any succoeding slot?
This number is known as permutations of $n$ things taken $k$ at a time and is denoted by $P(n, k)$. It is easy to figure it out. There are $n$ choices for the first slot. For each choice for the fist slot, there remain $n-1$ choices for the second slot. Thus there are $n(n-1)$ ways to fill the first two slots. Now there remain $n-2$ ways to fill the third. Thus there are $n(n-1)(n-2)$ ways to fill the first three slots. Continuing this way, you see there are
$$
P(n, k)=n(n-1)(n-2) \cdots(n-k+1)
$$
ways to do this.
Now define for $k$ a positive integer, $k ! \equiv k(k-1)(k-2) \cdots 1,0 ! \equiv 1$. This is called $k$ factorial. Thus $P(k, k)=k !$ and you should verify that $P(n, k)=\frac{n !}{(n-k) !}$. Now consider the number of ways of selecting a set of $k$ different numbers from $S_{n}$. For each set of $k$ numbers there are $P(k, k)=k !$ ways of listing these numbers in order. Therefore, denoting by $\left(\begin{array}{l}n \ k\end{array}\right)$ the number of ways of selecting a set of $k$ numbers from $S_{n}$, it must be the case that
$$
\left(\begin{array}{c}
n \
k
\end{array}\right) k !=P(n, k)=\frac{n !}{(n-k) !}
$$
Therefore, $\left(\begin{array}{c}n \ k\end{array}\right)=\frac{n !}{k !(n-k) !}$. How many ways are there to seloct no numbers from $S_{n}$ ?

如果 $2 x-1>2$ 或 $2 x-13 / 2$ 或 $x-1$ 和 $|x+1|=-1-x$ 当 $x \leq-1$。因此，绘制它的图形并不难。类似的考虑适用于其他关系。函数和它们的图表稍后会正式讨论，但我假设读者已经看到了这些东西。结果是如上例所示，当 $x=3$ 或 $x=\frac{1}{3}$ 时，等式完全成立。考虑 $\frac{1}{3}$ 和 3 之间的 $x$。您可以看到 $x$ 的这些值并不能解决不等式。例如 $x=1$ 不起作用。因此，必须排除 $\left(\frac{1}{3}, 3\right)$。大于 3 的 $x$ 的值不会产生相等性，因此对于这些点，要么 $|x+1|0$ 是一个给定的正数。得到一个数，$\delta>0$，使得如果 $|x-1|2$。 2. 求解$(3 x+2)(x-3)>0$。 7. 求解 $x^{2}-2 x \leq 0$。 3. 求解 $\frac{x+2}{3 x-2}b^{-1}$。就实线上的间隔而言。 14. 在数轴上画出溶液 25。证明如果 $|x-6|2$。 $7 .$ 15. 在数轴上画出解决方案 26. 假设 $|x-8|0$，这样如果 17.求解$|x+2|0$，使得如果 gle 不等式。 $|x-4|0$ 是一个给定的正数 20. 求解 $(x+1)(2 x-2) x \geq 0$。数字。获得一个数 $\delta>$ 0 ，使得如果 $|x-1|1$。 $|\sqrt{x}-1|2$。以某种方式挂在 $\varepsilon$ 上。你需要告诉如何。 $2.6$ 二项式定理考虑以下问题：您有整数 $S_{n}=\{1,2, \cdots, n\}$ 并且 $k$ 是不大于 $n$ 的整数。如果每当使用来自 $S_{n}$ 的整数时，有多少种方法可以用这些整数从左到右填充 $k$ 插槽，它不能在任何后续插槽中重新使用？这个数字被称为每次取 $k$ 的 $n$ 个事物的排列，用 $P(n, k)$ 表示。很容易弄清楚。第一个插槽有 $n$ 个选项。对于第一个插槽的每个选择，第二个插槽仍有 $n-1$ 个选择。因此，有 $n(n-1)$ 种方式来填充前两个插槽。现在还有 $n-2$ 种方式来填充第三个。因此，有 $n(n-1)(n-2)$ 种方式来填充前三个插槽。继续这样下去，你会看到有 $$ P(n, k)=n(n-1)(n-2) \cdots(n-k+1) $$ 方法来做到这一点。现在为 $k$ 定义一个正整数 $k ！ \equiv k(k-1)(k-2) \cdots 1,0 ！ \相当于1美元。这称为$k$ 阶乘。因此 $P(k, k)=k !$ 并且你应该验证 $P(n, k)=\frac{n !}{(n-k) !}$。现在考虑从 $S_{n}$ 中选择一组 $k$ 个不同数字的方法的数量。对于每组 $k$ 个数字，有 $P(k, k)=k !$ 种按顺序列出这些数字的方法。因此，用 $\left(\begin{array}{l}n \\ k\end{array}\right)$ 表示从 $S_{n}$ 中选择一组 $k$ 个数的方式的数量，一定是这样的 $$ \left(\begin{数组}{c} n \\ ķ \end{数组}\right) k !=P(n, k)=\frac{n !}{(n-k) !} $$ 因此，$\left(\begin{array}{c}n \\ k\end{array}\right)=\frac{n !}{k !(n-k) !}$。有多少种方法可以从 $S_{n}$ 中选择没有数字？