### 数学代写|微积分代写Calculus代写|Absolute value

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## 数学代写|微积分代写Calculus代写|Absolute value

The idea of absolute value can be visualized as the distance from the number to zero on the number line (figure 12 ).

Figure 12 Absolute value as the distance on the number line between the number and zero

Because the number 3 is located 3 units away from zero on the number line, $|3|=3$. Because the number $-2$ is located 2 units away from zero on the number line, $|-2|=2$.

Alternately, we can think of folding the number line at the number 0 , folding the left side of the number line onto the right side. The absolute value of a number is where it is located after the folding. The formal definition of absolute value says that we leave positive numbers alone but “reflect” or “fold” the negative numbers by taking their negatives.

How can we simplify $\sqrt{x^{2}}$ ? Many assume the answer is $x$, but this is not correct, because the square root operation always gives us the nonnegative number with the square that is inside. For instance, $\sqrt{16}=$ 4 and not $-4$ and not $\pm 4$. Therefore,
$$\sqrt{4^{2}}=\sqrt{16}=4$$
whereas
$$\sqrt{(-4)^{2}}=\sqrt{16}=4(\operatorname{not}-4) .$$
This illustrates the following fact:
$$\sqrt{x^{2}}=|x| .$$

## 数学代写|微积分代写Calculus代写|Absolute value equations and inequalities

Which numbers satisfy $|x|=3$ ? The two numbers $x=3$ and $x=-3$ are the only solutions to this equation. See figure $13 .$
ABSOLUTE VALUE EQUATIONS
If $a \geq 0$, then
$$|x|=a \text { if and only if } x=\pm a .$$
Example 5 Solve $x^{2}=16$
Solution We begin by taking the square root of both sides of the equation and simplifying:
\begin{aligned} \sqrt{x^{2}} &=\sqrt{16} \ |x| &=4 \end{aligned}
Next we solve the absolute value equation:
$$x=\pm 4$$

It is customary to skip the middle two steps when solving the equation of example 5 , writing
$$\begin{array}{r} x^{2}=16, \ x=\pm 4 . \end{array}$$
ABSOLUTE VALUE INEQUALITIES, < If $a>0$, then
$$|x|<a \text { if and only if }-a<x<a \text {. }$$
Example 6 Solve $|4 x-1|<3$.
Solution First we rewrite the equation as a compound inequality:
Then we proceed to isolate $x$ in the middle of the inequality by adding 1 to all three parts of the compound inequality and dividing all parts by 4 :
$$\begin{array}{r} -3<4 x-1<3 \ -2<4 x<4 \ -\frac{1}{2}<x<1 \end{array}$$
The solution is $-\frac{1}{2}<x<1$.
If the variable is both inside and outside the absolute values, then the situation can quickly become more complicated. The solution method of example 6 needs to be modified and expanded to solve an inequality such as $|2 x-3|<5 x$.

## 数学代写|微积分代写Calculus代写|Distance on the number line

Consider the distance on the number line between the numbers 3 and 7 (figure 14). Subtracting the two numbers may or may not give that distance:
\begin{aligned} &7-3=4 \ &3-7=-4 \end{aligned}

However, taking the absolute value of the difference between the numbers does the trick:
\begin{aligned} &|7-3|=|4|=4 \ &|3-7|=|-4|=4 \end{aligned}
The absolute value of the difference still works even if one number is positive and the other is negative (see figure 15 ):
\begin{aligned} &|(-5)-3|=|-8|=8 \ &|3-(-5)|=|8|=8 \end{aligned}
Figure 15 The distance between the numbers 3 and $-5$
The calculation works equally well when both numbers are negative (try it).

Often in calculus, formulas for intuitive ideas such as length, area, volume, average value, and many others can be developed in a manner similar to the development of the formula for distance on the number line. Rather than offer a “proof” of such formulas, which would require us to have already defined (or described axiomatically) the term in question, we offer the formula as the definition of the term instead.

## 数学代写|微积分代写Calculus代写|Absolute value

42=16=4

(−4)2=16=4(不是−4).

X2=|X|.

## 数学代写|微积分代写Calculus代写|Absolute value equations and inequalities

|X|=一个 当且仅当 X=±一个.

X2=16 |X|=4

X=±4

X2=16, X=±4.

|X|<一个 当且仅当 −一个<X<一个.

−3<4X−1<3 −2<4X<4 −12<X<1

## 数学代写|微积分代写Calculus代写|Distance on the number line

7−3=4 3−7=−4

|7−3|=|4|=4 |3−7|=|−4|=4

|(−5)−3|=|−8|=8 |3−(−5)|=|8|=8

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