When we take the absolute value of a function, any negative values get changed into positive values. Essentially, |f(x)| will take all of the negative values off(x) and reflect them across thex-axis. However, any values off(x) that are positive or equal to zero will not be changed, because the absolute value of a positive number (or zero) is still the same number.

If we can show thatf(x) has negative values, then |f(x)| will be different fromf(x) at some points, because its negative values will be changed to positive values. In other words, our answer will consist of the function that never has negative values.

Let's look atf(x) = 2x+ 3. Obviously, this equation of a line will have negative values. For example, wherex= –4,f(–4) = 2(–4) + 3 = –5, which is negative. Thus,f(x) has negative values, and if we were to graph |f(x)|, the result would be different fromf(x). Therefore,f(x) = 2x+ 3 isn't the correct answer.

Next, let's look atf(x) =x²– 9. If we letx= 1, thenf(1) = 1 – 9 = –8, which is negative. Thus |f(x)| will not be the same asf(x), and we can eliminate this choice as well.

Now, let's examinef(x) =x²– 2x. We know thatx²by itself can never be negative. However, ifx²is really small, then adding –2xcould make it negative. Therefore, let's evaluatef(x) whenxis a fractional value such as 1/2.f(1/2) = 1/4 – 1 = –3/4, which is negative. Thus, there are some values onf(x) that are negative, so we can eliminate this function.

Next, let's examinef(x) =x⁴+x. In general, any number taken to an even-numbered power must always be non-negative. Therefore,x⁴cannot be negative, because if we multiplied a negative number by itself four times, the result would be positive. However, thexterm could be negative. If we letxbe a small negative fraction, thenx⁴would be close to zero, and we would be left withx, which is negative. For example, let's findf(x) whenx= –1/2.f(–1/2) = (–1/2)⁴+ (–1/2) = (1/16) – (1/2) = –7/16, which is negative. Thus, |f(x)| is not always the same asf(x).

By process of elimination, the answer isf(x) =x⁴+ (1 –x)². This makes sense becausex⁴can't be negative, and because (1 –x)²can't be negative. No matter what we subtract from one, when we square the final result, we can't get a negative number. And if we addx⁴and (1 –x)², the result will also be non-negative, because adding two non-negative numbers always produces a non-negative result. Therefore,f(x) =x⁴+ (1 –x)²will not have any negative values, and |f(x)| will be the same asf(x所有的值)x.

The answer isf(x) =x⁴+ (1 –x)².

We need to solve the two equations |2a – 3| = 5 and |3 – 4b| = 11, in order to determine the possible values of a and b. When solving equations involving absolute values, we must remember to consider both the positive and negative cases. For example, if |x| = 4, then x can be either 4 or –4.

Let's look at |2a – 3| = 5. The two equations we need to solve are 2a – 3 = 5 and 2a – 3 = –5.

2a – 3 = 5 or 2a – 3 = –5

Add 3 to both sides.

2a = 8 or 2a = –2

Divide by 2.

a = 4 or a = –1

Therefore, the two possible values for a are 4 and –1. However, the problem states that both a and b are negative. Thus, a must equal –1.

Let's now find the values of b.

3 – 4b = 11 or 3 – 4b = –11

年代ubtract 3 from both sides.

–4b = 8 or –4b = –14

Divide by –4.

b = –2 or b = 7/2

因为b也必须负,b必须等于2。

We have determined that a is –1 and b is –2. The original question asks us to find |b – a|.

|b – a| = |–2 – (–1)| = | –2 + 1 | = |–1| = 1.

The answer is 1.

In order the find the answer, you must first solve what is inside the absolute value signs.

Following order of operations, you must first multiplywhich equals.

Then you must subtractfromas shown below:

Now, you must take the absolute value ofwhich is positive, the correct answer.

年代ubstitute 0.6 for:

年代ubstitute.

is the absolute value of, which in turn is the sum of a number and seven and a number. Therefore,可以写成“的绝对值之和啊f a number and seven". Since it is equal to, it is three less than the number, so the equation that corresponds to the sentence is

"The absolute value of the sum of a number and seven is three less than the number."

, or, equivalently,