Posts tagged with: absolute value

Mike,
I do not understand number 6 in the absolute value section of your book. Why can’t |0| + |7| = 7? Zero is even and 7 is odd.

I think you may be misinterpreting the question. Here’s #6:

The question is asking which choice could not equal the sum of a and b when |a| + |b| = 7. The value of a or b could be zero, as you point out, but that’s not what the question is asking.

Mike, in your book you wrote |a| = -a ? I thought the absolute value could only be a positive number.

I think you’re referring to this table:

All I’m really trying to do there is remind you that when a number is less than zero, taking its absolute value means changing its sign. For example, say the a in the number line above is –5. Obviously you know that |–5| = 5, but that also means that if a = –5, then |a| = –a, because –a = –(–5) = 5.

Does that help?

Test 6 calc section number 28

There are two equally valid ways to go here. If you’ve read enough of me, though, you won’t be surprised to learn that I’ll suggest the simpler, less mathy way first.

The question tells you that two points are both 3 units away from –4. You don’t need to make any equations to know what those points are! They’re –1 and –7! So just work through the answer choices until you find one that is true for both x = –1 and x = –7. Start with choice A:

    \begin{align*}|x+4|&=3\\\\|-1+4|&=3\\|3|&=3\\3&=3\\\\|-7+4|&=3\\|-3|&=3\\3&=3\end{align*}

Lookit that: already done!

Algebraically, what’s happening here is that you’re looking to write an equation for “the positive difference between x and –4 is 3.” The way you write “the positive difference” algebraically when you don’t know which of the values you’re subtracting is bigger is you put the subtraction in absolute value brackets. Therefore, “the positive difference between x and –4″ is written |x-(-4)|=|x+4|.

The reason I love the simpler way on this question is that you can just as easily write |-4-x| for the positive difference between x and –4. That’s equivalent because of the absolute value brackets, of course, but if that’s what you start out writing, you might not recognize the correct answer choice right away. If, on the other hand, you just start out by finding the answer choice that works for –7 and –1, you can’t go wrong!

Hi Mike, I find these Qs confusing and I lose valuable time trying to think my way through them (usually get them wrong anyway!) What’s a good stepwise approach? Thanks!

If 6 < |x-3| < 7 and x < 0, what is one possible value of |x| ?

I think it’s helpful on questions like this to plug in—to think in terms of actual numbers as much as possible. In this case, I’d look at the inequality, which says that |x-3| must be between 6 and 7, and I’d simplify that by saying that 6.5 is between 6 and 7, so why not just say |x-3|=6.5?

From there, you can say that x-3=6.5, in which case x=9.5, or x-3=-6.5, in which case x=-3.5. Since the question says that x<0, we must choose x=-3.5, which means |x|=3.5. So that’s what we grid.

Algebraically this is a bit trickier, but if you want to see a step-by-step solution, you have to begin by noting that removing the absolute value brackets from the original inequality range results in two possible inequality ranges:

    \begin{align*}6&<x-3<7\\&-or-\\-6&>x-3>-7\end{align*}

Again, we know that x<0, so we only need to deal with the second inequality. Solve for x by adding 3 to each part:

    \begin{align*}-6+3&>x>-7+3\\-3&>x>-4\end{align*}

So x must be between –3 and –4. That means the absolute value of x must be between 3 and 4.

Subject Test Question:

The area bound by the relationship |x|+|y|=2 is

A) 8
B) 1
C) 2
D) 4
E) there is no finite area.

How do you find this algebraically?

You need to capture all the combinations of negative and positive x and y. There are four equations to consider:

xy = 2
–xy = 2
x + (–y) = 2
x + (–y) = 2

Each of those is a line that can be represented in y = mx + b form. In the same order:

y = –x + 2
yx + 2
y = –x – 2
y = x – 2

Those make a square with vertices at (2, 0), (0, 2), (–2, 0), and (0, –2). The diagonals of the square are 4 long, making the sides of the square 2\sqrt{2} long. The area of a square with those side lengths is \left(2\sqrt{2}\right)^2=8.

Subject Test question:

|2x-1| = 4x+5 has how many numbers in its solution set?

A) 0
B) 1
C) 2
D) an infinite number
E) none of above

How do you find the answer algebraically without graphing it?

First of all, graphing this (and many questions on the Subject Tests) is a really good way to go. If you don’t have a graphing calculator of your own but you know how to use one (maybe your school has them?), talk to your teacher and see if you can borrow one for the test. The graphing solution, since I just can’t help myself, looks like this:

Since the slope of the red line is steeper than the slope of the blue line, there will only be one intersection.

To solve this algebraically, you have to solve two equations:

2x – 1 = 4x + 5
–6 = 2x
3 = x

–(2x – 1) = 4x + 5
–2x + 1 = 4x + 5
–4 = 6x
–2/3 = x

Then you have to plug both values back into the original equation to ensure neither is extraneous. Hint: one will be!

|2(–3) – 1| = 4(–3) + 5
|–6 – 1| = –12 + 5
7 = –7

Oops! That’s not true! –3 is an extraneous solution.

|2(–2/3) – 1| = 4(–2/3) + 5
|–4/3 – 1| = –8/3 + 5
7/3 = 7/3

Yep—that one worked just fine.

How to solve q19, pg 154, from your study guide. Thanks.

The thing I tell most students who struggle with this one to do is to graph every choice—that’ll work like a charm because it’ll not only show you exactly which is the right answer, but it’ll also be good practice recognizing how shifts inside and outside of functions work.

The other thing I tell people is that there are two functions that might produce the given graph, and the correct answer is perhaps the less obvious one. Here are both.

absvalgraph

 

For a dog that weighs w pounds and x inches long to qualify for a certain competition, the difference between w and 3/2 x can no more than 5. If a dog with length 24 inches qualify for the competition, what is one possible weight, in pounds, of this dog?

\lvert w-\dfrac{3}{2}(24)\rvert <5

\lvert w-36\rvert <5

-5<w-36<5

31<w<41

My question regarding making absolute value fall in a particular range. In your book you gave two example (pg 149) and my question is what value should I consider as the end? is it the highest value or the (highest value+1) . Because in question#19 pg149 you consideed 181 to be the end, when 181 isn’t included! & what if the inequality symbol is less than or equal to, what should I do in that case?

If you’re doing the problem algebraically (recommended), it doesn’t matter what the inequality symbols is, as long as you’re consistent. Converting, say |w-162|<19 to 143<w<181 is algebraically exactly the same as converting |w-162|\leq 19 to 143\leq w \leq 181. Just take your cue from the question’s answer choices and use the same symbol the question does.

If you’re solving by plugging in, as I do first on pages 149 through 151, then it also doesn’t matter much as long as you’re able to keep track of whether the number you’re testing should be producing true inequalities or false ones.  I used 142 and 182 as examples just to make it a bit more obvious that those numbers SHOULD NOT produce true inequalities, although given the less-than symbol in play I could have used 143 and 181.

Hi mike! This question is from the May 2015 SAT.

(will post photo)

In the xy plane above, f and g are functions defined by f(x)=abs[x] and g(x)=-abs[x] + 3 for all values x. What is the area of the shaded region bounded by the graphs of the two functions?

You don’t even need to post the picture—I can graph those functions:

First, note that the bounded region is a square. You know this if you remember that the absolute value graphs will create 45º angles with the axes, just like the graph of yx does.

Once you’ve realized that, all you need to do is find the length of one of its sides. Best way to go there is to recognize that the diagonal of the square goes from (0, 0) to (0, 3), so it has a length of 3. A square with a diagonal of length 3 will have sides of length \dfrac{3\sqrt{2}}{2}. (Brush up on your 45º-45º-90º triangles if you’re not clear why.)

So the area of the square is \dfrac{3\sqrt{2}}{2}\times\dfrac{3\sqrt{2}}{2}=\dfrac{9}{2}.

I’m just going to make up a symbol for better visualization. The symbol will look like this: #

For all numbers x and y, let x # y be defined by x # y= |x^2-y^2| + 2. What is the smallest possible value of x # y?

This was a 2/5 on the difficulty scale yet I somehow didn’t understand this and still got it wrong. I tried to do some weird algebra that got me nowhere so I moved on. Funny thing this was the only question I got wrong in the section.

A 0
B 1
C 2
D 3
E 4

Think about the least possible value you can have inside absolute value brackets, using a simpler expression. What’s the least possible value of |x|? When x = 0, then |x| = 0. For any other value of x, |x| will be positive, so the least possible value of |x| is 0.

So far, so good? If the least possible value of |x| is 0, then what’s the least possible value of |x| + 2? It’s 2, right? It’s gotta be.

The same thing is going on here. The question says for all values of x and y, so that means x and y can be equal, which would make x^2-y^2 equal 0. So it’s possible to have a 0 in the absolute value brackets, and that’s by definition the least value you can have in any absolute value brackets. From there, the least possible value of the whole expression \left|x^2-y^2\right|+2 is 2.

You know the drill by now. The prize this week will, once again, be beta access to the Math Guide, which is really coming along nicely, if I do say so myself. I kinda can’t believe it’s going to top 300 pages, but at this rate we’re definitely heading in that direction.

Anyway, to get a look at it now and come along for the ride while I finish it up, be the first to answer the following question correctly in the comments, or send me $5 using the link above.

A stereo equipment store owner notices that all his customers spend between $110 and $298, inclusive, in his store when they come in. For no discernible reason other than that I need a difficult math problem, he decides to express the range of dollar amounts his customers spend, s, in an inequality of the form |s – j| ≤ k, where j and k are constants. What is jk?

I’ll post the solution Monday. Good luck!UPDATE: Nice work, Serplet. Solution below.

Questions like this aren’t super common on the SAT, but when they appear they always follow the same pattern:

|variable – middle of range| < distance from middle to ends of range

In this case, the range is 110 < s < 298, so the middle of the range is (110 + 298)/2 = 204 , and the distance from the middle to the ends of the range is (298 – 110)/2 = 94.

So j = 204, and k = 94. Their product is 19176.

For more practice with this kind of question, go here.

Absolute rock. Source.

Disclaimer: this is really minor stuff as far as how often it appears on the SAT, so if you’re looking for quick tips to really raise your score, I suggest you start elsewhere. This kind of question is pretty rare.

I trust you already know the very basics of absolute value: that |5| = 5, and |-5| = 5, etc. If you don’t, leave this page open and view a quick tutorial here before continuing. Ok, all caught up? Let’s do this.

Absolute values and inequalities

Remember that |x| = 5 means that x = 5, or x = -5. You can draw similarly simple conclusions with inequalities. If I told you that |y| < 3, and y is an integer, then what are the possible values for y? There aren’t many: y could equal 2, 1, 0, -1, or -2. In other words, y has to be less than 3, and greater than -3. Like so:

|x| < 5

< 5 and x > -5

-5 < x < 5

Here’s an example of the kind of question you’ll usually get on the SAT:

  1. In order to be considered “good for eating” by the La’Urthg Orcs of Kranranul, a human must weigh between 143 and 181 pounds. Which of the following inequalities gives all the possible weights, w, that a human in Kranranul should NOT want to be?
     
    (A) |w – 143| < 38
    (B) |w – 162| < 19
    (C) |w + 38| < 181
    (D) |w – 181| < 22
    (E) |w – 162| < 24

Answer, explanation, and so much more below the cut:

First of all, it’s possible to plug in here, although it matters greatly what numbers you choose. Picking a number right in the middle of the range (like 165) probably won’t help you out much. To plug in successfully, choose weights that should just barely be within the range (like 144 or 180) and then, if you still haven’t eliminated all the answers, choose weights that should just barely be outside the range (like 142 or 182). Be careful with this second part! You’re looking to eliminate any choices that WORK when you know you picked a number that SHOULDN’T WORK. Let’s see what happens when we pick w = 144.

A person who weighs 144 pounds should try to lose weight, pronto, lest he/she become a meal for the orcs. So we’re looking for choices that will give us TRUE inequalities when we plug in 144:

(A) |144 – 143| < 38
|1| < 38
1 < 38 — TRUE

(B) |144 – 162| < 19
|-18| < 19
18 < 19 — TRUE

(C) |144 + 38| < 181 
 |182| < 181
 182 < 181 — FALSE

(D) |144 – 181| < 22 
 |-37| < 22 
 37 < 22 — FALSE

(E) |144 – 162| < 24
|-18| < 24
18 < 24 — TRUE

So, bummer. our first plug-in only eliminated 2/5 choices. What happens when we plug in a number that shouldn’t work, like 141? Remember, now we’re looking to eliminate anything that gives us TRUE, because a 141 pound person is NOT considered delectable by the orcs. Note that I’m not bothering with (C) and (D) since we’ve already eliminated them.

(A) |141 – 143| < 38 
 |-2| < 38 
 2 < 38 — TRUE

(B) |141 – 162| < 19
|-21| < 19
21 < 19 — FALSE

(E) |141 – 162| < 24 
 |-21| < 24 
 21 < 24 — TRUE

Only (B) was true when we needed it to be true, and false when we needed it to be false, so that’s our answer. But if plugging in here seems cumbersome to you, you’re not alone. I actually prefer to do this question another way.

Let’s have a look at (B), our correct answer, and convert it like we did at the beginning of this post:

|w – 162| < 19

-19 < w – 162 < 19

Here’s where things get awesome. Add 162 to all 3 sections of the inequality now:

-19 + 162 < w – 162 + 162 < 19 +162
143 < w < 181

 Wow, that’s exactly what we were looking for. I mean, I knew that was coming, and I’m still amazed. I can’t even imagine how you must feel.

So on a question like this, you can just convert every answer choice in this way to see which one gives you what you want, or you can even take it one step further: in the correct answer, the part inside the absolute value brackets will be the middle of the range, and the bit on the other side of the less-than sign will be the distance from the middle to the ends of the range, or half the range.  

|w – 162| < 19
|variable – middle of range| < half of range

Not bad, right? This question, as I said above, is pretty rare, but if it appears on your test you know exactly what to do. GET SOME.

Absolute values and functions

Of course, if you have absolute value brackets around an expression, you don’t perform the absolute value operation until you’ve completed all the operations inside. For example:

GOOD: |-8 + 5| = |-3| = 3
BAD: |-8 + 5| = 8 + 5 = 13

Seriously, don’t do that second one. Don’t.

The same is true if you have absolute value brackets around a function, like |f(x)|. The absolute value brackets don’t take effect until the function has done its thing inside. If the function comes out positive on its own, the brackets have no effect. If the function comes out negative, it becomes positive:

See? Now here’s the important part: what happens to the graph of a function when you take the absolute value of the function? Well, when it’s positive, nothing. When it’s negative, it reflects off the x-axis. In other words, it bounces. BOING! Observe:

Cool, right?

ZOMG I can’t wait to practice!

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