PreCalculus | Two absolute value inequalities.

We solve a few "trickier" inequalities involving absolute values.

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[ThAlEdison]

2:56 Simple oversight, -15/3 is -5 not -3.

[kkanden]

feels good to finally be able to understand every step of a problem on this channel

[MathMadeEasy]

Hey Michael, I’ve always loved your unique style of teaching math content and your wide variety of concept. I’ve tried to incorporate some of your methods into my videos. I know you probably won’t read this but do you have any tips on how to grow a small math channel like mine. I’ve asked the same question to other YouTubers but they haven’t responded so I hope you do. Thanks Michael!

[denny]

For the second problem, there's a nice bit of intuition I like to use, starting from the step 2<|x-1|<8.
The |x-1| can be interpreted as 'the distance between x and 1', which is between 2 and 8.
1±2 = -1 and 3, 1±8 = -7 and 9. The distance is greater than 2, so the inner region [-1, 3] is excluded, and the distance is less than 8, giving the same answer.

I find this intuition useful when working with complex numbers especially.

[skwbusaidi]

2<=|3x+5|<=10
Case 1
2<=3x+5<=10
-3<=3x<=5
-1<=x<=5/3

Case 2
2<=-(3x+5)<=10
-2>=3x+5>=-10
-7>=3x>=-15
-7/3>=x>=-5
The final solution is the union of the two inteval

Same can be done for second question

[goodplacetostop]

3:58 5/3 seems to be a good place to stop
9:03

[phamnguyenductin]

A trick for these types of equations/inequalities that I learned in high school was to rewrite the absolute value as |A| = √(A²) to turn them into quadratics. In this case, as both sides of the equation/inequality are non-negative, we can square both sides.

[spartacus]

Thanks for the videos, you deserve public recognition for your contribution to science ... successes



[MyOneFiftiethOfADollar]

I used abs(?)=sqrt(?^2) twice to produce a 4rth degree polynomial with zeros {-7, -1, 3, 9}.

Sign analysis based on these 4 zeros also yields (-7, -1) U (3, 9) as presented in video.

This alternate definition of absolute value is slightly less cumbersome than plowing through the piece wise definition of absolute value.

[stewartcopeland]

07:38
the magician is discovered: he uses a tool to erase the board

[alexfujita]

Thank you, needed to review that stuff

[armacham]

from 4:58 to 5:01 the problem changes . . . what a clever way to punish people who try to do the problems before they watch you solve them

[mohdkhalid]

@Michael Penn,
Respected sir, please solve the following question:
Find the value of arcsine(sin 23) and arcsine(sin 24) the interval -pi/2 to pi/2, where both 23 and 24 is in radian.Thank you!

[jonathan]

Well, Michael shows the 'standard' way to the problem, but in terms of problem solving techniques, I would say drawing out the graph by beginning with the innermost function and working the way out, and look at the intersection with the constant function would be the fastest.

[skwbusaidi]

Second question
| |x-1| -5|<3
-3<|x-1| -5<3
2<|x-1|<8
Case 1
2<x-1<8
3<x<9
Case 2
2<-(x-1)<8
-2>x-1>-8
-1>x>-7
The final solution (-7, -1) + (3, 9)

[CM_France]

Hi,

The camera is more and more down, we can't see the top of the board when we stop for reading, it is hiden by text that appears (title of the video or some thing). You can rise up the camera, you have room at the bottom.

For fun:
0 "and so on and so forth",
3 "let's go ahead and"
2 "let's may be go ahead and", including 1 "let's may be go ahead and do that",
5 "great", including 2 "ok, great" <-- may be 7 in all, if I believe Mr Jones
and the "and that's a good place to stop" at the end.

[physicsnovice]

Comment I posted somewhere else and decided to post here also:

Here's bunch of examples I've encountered in different places (my highschool, forums, textbooks, other videos) that might be helpful.

I'll skip the definition and list some properties. The proofs can done by cases or some tricks using previously proved properties, they are available online.

a) |ab|=|a||b|
b) |a|=b <=> b>=0 and (a=-b or a=b)
c) |x| < a <=> -a < x < a (when a is negative also works)
d) |x| > a <=> x>a or x <-a (when a is negative also works)
e) |a|=|b| <=> a = -b or a = b


Examples
1. |x-1|=2x+2
We can try to solve x-1=2x+2 or x-1 = -2x-2 and check for solutions
x=-3 or x =-1/3
but x=-3 is not solution because we arrive to situation when we have |4|=-4
So either we have to recheck solutions or use property |a|=b <=> b>=0 and (a=-b or a=b) (check that 2x+2 is above 0)
or by cases
x>=1 and x-1=2x+2 => x = -3 but contradiction with previous assumption
x<1 and -x+1=2x+2 => x = -1/3 fine

note squaring (x-1)^2=(2x+2)^2 only works when we assume 2x+2 >= 0


2. |x-1|<2x+2
|a|<b <=> -b < a < b (note no need to check if b >= 0)
-2x-2 < x - 1 < 2x + 2
-2x-2 < x - 1 and x - 1 < 2x + 2
x > -1/3 and x > -3
x > -1/3

3. |x-1|<|2x+2|
Methods
a) graphing
b) cases
c) squaring
d) and / or and properties c) d)
|x-1|<|2x+2| <=> -|2x+2| < x-1 < |2x+2|
x-1 < |2x+2|
x-1 < 2x+2 or -x+1 > 2x+2
x > -3 or x < -1/3 (all x)
and
-|2x+2| < x-1 <=> |2x+2| > -x+1
2x+2 > -x+1 or 2x+2 < x-1
x > -1/3 or x < -3
so x > -1/3 or x < -3


4. |x| + |x-1| = 2 (when sum of distances from 0 and 1 is equal to 2)
cases
x >=1 2x-1=2, <=> x = 3/2 ok
0<=x<1 x - x + 1 = 2 <=> 1=2 contradiction bad
x<0 -x-x+1 = 2 <=> x = -0.5 ok

We can also do bunch of squaring
x^2+(x-1)^2+2|x||x-1| = 4
2|x||x-1| = 3 + 2x - 2x^2
4x^2(x-1)^2=(3 + 2x - 2x^2)^2

12x^2-12x-9=0
(x-3/2)(x+1/2)=0

6. If |x|>|y| and x+y>0 show x>y
|x|>|y| <=> x^2>y^2 <=> (x-y)(x+y)>0 => because x+y>0, then x-y>0 <=> x>y

7. Graph |x|+|y|=1
By cases or squaring |y|=1-|x| or x+y=1 and noticing it's symmetrical to the lines x=0 and y=0
x>=0 y>=0
x+y=1
x<0 y>=0
...

8.
||x|-|y||<2
|x-y|+|x+y|<6
cases/symmetry/squaring/c)d) properties and or/ands (graph looks like x)
|x-y|+|x+y|=a notice it's rotated |x|+|y|=b (rotation transformation)

9.
System either by graphing or algebraically
|x|+|y|=1
|y|=|x-1|
Substitute |y|
|x|+|x-1|=1
all points between [0, 1]
so solution is
|y|=|x-1| when x [0, 1]
y = x-1
y = -x+1 for x [0, 1]

10. For what m equation |mx-m|-|m|=-3 has solution
a) graph eqaution on m, x plane and read values of m
b) algebraically using a) property
|m||x-1|=|m|-3
|x-1|=(|m|-3)/|m|
(|m|-3)/|m| >= 0
so (-oo, -3) or (3, oo)
c) squaring and having popular quadratic equations with parameters (when delta > 0 and stuff)
d) notice |m|-3 has to be positive, and factor near x can't be zero or if it's zero we have to arrive to 0=0

11. If there's right triangle and one side is 2sqrt(a), hypotenuse a+1, and a>0 what's the other side? Pythagorean theorem
4a + x^2 = a^2+2a+1
x = sqrt((a-1)^2) = |a-1|

12. If there are two lines m:2x+y+1=0 and n:x-2y+3=0, what are the bisector lines(divide angles of lines crossover by two)?
Points (x, y) on bisector lines are such that their distance is equal from lines m and n so (distance formula |Ax+By+C|/sqrt(A^2+B^2))

|2x+y+1|=|x-2y+3|
Now either solve by cases, or recall |a|=|b| <=> a=b or a=-b, or square |a|=|b| <=> a^2=b^2 <=> a^2-b^2=0 <=> (a-b)(a+b)=0 <=>

13. When the line going through point (2, 0) is perpendicular to circle x^2+y^2=2
Line a(x-2)=y <=> ax-y-2a=0
Now either x^2+y^2=2 and ax-y-2a=0 equations have to have one solution or distance from line to point (0, 0) is equal to circle radius.
|a * 0 - 1 * 0 -2a|/sqrt(a^2+1)=sqrt(2)
|-2a|=sqrt(2)*sqrt(a^2+1)
4a^2=2a^2+2
a^2=1
a=+-1

14. if xy = n and x+y=m what's the value of |x|+|y|

[stephenbeck]

Do you have a list of more interesting abs value problems or a textbook with these? My precalculus book has some pretty boring ones.

[ammaryildirim]

why did we take 'or' for the first part of the first inequality, but 'and' for the second part of the first inequality?

[savagezerox]

Yo Hold UP. Why did you change the 2nd question?