Question

does anyone how to do piecewise

Answer 1

Yeah

Answer 2

can u please help us we need to write the following absolute value ecpressions as piece wise expressions.

Answer 3

yeah,post it

Answer 4

\[\left| x ^{2}+x-12 \right|\]

Answer 5

Start by factoring the part in the absolute values.

Answer 6

ok so x - 3 and x +4

Answer 7

straight off the bad you can see that there are not restrictions, so all values of x are allowed

Answer 8

This means that Domain is ALL REAL VALUES OF X, or (-inf, inf)

Answer 9

so:

|(x-3)(x+4)|

Now, we know that |a| = b if a >= 0
And |a| = -b if a < 0

So:

|(x-3)(x+4)| = (x-3)(x+4) when (x-3)(x+4) >= 0
AND
|(x-3)(x+4)| = -(x-3)(x+4) when (x-3)(x+4) < 0

So you just have to find where that product is positive, and where it's negative.

Answer 10

oh, is this a domain question?

Answer 11

no its not domain. its piecewise expressions.

Answer 12

my bad

Answer 13

set X^2 +X -12 = 0 so you can see where it crosses the x-axis
x^2 + x - 12 = 0

(x + 4)(x -3) = 0

x = {-4, 3}

and we can tell since x^2 is positive that the graph opens up

so below -4 and above 3 the graph will be positive (which is a necessity for Absolute value functions)

so in between -4 and 3 you will need the negative of x^2 + x - 12

-1(x^2 + x - 12) = -x^2 - x + 12

so finally

x^2 + x - 12 when x<-4 and x>3
-x^2 - x + 12 when -4<x<3

Answer 14

i think this is what you are looking for:)

Answer 15

satisfied??

Answer 16

yeah i am satisfied. now tell me the piecewise of \[|x ^{2}-1|\] and i know thats (x-1)(x+1)

Answer 17

good

Answer 18

what class is this for? precalc

Answer 19

no AP Calc

Answer 20

na, come on. Did you just start the class

Answer 21

its a summer assignment before our senior year

Answer 22

so this is work before you start calculus..so it mus be precalc ;)

Answer 23

so whats the piecewise of that

Answer 24

okay hold on buddy

Answer 25

noo its a summer assignment thank you very much. so

Answer 26

then why don't you do it then ?

Answer 27

i think you should be a fan of mine, that way when you do start calc, you can ask me questions.

Answer 28

i will tell you the answer to your question, if you become a fan

Answer 29

it will benefit me as well, because that way i wont forget my calculus:)

Answer 30

because i dont know how to do some of it why else would we be on here, and lagrange we dont know how to become a fan , do we just go to your page??? and so would the piece wise be x square - 1 when x is less than -1, x greater then 1, and then -x square plus one when -1 is less than x less than 1

Answer 31

here what your looking for:
For |x^ - 1| we need to know when x^2 - 1 >= 0 and when x^2 - 1 < 0. It is easy to see that x^2 - 1 >= 0 when x < -1 or when x > 1, and x^2 - 1 < 0 when -1 < x < 1. Therefore we have

|x^2 - 1| = x^2 - 1 when x < -1,
= -(x^2 - 1) = 1 - x^2 when -1 < x < 1,
= x^2 - 1 when x > 1.

Answer 32

olay , so what about this one\[\left| x ^{2}+4x+4 \right|\]

Answer 33

are you just writing down what Lagrange is typing or are you actually thinking about what he wrote?

Answer 34

no, way i just did one like that. Follow the same procedure as before. Tell you what, try and do it your self, and when you get stuck i will help. Okay, that way you learn how to do them. I would start by factoring:)

Answer 35

and if we want to ask just you questions do we go to your profile?

Answer 36

no, right here, just post your questions

Answer 37

i did factor its x+2 and x+2 , i am only confused after tht i don't understand where it came from

Answer 38

i did factor its x+2 and x+2 , i am only confused after tht i don't understand where it came from

Answer 39

by the way its very weird that you keep referring to yourself as "we", it makes it sound as if you are possessed:)

Answer 40

Borg

Answer 41

its b/c im with a freind and "we" need help on it! i and what the heck is borg? and ok i factored now waht????m not possessed!

Answer 42

?

Answer 43

?

Answer 44

nkjouig