Question

consider the piecewise h(x) = -2/3(x+1) x<-1
-4x-4 -1≤ x≤ 0
x-8 1 11 years ago

Answer 1

@campbell_st

Answer 2

@tkhunny

Answer 3

You ARE going to have to graph those three lines. Just cross out the parts you don't need.

Answer 4

http://awesomescreenshot.com/0303lr6g9d

Answer 5

What do you think? Did we cover anything below y = -6?

Answer 6

i think [-7, negative infinity)

Answer 7

@tkhunny

Answer 8

@amistre64

Answer 9

are the pieces continuous at the joints?

Answer 10

no

Answer 11

http://awesomescreenshot.com/0303lr6g9d

Answer 12

well it cant go to negative infinity since there is nothing below -7

Answer 13

id say -7 and up works fine

Answer 14

so its [-7, positive infinity)

Answer 15

-2/3(x+1) x<-1
-4x-4 -1≤ x≤ 0
x-8 1<x<4


-2/3(x+1) -2(x+1)/3 > 0
-4x-4 0>= -4x-4 >= -4
x-8 -7<x-8<-4

Answer 16

-7 to -4 to 0 to inf

Answer 17

ah okay so [-7,-4] and [0, positive infinity)

Answer 18

-4 to 0 is still in the range

Answer 19

wait so now I'm confused.. how to write it in interval notation? i thought it was [-7, positive infinity)

Answer 20

it is

Answer 21

oh okay

Answer 22

wait, didnt see the lack of equailty on the -7

(-7,inf)

Answer 23

but it includes 7 sp its [-7, pos inf)

Answer 24

how does it include -7?

Answer 25

includes -7

Answer 26

on the graph of h(x) = x-8

Answer 27

http://awesomescreenshot.com/0303lr6g9d

Answer 28

x-8 1<x<4 , x is never equal to 1

Answer 29

oh my bad. thanks!

Answer 30

good luck