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

http://awesomescreenshot.com/0303lr6g9d

Answer 2

@jim_thompson5910

Answer 3

@ganeshie8

Answer 4

So they want you to find

\[\Large h^{-1}(3)\]

right?

Answer 5

yep

Answer 6

if you can show me how to do one i can do the rest

Answer 7

that's effectively asking: what is the value of x when h(x) = 3

Answer 8

put another way, "what is the x value for the point (x,3)"

this point lies on the piecewise function

Answer 9

so you'd go to the graph and find where the point with a y coordinate of 3 is

it is on the left most piece, the red piece

so you'd solve -2/3(x+1) = 3 for x

Answer 10

so are the answers 7/2, -1, and 11?

Answer 11

@jim_thompson5910

Answer 12

also, is h(x) invertible?

Answer 13

how are you getting 7/2 ?

Answer 14

i solved for x. sorry i meant -3, 0, and 3 respectively
so i got 7/2, -1, and 11

Answer 15

-2/3(x+1) = 3

-2(x+1) = 3*3

-2(x+1) = 9

-2x-2 = 9

-2x = 9+2

-2x = 11

x = -11/2

Answer 16

So that effectively says: h(-11/2) = 3

Answer 17

you are correct in stating that h(-1) = 0

Answer 18

h(11) is not defined because the function only goes up to x = 4, but it doesn't include 4 itself. This is the furthest you can go to the right.

Answer 19

the question says evaluate h^-1(-3), h^-1(0), h^-1(3) so it has to be h^-1(3)=7/2, not h^-1(7/2)=3

Answer 20

h^-1(-3)=7/2, not h^-1(7/2)=-3

Answer 21

oh we're talking about two different things

I'm saying that because h(-11/2) = 3, this means h^-1(3) = -11/2

Answer 22

but h^-1(3) = 7/2, not -11/2

Answer 23

7/2 = 3.5 isn't in the domain of h(x) though

Answer 24

I'm really confused so when i evaluated h, i got h(-3)=4/3, h(0)=-4, and h(3)=-5

Answer 25

so i have to find the inverse of h(-3), h(0), and h(3)

Answer 26

when you find the inverse of a function, like y = x+2, you are simply solving for y to get

x = y - 2

Answer 27

but when i find the inverse of h(-3), i should get 4/3, but i get 7/2

Answer 28

when they ask: "find the inverse of f^-1(2)" they mean plug in y = 2 and solve for x

x = y-2

x = 2-2

x = 0

so that tells us f(0) = 2

Answer 29

so the book is saying the answer is 4/3?

Answer 30

there is no answer key

Answer 31

but my graphic and algebraic answer is different

Answer 32

ok let's do h^-1(3)

Answer 33

step 1) locate 3 on the y axis

step 2) draw a horizontal line through 3 on the y axis

step 3) find out where this horizontal line intersects with the graph of h(x). This happens at x = -11/2 and you can determine that by solving -2/3(x+1) = 3


Those steps tell us that h(-11/2) = 3

Answer 34

also, -11/2 = -6.5 so you should be exactly between -5 and -6 on the x axis (where the horizontal line and h(x) intersect)

Answer 35

but h^-1(3) is supposed to be h(x)=x-8

Answer 36

so its h^-1(x)=3+8=11

Answer 37

but that green piece is nowhere near y = 3

Answer 38

but according to h(x)=x-8, 1<x<4 and 1<3<4

Answer 39

x isn't 3, y is

Answer 40

if you evaluated h(3), then yes, you'd say

x = 3 is between 1 and 4 (1 < x < 4 ---> 1 < 3 < 4)

so h(3) = 3-8 = -5

Answer 41

but that's not what they're asking

Answer 42

wait so hold on:
for h(x)=-2/3(x+1), x=-2/3y-2/3, y=-3/2x-1

Answer 43

yes, then you would plug in x = 3

Answer 44

okay so h^-1(3)=11/2

Answer 45

-11/2

Answer 46

correct

Answer 47

okay then i get h(x)=-4x-4 so h^-1(x)=-1/4x-1

Answer 48

so do i plug in 0

Answer 49

and then h^-1(0)=-1

Answer 50

correct

Answer 51

yes

Answer 52

so, h^-1(x)=x+8, so h^-1(-3)=5

Answer 53

notice how 5 isn't in the domain though

Answer 54

yea

Answer 55

so h^-1(-3) is undefined

Answer 56

it would be h^-1(-3) = 5 if you extended the domain out to x = 5

Answer 57

so is h(x) invertible

Answer 58

yes because it is one-to-one (it passes the horizontal line test)