Question

Function question

Answer 1

SUppose h(4z+3)=(5+2z)/(4-z)

h(z)=?
h^-1(4)=?

Answer 2

if u=4z+3
then z=(u-3)/4
see if you can use this to write h(4z+3)=(5+2z)/(4-z) in terms of u

Answer 3

im getting (-4+8u)/(28-4u)

Answer 4

maybe I did something wrong I got something a bit different then that
\[h(u)=\frac{5+2(\frac{u-3}{4})}{4-\frac{u-3}{4}} \\ \text{ multiply top and bot by } 4 \\ h(u)=\frac{20+2(u-3)}{16-(u-3)}\]

Answer 5

you could distribute on top and bottom
and then combine like terms

Answer 6

but anyways you can find h(z) just by now replacing u with z

Answer 7

the top part wouldnt be 8(u-3)?

Answer 8

how did you get that

Answer 9

oh wait nevermind unless it was 2+(u-3/4), it will remain a 2 i see what i did wrong

Answer 10

sooo final answer=

(14+2u)/(19-u)?

Answer 11

that is what i have for h(u)
so this means
\[h(z)=\frac{14+2z}{19-z}\]

Answer 12

\[h^{-1}(4)=z \\ h(z)=4 \text{ so solve } 4=\frac{14+2z}{19-z} \text{ for } z\]

Answer 13

quick question why is u replaced by z?

Answer 14

because we wanted to find h(z)

Answer 15

just like if we wanted to find h(5) we would replace u with 5
or if we wanted to find h(fish) we would replace u with fish

Answer 16

i guess i was a little confused but if we put (u-3)/4 back into the z's, we would get our previous work so it makes sense

Answer 17

as for the next question why is h^-1(4)=z

Answer 18

I assigned a value to h^-1(4)
I called that value z
you could have chosen a different letter if you like

Answer 19

\[h^{-1}(4)=x \text{ then } h(x)=4 \\ \text{ so we need to solve } \frac{2x+14}{19-x}=4 \text{ for } x\]

Answer 20

oh, im starting to see a substitution pattern here

Answer 21

im still a bit confused on where u got 2x+14/19-x, is that just the inverse?

Answer 22

\[\text{ remember we got } h(z)=\frac{2z+14}{19-z}\]

Answer 23

\[\text{ if } (a,b) \text{ is on the graph of } h \text{ then } (b,a) \text{ is on the graph of } h^{-1} \\ \text{ so if } h(a)=b \\ \text{ then } h^{-1}(b)=a \\ \text{ or you can read this the other way also } \\ if (b,a) \text{ is on the graph of } h^{-1} \text{ then } (a,b) \text{ is on the graph of } h \\ \text{ so if } h^{-1}(b)=a \text{ then } h(a)=b \]

Answer 24

so we have
\[h^{-1}(4)=a \text{ which means } h(a)=4\]

Answer 25

\[\text{ so solving the following for } a \\ \frac{2a+14}{19-a}=4 \text{ will give us } h^{-1}(4) \text{ since } a=h^{-1}(4)\]

Answer 26

starting to make sense now. but if they gave me the question h^-1(4) without asking h(z), is there another way to solve or do i have to get h(z) first?

Answer 27

\[h(4z+3)=\frac{5+2z}{4-z} \\ \text{ we want to find } h^{-1}(4) \\ \text{ Let's call } h^{-1}(4)=a \\ \text{ so } h(a)=4 \\ \text{ yep we still basically have \to find that same left expression since this gives us } \\ h(a)=\frac{5+2 \cdot \frac{a-3}{4}}{4-\frac{a-3}{4}}=4\]

Answer 28

what I'm saying is I don't see a way around it

Answer 29

but that doesn't mean there isn't a way around it

Answer 30

it doesn't take too long with your method, ill stick with it. anyhow, the answer is 10 (1/3)?

Answer 31

\[4=\frac{14+2x}{19-x} \\ 4(19-x)=14+2z \\ 76-4x=14+2x \\ 76-14=4x+2x \\ 62=6x \\ x=\frac{62}{6}=\frac{31}{3}\]
yep seems great

Answer 32

oh yeah this also works
\[h(4z+3)=\frac{5+2z}{4-z} \implies 4z+3=h^{-1}(\frac{5+2z}{4-z}) \\ \text{ we want } \frac{5+2z}{4-z}=4 \text{ so this gives } 5+2z=16-4z \\ 4z+2z=16-5 \\ 6z=11 \\ z=\frac{11}{6} \\ \text{ so we have } h^{-1}(4)=4(\frac{11}{6})+3 =\frac{44}{6}+3=\frac{44}{6}+\frac{18}{6}=\frac{62}{6}=\frac{31}{3}\]

Answer 33

you know if you wanted to skip finding h(z)

Answer 34

i'll look into that later. thanks for your help.

Answer 35

np

Answer 36

by the way...

Answer 37

I know it was confusing calling z=(u-3)/4
then just replacing u with z
but you just have to remember h(u)=(2u+14)/(19-u) is just a function
so if I wanted to find h(z) you can do this just by replacing the old input u with the new input z