Function question

- dtan5457

Function question

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- dtan5457

SUppose h(4z+3)=(5+2z)/(4-z)
h(z)=?
h^-1(4)=?

- freckles

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

- dtan5457

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

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## More answers

- freckles

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

- freckles

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

- freckles

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

- dtan5457

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

- freckles

how did you get that

- dtan5457

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

- dtan5457

sooo final answer=
(14+2u)/(19-u)?

- freckles

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

- freckles

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

- dtan5457

quick question why is u replaced by z?

- freckles

because we wanted to find h(z)

- freckles

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

- dtan5457

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

- dtan5457

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

- freckles

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

- freckles

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

- dtan5457

oh, im starting to see a substitution pattern here

- dtan5457

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

- freckles

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

- freckles

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

- freckles

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

- freckles

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

- dtan5457

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?

- freckles

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

- freckles

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

- freckles

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

- dtan5457

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

- freckles

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

- freckles

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

- freckles

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

- dtan5457

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

- freckles

np

- freckles

by the way...

- freckles

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

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