Function question

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Function question

Mathematics
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SUppose h(4z+3)=(5+2z)/(4-z) h(z)=? h^-1(4)=?
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
im getting (-4+8u)/(28-4u)

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maybe I did something wrong I got something a bit different then that \[h(u)=\frac{5+2(\frac{u-3}{4})}{4-\frac{u-3}{4}} \\ \text{ multiply top and bot by } 4 \\ h(u)=\frac{20+2(u-3)}{16-(u-3)}\]
you could distribute on top and bottom and then combine like terms
but anyways you can find h(z) just by now replacing u with z
the top part wouldnt be 8(u-3)?
how did you get that
oh wait nevermind unless it was 2+(u-3/4), it will remain a 2 i see what i did wrong
sooo final answer= (14+2u)/(19-u)?
that is what i have for h(u) so this means \[h(z)=\frac{14+2z}{19-z}\]
\[h^{-1}(4)=z \\ h(z)=4 \text{ so solve } 4=\frac{14+2z}{19-z} \text{ for } z\]
quick question why is u replaced by z?
because we wanted to find h(z)
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
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
as for the next question why is h^-1(4)=z
I assigned a value to h^-1(4) I called that value z you could have chosen a different letter if you like
\[h^{-1}(4)=x \text{ then } h(x)=4 \\ \text{ so we need to solve } \frac{2x+14}{19-x}=4 \text{ for } x\]
oh, im starting to see a substitution pattern here
im still a bit confused on where u got 2x+14/19-x, is that just the inverse?
\[\text{ remember we got } h(z)=\frac{2z+14}{19-z}\]
\[\text{ if } (a,b) \text{ is on the graph of } h \text{ then } (b,a) \text{ is on the graph of } h^{-1} \\ \text{ so if } h(a)=b \\ \text{ then } h^{-1}(b)=a \\ \text{ or you can read this the other way also } \\ if (b,a) \text{ is on the graph of } h^{-1} \text{ then } (a,b) \text{ is on the graph of } h \\ \text{ so if } h^{-1}(b)=a \text{ then } h(a)=b \]
so we have \[h^{-1}(4)=a \text{ which means } h(a)=4\]
\[\text{ so solving the following for } a \\ \frac{2a+14}{19-a}=4 \text{ will give us } h^{-1}(4) \text{ since } a=h^{-1}(4)\]
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?
\[h(4z+3)=\frac{5+2z}{4-z} \\ \text{ we want to find } h^{-1}(4) \\ \text{ Let's call } h^{-1}(4)=a \\ \text{ so } h(a)=4 \\ \text{ yep we still basically have \to find that same left expression since this gives us } \\ h(a)=\frac{5+2 \cdot \frac{a-3}{4}}{4-\frac{a-3}{4}}=4\]
what I'm saying is I don't see a way around it
but that doesn't mean there isn't a way around it
it doesn't take too long with your method, ill stick with it. anyhow, the answer is 10 (1/3)?
\[4=\frac{14+2x}{19-x} \\ 4(19-x)=14+2z \\ 76-4x=14+2x \\ 76-14=4x+2x \\ 62=6x \\ x=\frac{62}{6}=\frac{31}{3}\] yep seems great
oh yeah this also works \[h(4z+3)=\frac{5+2z}{4-z} \implies 4z+3=h^{-1}(\frac{5+2z}{4-z}) \\ \text{ we want } \frac{5+2z}{4-z}=4 \text{ so this gives } 5+2z=16-4z \\ 4z+2z=16-5 \\ 6z=11 \\ z=\frac{11}{6} \\ \text{ so we have } h^{-1}(4)=4(\frac{11}{6})+3 =\frac{44}{6}+3=\frac{44}{6}+\frac{18}{6}=\frac{62}{6}=\frac{31}{3}\]
you know if you wanted to skip finding h(z)
i'll look into that later. thanks for your help.
np
by the way...
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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