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dtan5457
 one year ago
Function question
dtan5457
 one year ago
Function question

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dtan5457
 one year ago
Best ResponseYou've already chosen the best response.0SUppose h(4z+3)=(5+2z)/(4z) h(z)=? h^1(4)=?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2if u=4z+3 then z=(u3)/4 see if you can use this to write h(4z+3)=(5+2z)/(4z) in terms of u

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.0im getting (4+8u)/(284u)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2maybe I did something wrong I got something a bit different then that \[h(u)=\frac{5+2(\frac{u3}{4})}{4\frac{u3}{4}} \\ \text{ multiply top and bot by } 4 \\ h(u)=\frac{20+2(u3)}{16(u3)}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you could distribute on top and bottom and then combine like terms

freckles
 one year ago
Best ResponseYou've already chosen the best response.2but anyways you can find h(z) just by now replacing u with z

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.0the top part wouldnt be 8(u3)?

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.0oh wait nevermind unless it was 2+(u3/4), it will remain a 2 i see what i did wrong

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.0sooo final answer= (14+2u)/(19u)?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2that is what i have for h(u) so this means \[h(z)=\frac{14+2z}{19z}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[h^{1}(4)=z \\ h(z)=4 \text{ so solve } 4=\frac{14+2z}{19z} \text{ for } z\]

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.0quick question why is u replaced by z?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2because we wanted to find h(z)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2just 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
 one year ago
Best ResponseYou've already chosen the best response.0i guess i was a little confused but if we put (u3)/4 back into the z's, we would get our previous work so it makes sense

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.0as for the next question why is h^1(4)=z

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I assigned a value to h^1(4) I called that value z you could have chosen a different letter if you like

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[h^{1}(4)=x \text{ then } h(x)=4 \\ \text{ so we need to solve } \frac{2x+14}{19x}=4 \text{ for } x\]

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.0oh, im starting to see a substitution pattern here

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.0im still a bit confused on where u got 2x+14/19x, is that just the inverse?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\text{ remember we got } h(z)=\frac{2z+14}{19z}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\text{ if } (a,b) \text{ is on the graph of } h \text{ then } (b,a) \text{ is on the graph of } h^{1} \\ \text{ so if } h(a)=b \\ \text{ then } h^{1}(b)=a \\ \text{ or you can read this the other way also } \\ if (b,a) \text{ is on the graph of } h^{1} \text{ then } (a,b) \text{ is on the graph of } h \\ \text{ so if } h^{1}(b)=a \text{ then } h(a)=b \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so we have \[h^{1}(4)=a \text{ which means } h(a)=4\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\text{ so solving the following for } a \\ \frac{2a+14}{19a}=4 \text{ will give us } h^{1}(4) \text{ since } a=h^{1}(4)\]

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.0starting 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
 one year ago
Best ResponseYou've already chosen the best response.2\[h(4z+3)=\frac{5+2z}{4z} \\ \text{ we want to find } h^{1}(4) \\ \text{ Let's call } h^{1}(4)=a \\ \text{ so } h(a)=4 \\ \text{ yep we still basically have \to find that same left expression since this gives us } \\ h(a)=\frac{5+2 \cdot \frac{a3}{4}}{4\frac{a3}{4}}=4\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2what I'm saying is I don't see a way around it

freckles
 one year ago
Best ResponseYou've already chosen the best response.2but that doesn't mean there isn't a way around it

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.0it doesn't take too long with your method, ill stick with it. anyhow, the answer is 10 (1/3)?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[4=\frac{14+2x}{19x} \\ 4(19x)=14+2z \\ 764x=14+2x \\ 7614=4x+2x \\ 62=6x \\ x=\frac{62}{6}=\frac{31}{3}\] yep seems great

freckles
 one year ago
Best ResponseYou've already chosen the best response.2oh yeah this also works \[h(4z+3)=\frac{5+2z}{4z} \implies 4z+3=h^{1}(\frac{5+2z}{4z}) \\ \text{ we want } \frac{5+2z}{4z}=4 \text{ so this gives } 5+2z=164z \\ 4z+2z=165 \\ 6z=11 \\ z=\frac{11}{6} \\ \text{ so we have } h^{1}(4)=4(\frac{11}{6})+3 =\frac{44}{6}+3=\frac{44}{6}+\frac{18}{6}=\frac{62}{6}=\frac{31}{3}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you know if you wanted to skip finding h(z)

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.0i'll look into that later. thanks for your help.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I know it was confusing calling z=(u3)/4 then just replacing u with z but you just have to remember h(u)=(2u+14)/(19u) 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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