dtan5457 one year ago Function question

1. dtan5457

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

2. 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

3. dtan5457

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

4. freckles

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

5. freckles

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

6. freckles

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

7. dtan5457

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

8. freckles

how did you get that

9. dtan5457

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

10. dtan5457

11. freckles

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

12. freckles

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

13. dtan5457

quick question why is u replaced by z?

14. freckles

because we wanted to find h(z)

15. 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

16. 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

17. dtan5457

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

18. freckles

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

19. freckles

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

20. dtan5457

oh, im starting to see a substitution pattern here

21. dtan5457

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

22. freckles

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

23. freckles

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

24. freckles

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

25. freckles

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

26. 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?

27. freckles

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

28. freckles

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

29. freckles

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

30. dtan5457

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

31. 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

32. freckles

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

33. freckles

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

34. dtan5457

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

35. freckles

np

36. freckles

by the way...

37. 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