@Jhannybean

@AaronAndyson

@ganeshie8

4. IrishBoy123

$$y = 2(n-2)^3$$ $$(\dfrac{y}{2})^\frac{1}{3} = n-2$$ etc

how did you do that

6. IrishBoy123

$$f(n) = 2(n-2)^3$$ $$y = 2(n-2)^3$$ yes?

7. IrishBoy123

$$\frac{y}{2} = (n-2)^3$$ ok?

why would you do y/2, isn't it |dw:1443341707529:dw|

9. IrishBoy123

$\dfrac{y}{2} = \dfrac{2(n-2)^3}{2}$

10. IrishBoy123

you want the inverse, is that right, ie $$n = n(y)$$

11. IrishBoy123

$\dfrac{y}{2} = \dfrac{2(n-2)^3}{2} \implies \dfrac{y}{2} = (n-2)^3$

f(n) = 2(n-2)^3

we need to inverse this function

14. IrishBoy123

yes, indeed label the $$f(n)$$ as y just because it makes it easier to write out. then we need to find and equation in form $$n = n(y) = f^{-1}(n)$$ so $$y = 2(n-2)^3$$ $$\frac{y}{2} = (n-2)^3$$ like so.

but isn't it like n= 2(y-2)^3

i like this way

@Miracrown

@insa

19. IrishBoy123

ok let's go that way $$n= 2(y-2)^3$$ $$\frac{n}{2} = (y-2)^3$$

alright, next

21. IrishBoy123

$(\frac{n}{2})^{\frac{1}{3}} = y-2$

22. IrishBoy123

yes?

wait wait, where did 1/3 come from

24. IrishBoy123

take cube root of each side

why cant it be |dw:1443342930062:dw|

26. IrishBoy123

$$a^3 = b$$ $$a = b^{\frac{1}{3}} = \sqrt[3]{b}$$ same thing

$\frac{ \sqrt[3]{n} }{ 2}$

lets go this way if you dont mind

29. IrishBoy123

if that way, than $\large (\frac{n}{2})^{\frac{1}{3}} = \sqrt[3]{\frac{n}{2}}= y-2$

yup, then $\frac{ \sqrt[3]{n} }{ 2 } + 2$

31. IrishBoy123

be careful the cube root must include the 1/2 its $y=\sqrt[3]{\frac{ n} { 2 }} + 2$ NOT $y=\frac{ \sqrt[3]{n} }{ 2 } + 2$

yeah

so is that it

34. IrishBoy123

good!

36. IrishBoy123

shoot but if it's a new question, new thread is best

find the inverse of g(x) = 1/x - 2

i know that x = 1/y - 2

@IrishBoy123

40. IrishBoy123

you like to do it this way $x = \frac{1}{y} - 2$ $\frac{1}{y} = x + 2$ $y = \frac{1}{x+2}$

and the inverse of -4x + 1 is $\frac{ x-1 }{ 4 }$ right

42. IrishBoy123

|dw:1443344105564:dw|