Question

How are these functions inverses: f(n)=2(n-2)^3 and g(n)=4+(4n^1/3)/2 (or cubic root: 4n)

Answer 1

if f(g(n)) = g(f(n)) = n they is inverses

Answer 2

\[f(n)=2(n-2)^{3} \]
\[g(n)= (4+\sqrt[3]{4n})\div2 \]

Answer 3

but can you show me how?

Answer 4

I've tried matching f(n) to g(n) but I just can't D: eek

Answer 5

and I don't know how to for this type

Answer 6

plug g(n) into f(n) and calulate it out; if it reduces to "n" its an inverse in that direction

Answer 7

how about using this method:

y=2(n-2)^3
n=2(y-2)^3
.....

Answer 8

that method might be useful if they both equate in the end
\[f(n)=2(n-2)^{3}\]
\[\frac{f(n)}{2}=(n-2)^{3}\]
\[\sqrt[3]{\frac{f(n)}{2}}=n-2\]
\[\sqrt[3]{\frac{f(n)}{2}}+2=n\]

Answer 9

but don't they equate since they're both inverses of each other?

Answer 10

I keep getting:

\[y=(\sqrt[3]{n+8})\div2\]

Answer 11

f(n)=2(n-2)^3
g(n)=4+(4n^1/3)/2

f(g(n)) = 2((4+(4n^1/3)/2)-2)^3

f(g(n)) = 2((2+(4n^1/3)/2))^3

f(g(n)) = 2((2+(4n^1/3)/2))^3

1 3 3 1
8 4 2 1
1 4'1/3n 4'2/3n^2 4'3/3n^3
8 12


im going blind with this thing

Answer 12

who says they are inverses of each other?

Answer 13

on my worksheet it says these two are inverses of each other:

http://www.kutasoftware.com/FreeWorksheets/Alg2Worksheets/Function%20Inverses.pdf

Answer 14

#6

Answer 15

"State if the given functions are inverses."

that does not say that they are indeed inverses

Answer 16

oh the answers are at the bottom

Answer 17

.... :/ thnx lol

Answer 18

it might be easier to test g(f(n)) to start with

Answer 19

ohh, okay

Answer 20

so I have: \[f(g(n)) = 2((4+\sqrt[3]{4n}\div2)-2)\]

Answer 21

f(n)=2(n-2)^3
g(n)=(4+(cbrt(4n))/2


g(f(n)) = (4+(cbrt(4(2(n-2)^3)))/2

g(f(n)) = (4+(cbrt(8 (n-2)^3))/2

g(f(n)) = (4+2cbrt((n-2)^3)/2

g(f(n)) = (4+2(n-2))/2

g(f(n)) = 2+(n-2) = n

Answer 22

^3

Answer 23

where did you get the 2 from: g(f(n)) = (4+2cbrt((n-2)^3)/2

Answer 24

ohh, nevermind: 8^1/3

Answer 25

trying to determine the other route is a pain

Answer 26

I'm sorry D:
Question: What happened to the cubic root here: g(f(n)) = (4+2(n-2))/2

Answer 27

cbrt..^3 cancel each outher out

f(-1) = 2(-1-2)^3 = -54

g(-54)=(4+cbrt(4(-54)))/2 does not equal -1

Answer 28

ohhh

Answer 29

i believe you have a restricted domain inverse with these

Answer 30

f(5) = 54
g(54) = 5

f(-1) = -54
g(-54) not= -1

Answer 31

does the "n" signify natural numbers?

Answer 32

ohh... so I have to use restrictions?
And last question: Why is it: g(f(n)) = 2+(n-2) = n
In the end I got: 4+2(n-2)/2=6(n-2)/2...

Answer 33

I don't know because they don't state anything about what 'n' means... I just thought it's just a variable like x

Answer 34

x is traditioanlly used to define any real variable; and n is used to define any natural number

Answer 35

ohh.. so what happens? o.O

Answer 36

g(f(n)), for n>=2, these are inverses

Answer 37

so I use restrictions? How did you know it would be n>=2? Is it because of the (n-2)?

Answer 38

\[g(f(n)) = 2+\frac 12 \sqrt[3]{4(2(n-2)^3)}\]
\[g(f(2)) = 2+\frac 12 \sqrt[3]{4(2(2-2)^3)}=2+\frac 12 \sqrt[3]{0}=2\]

\[g(f(3)) = 2+\frac 12 \sqrt[3]{4(2(3-2)^3)}=2+\frac 12 \sqrt[3]{8}=2+1=3\]

\[g(f(1)) = 2+\frac 12 \sqrt[3]{4(2(1-2)^3)}\]\[g(f(1)) = 2+\frac 12 \sqrt[3]{4(2(-1)^3)}\]\[g(f(1)) = 2+\frac 12 \sqrt[3]{-8}\]\[g(f(1)) = 2+\frac 12 (-2)=1\]
harumpf!! works for 1

what about 0?
\[g(f(0)) = 2+\frac 12 \sqrt[3]{4(2(0-2)^3)}\]\[g(f(0)) = 2+\frac 12 \sqrt[3]{4(2(-8))}\]\[g(f(0)) = 2+\frac 12 \sqrt[3]{8(-8))}\]\[g(f(0)) = 2+\frac 12 (2)(-2)\]\[g(f(0)) = 2-2=0\]
-1 perhaps?
\[g(f(-1)) = 2+\frac 12 \sqrt[3]{4(2(-1-2)^3)}\]\[g(f(-1)) = 2+\frac 12 \sqrt[3]{8(-27)}\]\[g(f(-1)) = 2+\frac 12 (2)(-3)\]\[g(f(-1)) = 2-3=-1\]

well, it appears f(n) is at least the inverse of g(n)

Answer 39

do the same setup for f(g(n)) and see if you get "n" back

Answer 40

ohh by trial and error of substituting the 'n' with a number?

Answer 41

btw, thank you so much!

Answer 42

i was testing out some stuff, the n>=2 might be wrong; but try it out for f(g(n)) to see if it applies