Question

@3mar

Answer 1

@3mar

Answer 2

Yes, I am here

Answer 3

ok do you see the attachment

Answer 4

yes

Answer 5

first of all: extract the root in one side

Answer 6

idk how to do that

Answer 7

ok

Answer 8

can you help?

Answer 9

let f(x)=y
\[y=-5\sqrt[3]{-4x+4}+2\]
We will separate xi= in one side; we get
\[y-2=-5\sqrt[3]{-4x+4}\]

raise both sides to the power 3
\[y-2=-5\sqrt[3]{-4x+4}\]
\[(y-2)^3=(-5\sqrt[3]{-4x+4})^3\]
\[(y-2)^3=-125(-4x+4)\]
\[(y-2)^3=-500(1-x)\]

until here is good?

Answer 10

yes

Answer 11

then
\[\frac{ (y-2)^3 }{ -500 }=1-x\]
\[1-x=\frac{ (y-2)^3 }{ -500 }\]
\[x=\frac{ (y-2)^3 }{ 500 }+1\]
Now you can swap x with y

Answer 12

are you with me?

Answer 13

It will be
\[y=\frac{ (x-2)^3 }{ 500 }+1\]
this is the inverse function

Answer 14

that doesnt match any if the answers

Answer 15

@3mar

Answer 16

some touches
\[y=\frac{ (x-2)^3+500 }{ 500 }\]
\[y=\frac{ x^3-6x^2+12x-8+500 }{ 500 }\]

Answer 17

b

Answer 18

I know
I am just simplifying it
\[f^{-1}(x)=\frac{ x^3-6x^2+12x+492 }{ 500 }\]
This is the second choice

Hope that helps

Answer 19

Are you there?

Answer 20

ok sorry thank you

Answer 21

how would i answer this
Explain in your own words how to find an inverse of a function. How can you use a graph to check to see if your two functions are in fact inverses?

Answer 22

Thank you for the medal!
Any time.

Answer 23

I will go now and when I am back I will answer you, excuse me.
@Kevin may know. ask him

Answer 24

The method is like what 3 Mar said on the top..
You can constitute the power 3 like this:
3y^3 + 3*(y)^2*(-2)^1 + 3*(y)^1*(-2)^2) + (-2)^3 = 500x - 500
3y^3−6y^2+12^y−8 = 500x - 500
3y^3−6y^2+12^y−8 + 500 = 500x
x = (3y^3−6y^2+12y+492)/500

Answer 25

Do invers x = y
y = (3x^3−6x^2+12x+492)/500

Answer 26

Thanks, Kevin.
I appreciate!
You are the man that I can depend on ;)
Thanks a lot.