Question

HELP ME!!!THANK YOU! GOD BLESS YOU!

Find the inverse of the function
f(x)=1/3 (x-4)



f^-1(x)=_____

Answer 1

Help!

Answer 2

@Destinymasha @poopsiedoodle @nickmifsud1223 @marylou004 @camerondoherty @caitlinnr14 @CloverRacer @dan815 @DrPepperx3 @FibonacciChick666

Answer 3

@kropot72 @Whitemonsterbunny17 @adrynicoleb @bibby

Answer 4

@arabpride @timaashorty @triciaal @kirbykirby @inkyvoyd @perl

Answer 5

Is it \[\large f(x) = \frac1{3(x-4)}\]
or\[\large f(x) =\frac13(x-4)\]
?

Answer 6

The second one

Answer 7

Okay. Here's how to do it ^^
First, replace f(x) with y.
\[\large y = \frac13(x-4)\]

NEXT, switch x and y.
\[\large x = \frac13(y-4)\]
Now just solve for y :)

Answer 8

How do I solve for y?

Answer 9

Well, if I might have an example...
Say you have \[\large x = \frac1{3y + 8}\]
First, multiply 3y+8 to both sides...
\[\Large x(3y+8) = \frac1{3y+8}\cdot (3y+8)\]
\[\Large x(3y+8) = \frac1{\cancel{3y+8}}\cdot \cancel{(3y+8)}\]
\[\Large x (3y+8) = 1\]

Now, divide both sides by x.
\[\Large 3y+8 = \frac1x\]
Next, subtract 8 from both sides to get
\[\Large 3y = \frac1x-8\]
\[\Large 3y = \frac{1-8x}{x}\]
Finally, divide both sides by 3.
\[\Large y = \frac{1-8x}{3x}\]

Mostly algebraic manipulation :D