Question

for the function f(x)= (3-4x)^2, find f^-1. determine whether f^-1 is a function. Can you show me how to do this ? Steps?

Answer 1

let f(x) be y
then simply solve for "x" this will give you the inverse function

Answer 2

f^-1(x) = [ (sq root x) -3 ] / (-4) << answer?

Answer 3

yes except the square root could be plus or minus

which also means that f^-1 is NOT a function because for any positive x you have 2 "y" values

Answer 4

So my answer would techinally be..? @dumbcow

Answer 5

@jdoe0001 can you show me how to do this btw I cant see anything you draw because something wrong on my end of open study

Answer 6

@agent0smith how is my answer incorrect or can you help me get a better understanding of what dumbcow said

Answer 7

y = (3-4x)^2

Solve that for x. Ugh... it's so hard to show w/o an equation editor.

Remember when you take square root of both sides, you have to add a plus or minus sign on the left.

Answer 8

it does not have an inverse

Answer 9

^Because of that plus or minus that you get when taking square roots.

Answer 10

what do you mean by the plus or minus that I will get when taking square roots? especially in this f^-1(x) = [ (sq root x) -3 ] / (-4) as my answer..

Answer 11

When you take the square root of both sides (very first step), you must add a plus or minus sign.

Answer 12

I can't show it w/o the equation editor

but if x^2 = 4

then x= 2 or x = -2

same thing here, except you just put a plus or minus sign in front of the sqrtx

Answer 13

+/-?

Answer 14

f ( f^( - 1 ) ( x ) ) = ( 3 - 4f^( - 1 ) (x ) )^2
x = ( 3 - 4f^( - 1 ) ( x ) )^2
+- sqrt ( x ) = 3 - 4f^( - 1 ) ( x )
- 3 + - sqrt ( x ) = - 4f^( - 1 ) ( x )
( 3 / 4 ) + - ( 1 / 4 ) sqrt ( x ) = f^( - 1 ) ( x )

Answer 15

right ?

Answer 16

I have no idea... this is far too hard to read like this but the last line looks right.

Answer 17

basically
x = (3 - 4y)^2
3-4y = +/- sqrtx
-4y = - 3 +/- sqrtx

y = 3/4 -/+ sqrtx

Answer 18

Yep.

Answer 19

so this this is not a function unless the domain of x is restricted so that the sign on the square root is unambiguous?

Answer 20

^exactly :)