Question

Quick question, anyone...

What process happens from 1/y=x+5
to get to f^-1(x)=1/5+5...

Answer 1

f^-1(x) = 1/x + 5

Answer 2

sorry

Answer 3

the first equation tells you how to calculate a value for y given a value for x:\[\frac{1}{y}=x+5\]

Answer 4

the second in the inverse

Answer 5

i.e. given a value for y, what value of x generated that value for y

Answer 6

you take the inverse of 1/y and the inverse of x+5 so they become y=1/x+5

Answer 7

so the inverse of x+5 is 1/x + 5.. simple as that

Answer 8

yep, its the inverse of both sides

Answer 9

if:\[\frac{1}{y}=x+5\]then:\[y=\frac{1}{x+5}\]

Answer 10

but:\[f^{-1}(x)\]has a completely different mening

Answer 11

like that :D

Answer 12

well the final answer is given to me as that.. f^-1(x) = 1/x + 5

Answer 13

is that:\[a) f^{-1}(x)=\frac{1}{x}+5\]or:\[b)f^{-1}(x)=\frac{1}{x+5}\]

Answer 14

so from where i am (1/y = x + 5) .. i just take the inverse of both sides.. and i have my answer

Answer 15

if its (b)

Answer 16

=O.. but its (a)

Answer 17

exactly what I thought

Answer 18

\(f^{-1}(x)\) represents the inverse function. let me try and explain how to get it

Answer 19

ok.. ha. i honestly that (a) and (b) were the same thing

Answer 20

(a) and (b) are very different

Answer 21

ah ok, so solve for y and then switch x and y to get the inverse in that case

Answer 22

you are given:\[\frac{1}{y}=x+5\]so if rearrange this to find x in terms of y, we get:\[x=\frac{1}{y}-5\]then replace \(x\) with \(f^{-1}(x)\) and \(y\) with \(x\) to get:\[f^{-1}(x)=\frac{1}{x}-5\]that is the inverse function

Answer 23

oh ok. i see. since x is in parentheses after f^-1... i ddnt think to do any re arranging.

Answer 24

but i understand what you are saying

Answer 25

note: this is not the same as the equation for \(f^{-1}(x)\) you listed in your question. I have worked it out to be with a minus sign before the 5 instead of a plus sign.

Answer 26

i understand ha. thanks

Answer 27

yw :)