Question

How to calculate the inverse function of f(x) = (exp(x))+2)/(exp(-x))?

Answer 1

so first, let me rewrite: \[f(x)=\frac{e^x+2}{e^{-x}}\]
Correct?

Answer 2

yes that is correct :)

Answer 3

ok, so now, in order to find the inverse, let's first talk about what the inverse is. The inverse "undo's" a function, but it also happens to be what happens when you swap the domain and range. So do you have any ideas for our first step?

Answer 4

Can you tell why the inverse exists?

Answer 5

Like you said, we exchange the domain and the range, so that's what I tried to do. So, we "swap" x for y and y for x and rearrange

Answer 6

where f(x) = y

Answer 7

then, first I wrote \[y = e ^{2x} + 2e ^{x}\]

Answer 8

Yup but before you do that, graph the function, as Abhishek pointed out, there does not have to be an inverse

Answer 9

hold off on that

Answer 10

I don't quite understand what Abhishek69 meant by that

Answer 11

technically* you should test for an inverse first, I just skipped the step because I knew the answer. My bad so in order for a fn to have an inverse, it must be 1-1 or pass the horizontal line test

Answer 12

aka for every x, there is a unique y

Answer 13

yep okay

Answer 14

does this eq pass?

Answer 15

yes it does

Answer 16

cool ok so now, your rewrite was correct, continue down the path

Answer 17

I then tried to take the natural log of both sides, and do the usual stuff (hopefully correct anyway). But in the end I got something that was quite different to the answer (it was a demonstration in the lecture I tried to redo). The answer I got was \[\frac{ 1 }{ 3 } \ln(\frac{ y }{ 2 })\]

Answer 18

so you made a fundamental error, ln(x+y) cannot be simplified at all

Answer 19

let's walk through what you did

Answer 20

so, \[\ln(y) = \ln(e ^{2x}) \times \ln(2e ^{x})\]

Answer 21

is that correct? I took the log of the whole of the right-hand side

Answer 22

no

Answer 23

ah.

Answer 24

they are not multiiplied

Answer 25

yes but i used the log rule

Answer 26

they are addition,

Answer 27

again ln(x+y) cannot be simplified

Answer 28

ln(xy)=(lnx)+(lny)

Answer 29

there is no rule for this situation

Answer 30

so rewrite from the beginning distribution you did

Answer 31

okay

Answer 32

SO i'll give you a hint. Your first step is going to be: complete the _____.

Answer 33

wait wait wait wait

Answer 34

hm?

Answer 35

I was being an idiot. I didn't do what I did, that was the "fail" attempt that i forgot to cross out

Answer 36

o.O ok?

Answer 37

yeah anyway, if i now take the log of each term, that's what i'd get to in the end. (with the 1/3 blah answer). now I complete the square?

Answer 38

no

Answer 39

there is no 1/3

Answer 40

you're right, it's wrong nonetheless

Answer 41

i was going to say, now if i let x = exp(x)

Answer 42

so let's start here \[y=e^2x+2e^x\]

Answer 43

and write y = x^2 + 2x, would that do something?

Answer 44

no

Answer 45

so let's start here
\[y=e^{2x}+2e^
x\]

Answer 46

just trust me ok, i'll walk you through

Answer 47

first step?

Answer 48

really not a clue, pardon

Answer 49

ok, so remember, swithc the variables to start

Answer 50

okay, so now we swap y and x?

Answer 51

yep type that in, you can copy and paste my LaTex commands using right click show math as

Answer 52

oh. I think this is it. I forgot to do the swapping from the start.

Answer 53

nope, please... please... I'm tired and losing patience, just do step by step with me please?

Answer 54

But I'm sorry, I have a lecture to go to in half an hour so I need to get ready. I'll try this question again later. Thank you for your time, and I'm sorry about this!

Answer 55

wait I have the answer
pleasssseeee

Answer 56

it's just messy if you do it with me we can be done in 10 min or less

Answer 57

I would love to but I really gotta dash :/ Thank you very much for the offer though

Answer 58

ugh, ok

Answer 59

just complete the square then simplify then ln then simplify then exp then simplify then ln again and done

Answer 60

okay okay, thanks a lot! will try again later after physics. have a nice day :)

Answer 61

you too. took me 8 steps, so when you do just try ok?

Answer 62

yep