Question

Find a formula for the inverse of the function.

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

y=ln(x+3)

y=x^2 - x

y=(1+e^-x)/(1+e^-x)

Answer 1

if the function is one-to-one,
you just solve the equation for x

Answer 2

\[y=\frac{ax+b}{cx+d} \\ \text{ multiply both sides by } cx+d \\ (cx+d)y=ax+b \\ \text{ distribute on the left } \\ cx y +dy=ax+b \\ \text{ now remember you want to solve for } x \\ \text{ get all of your terms with } x \text{ on one side } \\ \\ \text{ so subtract } ax \text{ and } dy \text{ on both sides } \\ cxy-ax=b-dy \\ \text{ now this allows you to do the following so you can solve for } x \\ x(cy-a)=b-dy \\ \text{ now divide both sides by } (cy-a)\]

Answer 3

Oh, ok. Thank you! :)

Answer 4

after solving for x,interchange x any y

Answer 5

y=(1+e^-x)/(1+e^-x)
How about the very last one? The negative exponent is confusing me.

Answer 6

if the negative exponent confused you can multiply top and bottom by e^x

Answer 7

\[y=\frac{1+e^{-x}}{1+e^{-x}} \cdot \frac{e^{x}}{e^{x}} \\ y=\frac{e^{x}+1}{e^{x}+1}\]

Answer 8

wait a minute

Answer 9

lol (1+exp(-x))/(1+exp(-x)) is 1

Answer 10

is there a type-o?

Answer 11

\[y=\frac{1+e^{-x}}{1+e^{-x}}=1 \]

Answer 12

and f(x)=1 is not one-to-one

Answer 13

I think the typo is that the numerator is (1-e^-x) while the denominator is (1+e^-x). But would that make that much of a difference?

Answer 14

Oh wait, it would. You can't simplify it to be 1.

Answer 15

yes! now the top is different from the bottom the function is not 1 anymore
now verify the function is one-to-one
and if it is solve for e^x and then solve for x
and if is not you are done because no inverse function exist

Answer 16

I said solve for e^x because you can still multiply top and bottom by e^x
giving you
\[f(x)=\frac{e^{x}-1}{e^{x}+1}\]

Answer 17

Ah I see how to solve it now. Thanks again!

Answer 18

are you looking for inverse relation or function?

Answer 19

Inverse function

Answer 20

so did you verify f(x)=(1-exp(-x))/(1+exp(-x)) is one-to-one?
you can use a graphing calculator and see if it passes the horizontal line test?

Answer 21

Oh ok. So I plugged it into the graphing calculator and the function doesn't pass he horizontal line test.

Answer 22

the*

Answer 23

Actually, it does. Nevermind, it looked like it didn't at first.

Answer 24

ok so that means you have to actually solve that equation above for e^x
then x

Answer 25

Alright, will do.