Question

I was looking at functions and their inverses: is this always true: A function and it's inverse are a reflection over the line y = x

Answer 1

yes because x and y swap places

Answer 2

so like what if you take a function and find it's inverse and they are not a reflection around the line y = x. that does not make them inverses right?

Answer 3

then you made a mistake somewhere in finding the inverse

Answer 4

hmm

Answer 5

This was a problem well kind of finding the inverse for

\[y = Ae^{-kx}\]

Answer 6

I just made this up lol

Answer 7

\[y = 2e^{-2x}\]

Answer 8

Swap x and y. Then solve for y

\[\Large y = Ae^{-kx}\]
\[\Large x = Ae^{-ky}\]
\[\Large \frac{x}{A} = e^{-ky}\]
\[\Large e^{-ky} = \frac{x}{A}\]
\[\Large -ky = \ln\left(\frac{x}{A}\right)\]
\[\Large y = -\frac{1}{k}*\ln\left(\frac{x}{A}\right)\]

Answer 9

you are the man

Answer 10

thanks I'm gonna graph this to see how they are related.

Answer 11

it's probably not 100% obvious, but the orange and purple curves are mirror images of each other

https://www.desmos.com/calculator/igmdna30pd

one way to check is to pick a point on the purple curve and reflect it over to find a corresponding point on the orange curve. Do this with a bunch of points to get a better sense of the reflection going on

Answer 12

I'm going to do that now

Answer 13

For instance (0,1) on the purple curve reflects over to (1,0) on the orange curve. Notice how x and y swapped places.

Answer 14

Yeah what I did was I plotted (2,0.0185) and then (0.0185,2) and it was on the inverse. so it seems that the x and y points are swapped.

Answer 15

so it seems like every point on f(x) (x,y) becomes (y,x) on f^-1(x)

Answer 16

correct

Answer 17

because of the coordinate swap, the domain and range also switch too