Question

PLEASE HELP

f^1(3) when f(x) = (2x+3)/5

Answer 1

Did you mean \(f^{-1}(3)\)?

Answer 2

Yes

Answer 3

First you want to set f(x) = y

Can you do that for me?

Answer 4

Y = f(x) (2x+3)/5 ?????

Answer 5

I meant replace the f(x) with y =...

Answer 6

Oh okay then it owuld be y = (2x+3)/5

Answer 7

Awesome :)

Now switch `y` with `x` , and rewrite the equation.

Answer 8

x = (2y+3)/5

Answer 9

Great :)

We want to solve for `y`, so the first thing we would need to do is multiply both sides of the equation by `5`.

Can you show me what the equation becomes when you do that?

Answer 10

5x = (2y+3)

Answer 11

Now we want to subtract -3 from both sides. What does your equation become?

Answer 12

5x-3 = 2y

Answer 13

And now divide both sides of your equation by 2.

Answer 14

5/2x-3/2 = y

Answer 15

And now we have found our inverse function :)\[f^{-1}(x) = \frac{5x-3}{2} \iff \frac{5x}{2} -\frac{3}{2}\]

Answer 16

Oh that makes so much sense, so is that it? Or are there more steps?

Answer 17

Now we want to find \(f^{-1}(3)\) , so all you have to do now is plug in `3` for x and solve :)

Answer 18

What do you get?

Answer 19

15/2 - 3/2
so that would be 12/2
so then it would be 6???

Answer 20

Yay! I got the same :D

Answer 21

Good job ^^

Answer 22

Okay awesome, so just to clarify would it be x = 6 or y = 6???

Answer 23

I believe it is x.

Answer 24

One thing I forgot to mention is we started out with \[f(x) = \frac{2x+3}{5}\]We rewrote it as \[y=\frac{2x+3}{5}\]Swapped variables.\[x=\frac{2y+3}{5}\]solved for y again.\[y=\frac{5x-3}{2}\]and then rewrote it once again as \(f^{-1}(x)\)\[f^{-1}(x) = \frac{5x-3}{2}\]To check which variable we are solving for, we could take the inverse of the original function, this will leave us with the function itself, either in x, or y. To check: \[f^{-1}(f(x)) = x\]