How to solve inverse functions?

[Chaim]

Find a formula f^-1(x). Give the domain of f^-1, including any restrictions "inherited" from f
So one of my problem is f(x)=3x-6

But I'm kind of confused on how to start off
At first I thought it was like having the equation all go to the power of -1
So I got y=(1/3)x-(1/6)

But in the answer book it says: y=(1/3)x + 2

 

[Deleted member 4993]

Find a formula f^-1(x). Give the domain of f^-1, including any restrictions "inherited" from f
So one of my problem is f(x)=3x-6

But I'm kind of confused on how to start off
At first I thought it was like having the equation all go to the power of -1
So I got y=(1/3)x-(1/6)

But in the answer book it says: y=(1/3)x + 2

y = 3x - 6

To find the inverse function switch 'x' and 'y'

x = 3y - 6

solve for 'y'

y = x/3 + 6/3 = 1/3 * x + 2

 

[Chaim]

y = 3x - 6

To find the inverse function switch 'x' and 'y'

x = 3y - 6

solve for 'y'

y = x/3 - 6/3 = 1/3 * x - 2

Oh wow!
That makes things seem simpler now!
Thanks
Now I'm getting the hang of these inverse functions

 

[lookagain]

y = 3x - 6

To find the inverse function switch 'x' and 'y'

x = 3y - 6

solve for 'y'

y = x/3 - 6/3 = 1/3 * x - 2 <-------- The subtraction should be an addition.

$\displaystyle x = 3y - 6$


$\displaystyle x + 6 = 3y - 6 + 6$


$\displaystyle x + 6 = 3y$


$\displaystyle 3y = x + 6$


$\displaystyle \frac{1}{3}(3y) = \frac{1}{3}(x + 6)$


$\displaystyle y = \frac{1}{3}x + \frac{1}{3}(6)$


$\displaystyle y = \frac{1}{3}x + \frac{6}{3}$


$\displaystyle y = \frac{1}{3}x + 2$


$\displaystyle f^{-1}(x) = \frac{1}{3}x + 2$

 

[mmm4444bot]

like having the equation all go to the power of -1

In the symbol $\displaystyle f^{-1}(x)$, the -1 is not an exponent, and the f is not a number.

Nothing is being raised to the power of -1. It's unfortunate that exponential notation is used in this way, but it is, so we must simply memorize the notation.

The entire symbol $\displaystyle f^{-1}(x)$ represents a number. It represents the number that comes out of the inverse function of f, when x is input.

 

[Chaim]

$\displaystyle x = 3y - 6$


$\displaystyle x + 6 = 3y - 6 + 6$


$\displaystyle x + 6 = 3y$


$\displaystyle 3y = x + 6$


$\displaystyle \frac{1}{3}(3y) = \frac{1}{3}(x + 6)$


$\displaystyle y = \frac{1}{3}x + \frac{1}{3}(6)$


$\displaystyle y = \frac{1}{3}x + \frac{6}{3}$


$\displaystyle y = \frac{1}{3}x + 2$


$\displaystyle f^{-1}(x) = \frac{1}{3}x + 2$

Oh right!
Thanks xD

 

[Chaim]

In the symbol $\displaystyle f^{-1}(x)$, the -1 is not an exponent, and the f is not a number.

Nothing is being raised to the power of -1. It's unfortunate that exponential notation is used in this way, but it is, so we must simply memorize the notation.

The entire symbol $\displaystyle f^{-1}(x)$ represents a number. It represents the number that comes out of the inverse function of f, when x is input.

Lol yeah, that makes so much more sense now, thanks xD
At first, when I saw the problem and made 3x go to the power of -1, I thought it made sense, but when the 6 came out all wrong, I was like 'What!?'
Lol thank you very much for that tip