# Thread: Solve determining inverse functions

1. ## Solve determining inverse functions

Determine wether f(x) = 3x + 8 and g(x) = x - 8 / 3 are inverse functions.

For f(x) = 3x - 19, find the inverse f-¹(x). Minus sign is supposed to be uppercase..

2. ## Re: Solve determining inverse functions

If $f(x)$ and $g(x)$ are inverse functions, then $f[g(x)]=g[f(x)] = x$

For the second problem, swap variables ...

$x=3y-19$

... solve for $y$

3. ## Re: Solve determining inverse functions

Is g(x)= x- 8/3 or is it (x- 8)/3? That is an important difference!

4. ## Re: Solve determining inverse functions

Originally Posted by HallsofIvy
Is g(x)= x- 8/3 or is it (x- 8)/3? That is an important difference!
While true you have managed so far to not sink to Plato's level of questioning the obvious.

5. ## Re: Solve determining inverse functions

Originally Posted by romsek
While true you have managed so far to not sink to Plato's level of questioning the obvious.
That's often a good thing...

-Dan

6. ## Re: Solve determining inverse functions

Originally Posted by topsquark
That's often a good thing...
-Dan
That is particularly true when one is qualified to comment.

7. ## Re: Solve determining inverse functions

Originally Posted by HallsofIvy
Is g(x)= x- 8/3 or is it (x- 8)/3? That is an important difference!
x - 8
3

8. ## Re: Solve determining inverse functions

I may have to "sink to Plato's level" (that would be a bad thing???). I don't think it was at all "obvious" that when the OP wrote "x- 8/3" he meant "(x- 8)/3".

9. ## Re: Solve determining inverse functions

Thanks, skeeter! Sorry, I'm kind of new to this.. but I'm kind of stuck with proving g[f(x)]=x. The first one should be the following, right?

1. Determine wether f(x) = 3x + 8 and g(x) = x - 8 / 3 are inverse functions.

f[g(x)] = f[(x - 8) / 3]
= 3x [(x - 8) / 3] + 8
= x - 8 + 8
= 8

g[f(x)] = g(3x + 8)
..and then how would you continue? Or would you've done it differently?

2. I understood this one.. should be the following?

f^-1(x) = (x + 19) / 3

Br,
Evan

10. ## Re: Solve determining inverse functions

Originally Posted by Griegevan
Thanks, skeeter! Sorry, I'm kind of new to this.. but I'm kind of stuck with proving g[f(x)]=x. The first one should be the following, right?

1. Determine wether f(x) = 3x + 8 and g(x) = x - 8 / 3 are inverse functions.

f[g(x)] = f[(x - 8) / 3]
= 3x [(x - 8) / 3] + 8
= x - 8 + 8
= 8

g[f(x)] = g(3x + 8)
..and then how would you continue? Or would you've done it differently?

2. I understood this one.. should be the following?

f^-1(x) = (x + 19) / 3

Br,
Evan
$\displaystyle f[g(x)] = 3g(x)+8 = \cancel{3}\left(\dfrac{x-8}{\cancel{3}}\right)+8 = (x-8)+8 = x$

$\displaystyle g[f(x)] = \dfrac{f(x)-8}{3} = \dfrac{(3x+8)-8}{3} = \dfrac{3x}{3} = x$

So, $\displaystyle f^{-1}(x) = g(x)$.

For the second one, yes, $\displaystyle f(x) = 3x-19 \Longrightarrow f^{-1}(x) = \dfrac{x+19}{3}$

11. ## Re: Solve determining inverse functions

Originally Posted by Griegevan
Thanks, skeeter! Sorry, I'm kind of new to this.. but I'm kind of stuck with proving g[f(x)]=x. The first one should be the following, right?

1. Determine wether f(x) = 3x + 8 and g(x) = x - 8 / 3 are inverse functions.

f[g(x)] = f[(x - 8) / 3]
= 3x [(x - 8) / 3] + 8
= x - 8 + 8
= 8

check this again ...

g[f(x)] = g(3x + 8)
..and then how would you continue? Or would you've done it differently?

g(3x+8) = [(3x+8) - 8]/3 = ... continue

2. I understood this one.. should be the following?

f^-1(x) = (x + 19) / 3

Br,
Evan
...