1. ## Function Question

Hi,

f(x) = 3x+4

So I found the inverse:

f(x) = 3x+4
x=3y+4
x-4=3y
y=x-4/3

Then it asks to find the domain and range. So...could someone help me? I'm not sure on how to find them.

2. Originally Posted by lax600
Hi,

f(x) = 3x+4

So I found the inverse:

$f(x) = 3x+4$
$x=3y+4$
$x-4=3y$
$y=\frac{x-4}{3}$

Then it asks to find the domain and range. So...could someone help me? I'm not sure on how to find them.
For your domain, do you see any restrictions for x? Can x be any real number?

For your range, do you see any restrictions for y given any real number x? Can y be any real number:

Both functions graph lines.

3. The inverse is simply:

f(x)^-1=(Y-4)/3

Try drawing a simple graph of the equation to see where the domain lies.

Note that when Y=4, x=0

4. Originally Posted by moolimanj
The inverse is simply:

f(x)^-1=(Y-4)/3 Mr F says: Wrong. The inverse is a function of x, not y.

Try drawing a simple graph of the equation to see where the domain lies.

Mr F says: From your note below, I assume you're suggesting that the graph of y = f(x) = 3x+4 be drawn. That being the case, it's worth noting that this graph will make it clear what dom f and ran f are. This is useful since ${\color{red}\text{dom} f^{-1} = \text{ran} f}$ and ${\color{red}\text{ran} f^{-1} = \text{dom} f}$.

Note that when Y=4, x=0
..

5. Good observation Mr F

Apologies, I meant to put the equation as a function of x after I rearranged.

Spot on with the point about the graph

6. Originally Posted by masters
For your domain, do you see any restrictions for x? Can x be any real number?

For your range, do you see any restrictions for y given any real number x? Can y be any real number:

Both functions graph lines.

so I could set x as 7?

7. Originally Posted by lax600
so I could set x as 7?
It would appear that you may choose any real number for x which would make the domain all real numbers.

And in doing so, the range (values for y) would also be any real number making the range all real numbers as well.

$D=\{x|x \in \Re \}$

$R=\{y|y \in \Re \}$