Thread: Showing a function is not surjective

1. Showing a function is not surjective

Hi, I've got a question here and I've got to show that a function is not surjective.

To do this I've got to find a value that the function cannot equal, but I'm not sure how to do this, is trail and error the right way or is there a more formal approach?

Here's the actual function:

$\displaystyle f : \mathbb{R}$ \$\displaystyle \frac{1}{3} \to \mathbb{R}$ given by $\displaystyle f(x) = \frac{5x+1}{3x-1}$

2. There's no real x such that f(x)=5/3, which is the horizontal asymptote.

3. Originally Posted by Black
There's no real x such that f(x)=5/3, which is the horizontal asymptote.
Ohh yeh thank you!

4. A direct way to do it is to set up the equation $\displaystyle \frac{5x+1}{3x-1}= y$ and try to solve for x.

$\displaystyle 5x+1= (3x+1)y= 3xy+ y$. 5x- 3xy= y - 1. (5- 3y)x= y- 1.

$\displaystyle x= \frac{y-1}{5- 3y}$ as long as the denominator is not 0.

If 5- 3y= 0, that is, if $\displaystyle y= \frac{3}{5}$, there is no such x.

5. Originally Posted by HallsofIvy
A direct way to do it is to set up the equation $\displaystyle \frac{5x+1}{3x-1}= y$ and try to solve for x.

$\displaystyle 5x+1= (3x+1)y= 3xy+ y$. 5x- 3xy= y - 1. (5- 3y)x= y- 1.

$\displaystyle x= \frac{y-1}{5- 3y}$ as long as the denominator is not 0.

If 5- 3y= 0, that is, if $\displaystyle y= \frac{3}{5}$, there is no such x.
Thanks for the reply, that's a good way of thinking of it.