Hello, Fiona!

I'll take a guess at what you meant . . .

For for what $\displaystyle x$-values is: $\displaystyle f(x) \:=\:\frac{2x-1}{x+3}$ a decreasing function?

If we are allowed to use Calculus, determine when $\displaystyle f'(x) \,<\,0.$

We have: .$\displaystyle f\:\!'(x) \:=\:\frac{2\cdot(x+3) - (2x-1)\cdot1}{(x+3)^2} \;=\;\frac{7}{(x+3)^2} $

And we see that $\displaystyle f\:\!'(x)$ is __always__ __positive__.

The function is **never** decreasing.

The graph is even more convincing . . . Code:

*: |
: |
* : |
* : |
* : |
* : |
* : |2
- - - - - - - - + - - - + - - - - - - -
: | *
-------------------+-------+-----*------------
-3: *
: * |
: * |
: |
:* |
: |