Question

How do you graph the inequality `5/(x+3) ≥ 3/x`?

Answer 1

The final graph will be the shape of `5/(x+3)` with `-3 <=x <=0; x>=4.5`

Explanation:

In order to graph the whole relationship, we need to graph both the left and right individually.

Here's the left side `(5/(x+3))`:

graph{5/(x+3) [-20, 20, -20, 20]}

And the right `(3/x)`:

graph{3/x [-20, 20, -20, 20]}

So now the question is, for what values of `x` is `5/(x+3)>=3/x`. We'll find that wherever the result of `5/(x+3)`, in essence the `y` value, is larger than that of `3/x`:

graph{(y-(5/(x+3)))(y-3/x)=0 [-20, 20, -20, 20]}

By observation, we can see that, approaching the graph from the left to the right, that the first value of `x` where this holds is at `x=-3` and continues until `x=0`.

At `x=4.5`, it holds again and does so until `+oo`:

graph{(y-(5/(x+3)))(y-3/x)=0 [4, 6, 0, 1]}

And so the final graph will be the shape of `5/(x+3)` with `-3 <=x <=0; x>=4.5`