# Thread: Range of y=3/(x^2+1)?

1. ## Range of y=3/(x^2+1)?

I want to know how to find the range of the equation y=3/(x^2+1) and y= (x-4)/x without using the graphical method?

Detailed explanation would be appreciated.
Please and thank you.

2. Originally Posted by Basketball
I want to know how to find the range of the equation y=3/(x^2+1)
For any value of x y will be positive. Do you know why?

Originally Posted by Basketball
y= (x-4)/4
This is a linear function. The range should be obvious!

3. $\displaystyle y = \frac{x - 4}{4} = \frac{1}{4}x - 1$.

This is a linear function. Linear functions have domain $\displaystyle \mathbf{R}$ and range $\displaystyle \mathbf{R}$ unless the domain is restricted.

$\displaystyle y = \frac{3}{x^2 + 1}$.

It should be clear that the denominator is never negative. In fact, the denominator is always $\displaystyle \geq 1$.

You should know that as a denominator gets larger, the fraction gets smaller. So that means that the largest possible value $\displaystyle y$ can take is $\displaystyle \frac{3}{1} = 3$, and the fraction will get closer to $\displaystyle 0$ as $\displaystyle x$ increases/decreases.

So the range is $\displaystyle y \in (0, 3]$.

4. Sorry I meant y=(x-4)/x, thank you for explaining the other equation to me.

5. Originally Posted by Basketball
Sorry I meant y=(x-4)/x, thank you for explaining the other equation to me.
$\displaystyle y = \frac{x - 4}{x} = 1 - \frac{4}{x}$.

This is a hyperbola. You should know that the standard hyperbola of the form $\displaystyle y = \frac{a}{x}$ has a horizontal asymptote at $\displaystyle y = 0$. So the standard hyperbolda's range would be $\displaystyle y \in \mathbf{R}\backslash\{0\}$.

In this case, the hyperbola has a vertical translation of $\displaystyle 1$ unit up.

So what do you think its range is?

6. Originally Posted by Prove It
$\displaystyle y = \frac{x - 4}{x} = 1 - \frac{4}{x}$.

This is a hyperbola. You should know that the standard hyperbola of the form $\displaystyle y = \frac{a}{x}$ has a horizontal asymptote at $\displaystyle y = 0$. So the standard hyperbolda's range would be $\displaystyle y \in \mathbf{R}\backslash\{0\}$.

In this case, the hyperbola has a vertical translation of $\displaystyle 1$ unit up.

So what do you think its range is?
y cannot be 1?

7. If I were to rewrite y=1-(4/x) in the format of y=(a/x), it would be y=-(4/x)+1. What does the negative in front of (4/x) do to the graph or what does it mean? Our teacher really didn't teach how the graphs would look like, I only know what y=(a/x) looks like. I don't really know how the numbers would affect the outcome of this graph.

8. Correct, the range is $\displaystyle y \in \mathbf{R}\backslash\{1\}$.

The negative gives a reflection in the $\displaystyle x$ axis. It doesn't affect the position of the asymptotes.

9. Originally Posted by Basketball
I want to know how to find the range of the equation y=3/(x^2+1) and y= (x-4)/x without using the graphical method?

Detailed explanation would be appreciated.
Please and thank you.
Personally, I think that generally speaking you are asking for trouble by trying to find the range of a function without using assistance from a graph. But that's just my opinion.

10. Originally Posted by mr fantastic
Personally, I think that generally speaking you are asking for trouble by trying to find the range of a function without using assistance from a graph. But that's just my opinion.
Unless of course you use your knowledge of functions to try to graph the function yourself

11. Another way to find the range of $\displaystyle y= \frac{x- 4}{x}$- find its inverse (solve for x): xy= x- 4, x- xy= 4, x(1- y)= 4, $\displaystyle x= \frac{4}{1- y}$.

If we think of that as a function of y, it is obvious that its domain is "$\displaystyle y\ne 1$. Since a function and its inverse "swap" domain and range, the range of the original function is "all y except 1".