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

**Basketball** · September 20th 2010, 05:18 PM

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.

**pickslides** · September 20th 2010, 05:36 PM

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!

**Prove It** · September 20th 2010, 05:37 PM

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

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

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

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

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

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

**Basketball** · September 20th 2010, 05:44 PM

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

**Prove It** · September 20th 2010, 05:47 PM

Originally Posted by Basketball

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

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

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

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

So what do you think its range is?

**Basketball** · September 20th 2010, 05:50 PM

Originally Posted by Prove It

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

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

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

So what do you think its range is?

y cannot be 1?

**Basketball** · September 20th 2010, 05:57 PM

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.

**Prove It** · September 20th 2010, 06:21 PM

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

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

**mr fantastic** · September 20th 2010, 06:32 PM

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.

**Prove It** · September 20th 2010, 10:52 PM

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

**HallsofIvy** · September 21st 2010, 02:14 AM

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

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

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