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September 20th, 2010, 05:32 PM   #1
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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?

Can someone explain how to find the range of functions without using graphs?

Detailed explanation would be appreciated.

Please and thank you.
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September 20th, 2010, 06:15 PM   #2
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Re: Range of y=3/(x^2+1)?

When you have

y = 3/(x + 1)

switch the x's and y's and solve for y, to get the inverse function, then find the domain of the resultant.

x = 3/(y + 1)

y = 3/x - 1

y = ?[(3 - x)/x]

The domain here is (0,3]

Thus the range of the original is (0,3].

Now, for y = (x - 4)/x

x = (y - 4)/y

y = 4/(1 - x)

The domain here is (-?,1) U (1,?)

Thus the range of the original is (-?,?)
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September 20th, 2010, 07:05 PM   #3
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Re: Range of y=3/(x^2+1)?

Quote:
Originally Posted by basketball041520
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?

Can someone explain how to find the range of functions without using graphs?

Detailed explanation would be appreciated.

Please and thank you.
There's an art and a science to it...

Notice that in y = 3/(x^2 + 1) that the top is always positive, and the bottom is always positive. In fact, the bottom has a minimum value of 1 (when x = 0), so the maximum value of y is 3/1 = 3. There is no "minimum", as the graph approaches zero as you move away from the y-axis.
The range is (0, 3]
This analysis required the knowledge of fractions and the range of "x^2", to which you can simply add the +1 to get the "range" of the bottom.

y = (x - 4)/x is the same as
y = x/x - 4/x
y = 1 - 4/x
The "-4/x" can take on any real value (positive OR negative) if you let x be very small or large. For this reason, the range is all real numbers.
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September 21st, 2010, 04:47 AM   #4
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Re: Range of y=3/(x^2+1)?

The "-4/x" can take on any real value (positive OR negative) if you let x be very small or large but 0. For this reason, the range is all real numbers but 1.

Hoempa
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September 21st, 2010, 04:52 AM   #5
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Re: Range of y=3/(x^2+1)?

http://www.youtube.com/watch?v=QgwZJ3zWnl0
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September 21st, 2010, 09:54 AM   #6
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Re: Range of y=3/(x^2+1)?

Quote:
Originally Posted by Hoempa
...For this reason, the range is all real numbers but 1.

Hoempa
Yep, I realized I failed to simply copy the domain of the inverse correctly after I went to bed last night.
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