# Homework Help: Range of quadratic equation.

1. Feb 21, 2012

### PrincePhoenix

The problem statement, all variables and given/known data

The following equation defines a rational function.Find its Range
f(x) = x2/(x-1)

The attempt at a solution

Let y = x2/(x-1)

=>y(x-1) = x2
=>xy-y = x2
=>x2-xy+y = 0

comparing with the general form of quadratic equation ax2+bx+c = 0,
a=1, b=-y, c=y.

Putting the values in quadratic formula,

x = y±√(y2 - 4y)/2

It is clear that f(x) will only be real when the term y2-4y is greater than or equal to zero. So,

y2-4y ≥ 0

y(y-4) ≥ 0 ---------(i)

divide both sides by y,
y-4 ≥ 0
y ≥ 4
---------------------------------------------------------------------------------
However just by looking at the discriminant it is clear that y ≤ 0. How do I get that? When I divide y-4 on both sides of (i), I get y≥0.

Last edited: Feb 21, 2012
2. Feb 21, 2012

### th4450

Can y-4 be negative?

3. Feb 21, 2012

### PrincePhoenix

I don't get what you're trying to say.

What I mean is that along with y-4 ≥ 0, there is another interval for values of range which is supposed to be y ≤ 0. How do I get that?

4. Feb 21, 2012

### genericusrnme

What happens when you multiply (or divide) both sides of an inequality like $a \le b$, what sign goes here $-a\ ?\ -b$

5. Feb 21, 2012

### PrincePhoenix

-a≥-b?

But isn't that because we multiplied '-' on both sides of the inequality??

6. Feb 21, 2012

### PrincePhoenix

Do you mean this is the way to do it??

y2-4y ≥ 0

y(y-4) ≥ 0 ---------(i)

multiply both sides by '-1'

-(y-4)y ≤ 0
divide both sides by -(y-4)

y≤0?

7. Feb 21, 2012

### genericusrnme

From here, take different cases;
y(y-4) ≥ 0

Suppose y-4=0 then;
Suppose y-4<0 then;
Suppose y-4>0 then;

8. Feb 21, 2012

### PrincePhoenix

y-4 = 0 => y=4
y-4 > 0 => y>4
y-4 < 0 => y<4?

Still I don't get you. I take the three different possible cases. What do I do with them? Do I have to check each to see which are correct? Or somehow use them to get y ≤ 0?

9. Feb 21, 2012

### genericusrnme

If (y-4) < 0, then (y-4) is negative

What happens now when you divide

y(y-4) ≥ 0

by the negative number (y-4)?

10. Feb 21, 2012

### PrincePhoenix

y ≤ 0.
Thank you.

11. Feb 21, 2012

### PrincePhoenix

Just a couple of questions to clear it up. When all the three possibilities were tried, how should I have known that the answer I got by y-4<0 one of the two intervals?
And can I find the second interval that I had found in my original solution using this method?

12. Feb 21, 2012

### genericusrnme

You know that it is correct because you're just checking equalities at all points, you're free to pick a value for y (or set of values for y) and check each of them.
When you set y-4<0 you are checking every y that's less than 4
When you set y-4=0 you are checking for y=0
When you set y-4>0 you are checking every y that's greater than 4
By doing this you have checked every possible value for y.

Using this method you will find ALL solutions

y(y-4) ≥ 0

Suppose y-4>0 then
y ≥ 0
Which just returns y>4

Suppose y-4=0 then
y*0=0
Which is obviously true, so y=4 is valid, combined with the first result we get
y ≥ 4

Suppose y-4<0 then
0 ≥ y

And so we arrive at the two sets of solutions using this method
0 ≥ y
y ≥ 4

13. Feb 21, 2012

### PrincePhoenix

Thanks a lot genericusername. . That cleared it up.

14. Feb 21, 2012

### genericusrnme

No problem buddy!

15. Feb 21, 2012

### checkitagain

It must be that $$y(y - 4) \ge 0.$$

Get critical numbers by setting y = 0 and y - 4 = 0 and solving for y in each respective equation.

So the critical numbers are y = 0, -4.

Test numbers to the left, in between, and to the right to see which
regions satisfy the inequality.

Get to your desired range from there.

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