# Real solutions

1. Dec 27, 2014

### Raghav Gupta

1. The problem statement, all variables and given/known data
To find number of real solutions of:
$\frac{1}{x-1}$ $+\frac{1}{x-2}$ + $\frac{1}{x-3}$ + $\frac{1}{x-4}$ =2

2. Relevant equations
It will form a 4th degree polynomial equation.

3. The attempt at a solution
The real solutions could be 0 or 2 or 4 as complex solutions always exist in pairs.
It is a 4th degree polynomial equation , the real roots could be 4 but how?

2. Dec 27, 2014

### Stephen Tashi

The given equation is not a 4th degree polynomial equation. When you do multiplications on it to form a 4th degree polynomial equation, you may create an equation that has a different solution set than the original equation. For example, the original equation does not have solutions x = 1, x = 2, x =3 or x = 4 since such values require division by zero. (For example, for $\frac {1}{x-2}$ to have a defined value, we cannot have $x = 2$.

Are you studying functions that have asymptotes? Perhaps you are expected to sketch a graph of the function $f(x) = \frac{1}{x-1} + \frac{1}{x-2} + \frac{1}{x-3} + \frac{1}{x-4} - 2$. The question asked for the number of roots ( the number of places the graph of the function crosses the x-axis) and not the particular values of the roots.

3. Dec 27, 2014

### Raghav Gupta

Sketching graph of that function seems difficult. I am not studying functions that have asymptotes. I only require number of real roots. Calculus can be used here I think but not able to apply. I know that the number of real roots maybe 0 or 2 or 4.

4. Dec 27, 2014

### BvU

If you find the combined sketch difficult, start with the first term. Then the second, then estimate what 1st + 2nd looks like. You will see the pattern quickly -- and have overcome your aversion to things that seem difficult as well :)

5. Dec 27, 2014

### Raghav Gupta

I was trying and found out that whenever value of x is negative y is always negative. So, there is no crossing of x axis when x -ve, so no real root till that point. Now the problem arises that around x=5 y value is coming +ve. So there is a crossing but there may be more. Can anyone help in that?

6. Dec 27, 2014

### Stephen Tashi

As I said before, you don't know that.

Let h be a small positive number, say h = 1/100,000. What is the approxomate value of the function at x = 1 + h ?

At that value, all terms except $\frac{1}{x-1}$ are of moderate size. The term $\frac{1}{x-1}$ has value 100,000. Now consider the value of the function at x = 2 - h.

7. Dec 27, 2014

### Raghav Gupta

At x = 2 - h , the term $\frac{1}{x-2}$ would be -100,000. It means the graph crossed x axis one time there.
Yeah , it seems I got it a bit.
Then, at x = 2 + h the value again becomes 100,000 second crossing
Then at 3-h , -100,000 third crossing
At 3+h fourth crossing,
4-h fifth crossing
4+h sixth crossing
Then at value between 5 and 6 seventh crossing and after that no crossing.
But from this the answer is coming 7.
I know the answer is 4 from calculators.
So, what is my mistake?

8. Dec 27, 2014

### Stephen Tashi

If you know the function was postive for some x < 2 then you could say there has been a crossing. But was f(x) ever positive for x < 2 ?

You have the basic idea. But to determine a crossing you must consider what happens between two places where f(x) "blows up", not just "at" one place.

Considering x = 1 + h and x = 2-h is what tells you that the graph crosses the x-axis between x = 1 and x = 2.

9. Dec 27, 2014

### Ray Vickson

I think you are mis-understanding what is meant by "crossing". When the graph of y = f(x) jumps from y = -∞ to y = +∞ as x passes from left to right through x = 1 (or x = 2, 3, or 4) we do not regard that as a "crossing", because there is no value of x in the interval (1-h,1+h) that gives f(x) = 0 exactly. Of course, if you draw the graph of y = f (x) and put in a vertical line at x = 1 (or x = 2, etc.) that vertical line does pass through the x-axis; but the line is not actually part of the graph.

Anyway, x-axis crossings are not the issue in this problem: y = 2 crossings are what matter.

10. Dec 27, 2014

### LCKurtz

@Ray Vickson

Hey Ray, have you ever seen this with Maple? I tried to plot the given function and the line y=2 and got this output:

10 + BB BB BB B B
+ BB BB BB B B
8 + BB BB BB B B
+ BB BB B B B B
6 + BB BB B B B B
+ BB B B B B B BB
4 + BB B B B BB B BB
+ B B B BB B B B BB
AAAAAAAAA2A*AAAAAAAAA*A*AAAAAAA*AA*AAAAAA*AA**AAAAA*AAAAAA******AAAAAAAAA
+ B BB B BB B BB B BBBBBBBBB
-+-+-+-+--+-+-+-+-+-+-*-+**-+-+-*-+-**+-+-*-+-+**-+-*-+-+-+-+-+-+-+--+-+-+-
+ B BB B B B BB B
-1BBBBBBBB + 1 BB 2 BB 3 BB 4 5 6
-2B*BBB B B B B B B B
+ BB B B B B B B B
-4 + BB B B B B B BB
+ B B B B B B BB
-6 + BB B B B BB BB
+ B B B B BB BB
-8 + B B BB BB BB
+ B B BB BB BB
-10 + B B BB BB BB

It's obviously some sort of ASCII output in a very rough shape of the graph. I'm using Maple 13.02 but I don't know how it got in this mode or how to get it out of it.

Last edited: Dec 27, 2014
11. Dec 27, 2014

### Raghav Gupta

Sorry Ray as I am not able to get it. Can you please elaborate.
It is seeming a hard puzzle to me what you said Stephen. Isn't it true that for x= 1 + h that is x < 2 f(x) is positive?

12. Dec 27, 2014

### Stephen Tashi

I didn't mean to imply that f(x) was never positive for x < 2. I meant that you must check that x was positive for some x < 2 before you conclude that there was a place where x = 0 between that place and x = 2 - h.

An example of what Ray is saying is that x = 2 is not a place where the graph crosses the x-axis, even though f(2-h) is negative and f(2+h) is positive. The value at x = 2 is a "singularity". It is a place where the graph does not exist.

13. Dec 27, 2014

### Raghav Gupta

So, how now to go for it?

14. Dec 27, 2014

### Raghav Gupta

15. Dec 27, 2014

### Stephen Tashi

For example, think about what happens between x = 1+h and x = 2 -h instead of thinking of something between x = 2-h and x = 2 + h.

After that think about what happens "at the ends". What happens between when x is very small and x = 1-h? What happens between x = 4+h and when x becomes very large.

16. Dec 27, 2014

### Raghav Gupta

Please can you refer my above post?
For my better understanding with diagram.

17. Dec 27, 2014

### Ray Vickson

No, I have never seen it, and when I give the following instructions (in Maple 11 on an HP Probook) everything works just fine---it even puts in the vertical line:
f:=1/(x-1):
pflot(f,x=-1..3,y=-10..10);

To plot it without the vertical line I just said
plot(f,x=-1..3, y=-10..10,discont=true);

Similarly,
g = add(1/(x-i),i=1..4): plot(g, x=-1..6, y=-10..10);
produces a graph with 4 vertical lines at x = 1,2,3,4.

18. Dec 27, 2014

### Stephen Tashi

The plot in your link isn't an informative plot. It doesn't clearly what happens between x = 0 and x = 5 and that interval is the important part of the graph. (I don't use Wolfram, so I can't advise you how to get a clearer plot. Can you plot only the interval between x = 0 and x = 5? Perhaps it will show up better that way.)

19. Dec 27, 2014

### Ray Vickson

If you want to understand what is happening, forget about trying to use Wolfram Alpha, and just make simple sketches done manually. (Yes, I am serious!) For $f(x) = 1/(x-1) + 1/(x-2) + 1/(x-3) + 1/(x-4)$, you can understand what happens near $x = 1$ just by looking at the simple function $1/(x-1)$. Slightly to the left of $x = 1$ this function is large negative and slightly to the right it is large positive. Basically, it jumps from $y = -\infty$ to $y = +\infty$ as $x$ passes through 1, from left to right. The other terms $1/(x-2) + 1/(x-3) + 1/(x-4)$ will just change things a bit near $x = 1$ but will not alter in any way the jump behavior at $x = 1$. (You should thing about why that is true!)

Similarly, you can understand what happens near $x = 2, 3, 4$ just by looking at the simpler functions $1/(x-2), \; 1/(x-3), \; 1/(x-4)$.

So, just to the right of $x = 1$ the function is large positive, and just to the left of $x = 2$ it is large negative, which means that between $x=1$ and $x=2$ there is a root of your equation $f(x) = 2$. Furthermore, you should think about why there is just one, single root in the interval $(1,2)$.

Keep going like that for other $x$-intervals.

20. Dec 27, 2014

### SammyS

Staff Emeritus
Try this in Wolfram Alpha.

http://m.wolframalpha.com/input/?i=...1/(x-3)+++1/(x-4)-2,+from+x=0+to+x=7&x=4&y=10

That takes x from 0 to 7.

If you simply plot $\ f(x) = 1/(x-1) + 1/(x-2) + 1/(x-3) + 1/(x-4)\$ without subtracting 2, you may get some idea regarding the methods Steven and Ray are suggesting.