Functions problem help needed

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Homework Statement


Two students are arguing over the following equations. One student claims that the graphs are the same, the other student claims they are different. The equations are as follows:


Homework Equations


y = x - 1 and y = x^2 - 1 / x + 1 (x + 1 is the divisor)

The Attempt at a Solution


I believe the graphs are different, but an algebraic solution must be included. Please help solving this problem algebraically. Any help is greatly appreciated. Thank you.
 

Answers and Replies

  • #2
eumyang
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Well, they are different. You'll have to simplify the 2nd function, but in doing so, an assumption has to be made. What is that assumption?
 
  • #3
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Well, when the second equation is simplified: x - 1 is left over, correct? I'm not sure what exact assumption you are referring to.
 
  • #4
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Don't simplify anything. Does a value for the second function exist if X = -1?

Simplifying a multidimensional equation to make it "easier" to solve as a linear one is not correct. However many dimensions an equation has, that many solutions you must also have. If you "simplify" to a lower dimension you will lose solutions and your entire solution would be incorrect.
 
  • #5
Ray Vickson
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Homework Statement


Two students are arguing over the following equations. One student claims that the graphs are the same, the other student claims they are different. The equations are as follows:


Homework Equations


y = x - 1 and y = x^2 - 1 / x + 1 (x + 1 is the divisor)

The Attempt at a Solution


I believe the graphs are different, but an algebraic solution must be included. Please help solving this problem algebraically. Any help is greatly appreciated. Thank you.
Never, never write what you wrote, which was
[tex] y = x^2 - \frac{1}{x} + 1[/tex]
when you mean
[tex] y = \frac{x^2 - 1}{x+1}.[/tex]
Use parentheses, like this: y = (x^2-1)/(x+1). It only takes an extra 1--2 seconds to type, and it makes everything clear and unambiguous.

Anyway, ask yourself: does the first form make sense for alll x? Does the second form make sense for all x? If not, what goes wront?
 
  • #6
epenguin
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Anyway, ask yourself: does the first form make sense for alll x? Does the second form make sense for all x? If not, what goes wront?
Well, they are different. You'll have to simplify the 2nd function, but in doing so, an assumption has to be made. What is that assumption?
Don't simplify anything. Does a value for the second function exist if X = -1?
Well you make me unsure what official doctrine is about whether f(x)g(x)/g(x) exists where g(x) = 0 .

But let me point out that that was not the question, the question was about graphs. Assuming every point satisfying y = the function has to be on a black line, are you able to show me graphs for these two functions, if there are two, which are different? :tongue2:
 
  • #7
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I apologize for not using parentheses. It's my first time using this forum. I have solved the problem if I simplify the multidimensional equation, but I'm not sure how that is not correct. Thank you, everyone.
 
  • #8
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I made a post earlier, but I didn't realize that I solved it for you. Be careful when you simplify, though. Revise the material that addresses continous functions and the conditions. You said you solved the problem if you simplify, well are you sure you didn't divide by 0?
 
  • #9
epenguin
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What is your, or anyone's, answer?
 
  • #10
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What is your, or anyone's, answer?
I guess the graphs would be different. The second function is not defined at x=-1 so I think the graph would be a straight line (it reduces to linear after factorisation) with a discontinuity at x=-1 (shown by a small circle?).
 
  • #11
epenguin
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I guess the graphs would be different. The second function is not defined at x=-1 so I think the graph would be a straight line (it reduces to linear after factorisation) with a discontinuity at x=-1 (shown by a small circle?).
I disagree, see #7.

But I indicate there how anyone can prove me wrong. :smile:
Graphically.
 
  • #12
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I disagree, see #7.

But I indicate there how anyone can prove me wrong. :smile:
Graphically.
See attachment. That's what I think the graph for second function would be but I would rather wait for someone else to confirm.
 

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  • #13
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This is a post that got deleted since it was considered as solving the problem. I hope it is alright to repost it now since people demand blood:
A function is continous if 3 conditions are met. Since one sides assumes they are the same function it must also mean they are both continuous because the 1st one is.
F(x) = x-1
G(x) = (x^2-1)/(x+1)
1) A function f(x) has a specific value for any argument value, for example x=a. do they? Any argument value, so x = -1.
F(-1) = -2 OK
G(-1) = N/a, since division by 0 occurs Not OK

2)A function has a specific limit in case of x->a. When a = -1
lim x->-1 F(x) = -2
lim x->-1 G(x) = no limit, because reduction is only allowed if the factors =/= 0, in this case they are 0 therefore operation not permitted.

3)The most obvious one - The value of f(x) in case of x=a is equal to the limit in case of x->a
The condition is met for f(x) in case of x=-1 and x->-1
The condition is Not met for g(x) for the same argument value.

For whatever else argument value these conditions are met, because then you are allowed to reduce the factors as they are not 0, but there occurs a discontinuity for g(x) which makes these 2 functions Different.

#13 Your graph looks like F(x) = 2x to me
 
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  • #14
epenguin
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See attachment. That's what I think the graph for second function would be but I would rather wait for someone else to confirm.
Elementary mistake there I think. Using the formula as written in #6 I think the slope should be 1 not 2. And also we should have the graph of the first condition for comparison.
 
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  • #15
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Elementary mistake there I think. Using the formula as written in #6 I think the slope should be 1 not 2. And also we should have the graph of the first gundtion for comparison.
Woops. Fixed it.
 
  • #16
epenguin
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Does the graph of y = (x - 1) look different?
 
  • #17
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Yes, it does not contain a discontinuity at x = -1.
 
  • #18
epenguin
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Yes, it does not contain a discontinuity at x = -1.
And your graph does? I can't see a discontinuity on your graph either.
 
  • #20
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After thinking about Riemann's theorem on removable singularities I get the feeling it's correc to state that they are the same functions or different, since Riemann doesn't say it's incorrect to imply a discontinuity, even though holomorphically extendable, makes the difference between our 2 functions. From hereon, I am a neutral party :D
 
  • #21
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I have figured the problem out. The graphs are different, as x =/= -1. Thus, using g(x) = (x^2-1) / (x+1) there is a point discontinuity at x = -1. Thank you everyone! Much appreciated.
 
  • #22
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I have solved the problem. The graphs are different because given g(x) = (x^2 - 1) / (x + 1) there is a point discontinuity, for x =/= - 1. Thank you, everyone! Much appreciated.
 
  • #23
epenguin
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I have figured the problem out. The graphs are different, as x =/= -1. Thus, using g(x) = (x^2-1) / (x+1) there is a point discontinuity at x = -1. Thank you everyone! Much appreciated.
And I claim that's wrong! No graph can show you a "point discontinuity" like that.

(to cut lng stry shrt). And the one we have doesn't.
 

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