A question that give wrong answer?

[matphysik]

I don't think this is true. Just because f fails to be defined at a couple of points, how does that show there exists y1 and y2 that don't get hit by some value of x?

In the joke solution I gave in post #15, I said to take a = x. Of course that's a "what color is the bear?" answer, not a math answer.

If you do that, you get f(x) = (x^3+6x-8) / (7x-8x^2) which blows up at two points. But if you graph it, you'll see that it's onto! The three connected components of the graph overlap to cover the y-axis.

In general, rational function can be onto even if the denominator has real zeros and therefore blows up at a couple of points. I didn't prove that, I just looked at a graph ... so it's possible I'm totally wrong here.

Here's the graph I got from OS X Grapher.

First, `a` (parameter) cannot by chosen to be the independent variable (x). Second, the graph of f(x) = (x^3+6x-8) / (7x-8x^2) has pole at x=0. So as you can see from the graph there is no `y`∈ℝ s.t. y=f(0).

When the denominator quadratic has no real roots then f (a quotient of two polynomials) is onto. Just the monomial term in the numerator quadratic guarantees that.

 

[matphysik]

Hello!
Today my maths teacher ask a question to all student in class. that was

f:R-->R
f(x) = (a*x[SUP]2[/SUP]+6x-8) / (a+6x-8*x[SUP]2[/SUP])
f is onto then find the value of [b]a[/b]

after 10 minutes some students answer. every body was wrong.
this question was also answered wrong in many Indian writer's books for many years. Now my teacher tell where we all students were wrong and know i know the correct answer. I just want to show this question to all those persons who love mathematical beauty.I am not intended to get answer for homework help.
So guys try it.
Please don't answer this if you already know this questions truth.

Just look at the essential part, and ignore the commentary

 

[SteveL27]

First, `a` (parameter) cannot by chosen to be the independent variable (x).

I said about three times that this is a joke solution ... but the resulting function is very interesting anyway.

Second, the graph of f(x) = (x^3+6x-8) / (7x-8x^2) has pole at x=0. So as you can see from the graph there is no `y`∈ℝ s.t. y=f(0).

Just look at the graph. There's clearly a zero at around x = 1 or so, where the right hand component of the graph crosses the x-axis. What do you make of that?

You keep saying that there's a pole at 0, which of course is true. But it does not therefore follow that NO value of x gives f(x) = 0.

 

[matphysik]

f(x) = (x^3+6x-8) / (7x-8x^2) has pole at x=0, and for `x` very close to zero f(x)~ -8/7x. Hence, as x->0- f-> +∞ and as x->0+ f-> -∞. If the resolution of your graph were better you could see this behaviour more clearly.

 

[SteveL27]

f(x) = (x^3+6x-8) / (7x-8x^2) has pole at x=0, and for `x` very close to zero f(x)~ -8/7x. Hence, as x->0- f-> +∞ and as x->0+ f-> -∞. If the resolution of your graph were better you could see this behaviour more clearly.

It has nothing to do with the behavior of f at x = 0. Better study the definition of onto. The graph crosses the x-axis at around x = 1. That means f takes the value zero for at least one value of x. So f is onto.

To see that f is onto, look at the y axis. You can see that you can pick any point on the y-axis and a horizontal line through it will intersect the graph of f.

You are fixating on the pole at 0 but that has nothing to do with whether f is onto. You are confusing domain and range.

 

[matphysik]

The original post says that the domain of f is all of ℝ.

 

[SteveL27]

The original post says that the domain of f is all of ℝ.

Well yes, you are correct about that. But the function f I gave is onto. I hope you see that. I ran it through the equation solver at http://www.quickmath.com/webMathematica3/quickmath/equations/solve/basic.jsp to find the real root.

This example shows that a rational function can have a pole (two, in fact) and still be onto.

 

[SteveL27]

The quadratic in the denominator must have complex roots, which occur when a<-9/8. Then f(x) will be analytic, and onto iff the domain of f is ℝ.

I tried a = -2, and the resulting function does not have a real zero. In fact there are two solutions, both complex. So you are right that with the denominator never zero, you get an analytic rational function, but not necessarily onto the reals when the domain is restricted to the reals.

Do you have a specific value of 'a' that you have shown to produce a function that is onto the reals when the domain is restricted to the reals?

 

[nickalh]

Why is VKash's original reasoning incorrect?
In my opinion, the test makers are being overly literal or legalistic.

I. See
Possible solution to NOT meet the criteria. This function IS ONTO. but does not answer what the question is asking.
Quick reasoning for "onto":
i. The Limit as x->(3-sqrt(33))/8 from left, the vertical tangent, is +oo & the limit from the right is -oo.
ii. Quotients of polynomial functions are continuous everywhere they are defined.
iii. So the example, f with a=3 is onto (-oo, +oo).
See II & III for why it's not an answer.

II. In many books, you'll find
f:R->R, g:R->R
f(x)= 1/(x-2) or g(x)= x(x-4)/(x-4) + sin x
of course we realize x=2 is not in the domain of f and x=4 is not in the domain of g, but it's a bit of mathematical shorthand or informality to call R, the domain.

III. So if a=3, would meet the criteria for onto. This is why it's acceptable usage in many textbooks, not so excessively formal.
However, under the exam question, they are apparently expecting & requiring a more formal (IMHO, excessively formal) interpretation.
The fact that the question askers require every single possible number in all of R to give a legitimate y value rules out Possible solution As far as I can tell, according to the test makers, any vertical tangent or removable discontinuity in the quotient means it's not a solution for f. So under the test makers interpretation, the denominator can never be zero ever for a removable discontinuity.

IV. I believe the previous posts make clear why if the denominator never reaches zero, it's not onto. Another time, I may elaborate on that.

 

[nickalh]

Errghh, is there way to edit a previous post?
These are clarifications to my July 17, 6:43am post.
To clear up some responses, to my post.

A. "function is bounded"
Remember all this in context, no vertical tangents, a<-9/8
I should say the function has a bound.
For example, y =x^2 has a bound; y>=0 for all x.
I can probably prove both bounded, meaning it has upper and lower bound based on both numerator and denominator have degree 2, but that is too complicated & unnecessary.

This is not true in general, consider f(x) = xsin(x). It will require some work to show that it is true for rational polynomial functions.

Of course, I know, but that goes beyond the scope of this thread.
Some work- not that much. Our specific situation is much simpler. The following is a ROUGH DRAFT OR SKETCH OF A PROOF also called a proof plan.

Pick some a <-9/8
for all a<-9/8 it can be shown the denominator is always negative.
Likewise, for any a<-9/8, it can be shown the numerator is always negative.
So for a<-9/8, f is always positive.
This establishes 0 as a lower bound for f, when a<-9/8.
Interested readers can fill in the details.

Response to SteveL,
I only see a link to the solver & am unsure exactly what you put in. For future reference, please post the link to your actual solution. WolframAlpha maybe more effective.
Also, the first mention I see of a=x as a joke, is quite a while after the mention of a =x.

When the denominator quadratic has no real roots then f (a quotient of two polynomials) is onto. Just the monomial term in the numerator quadratic guarantees that.

It's true for our original question, but in general, this is false
y=x^4/(x^2+1)
One rule of thumb, divide the highest two powered terms, pretend the "degree" of the rational is the degree of the result. This will give some insight into the behavior especially, when the denominator has no real roots.

To VKash,
I think all the talk of complex numbers so far, can safely be rephrased as "no real numbers".

 

[vkash]

Why is VKash's original reasoning incorrect?
I........
......
.

[size=+2]where am i wrong[/size]. I have used basic rules not any big trick.
According to SteveL27 a=x ((x^3+6x-8) / (7x-8x^2) ) is an answer. stevel does 0 comes in domain of function((x^3+6x-8) / (7x-8x^2)) Hu. as given in question 0 should in domain of function but that is not in your answer. so stop making unwanted comments on a function that is not valid for condition i propose.
You still think my explanation is incorrect(in this post about stevel) then find domain range of this function f(x) = 1/x does 0 in it's domain.
sorry that i make post little late.
keep talking on this question because debate make us better.

-------------------------
Mathematics is an assumption[/color]

 

[SteveL27]

Response to SteveL,
I only see a link to the solver & am unsure exactly what you put in. For future reference, please post the link to your actual solution. WolframAlpha maybe more effective.
Also, the first mention I see of a=x as a joke, is quite a while after the mention of a =x.

Here's a Wolfram Alpha link for the case of a = -2. Note that -2 is less than -9/8.

http://www.wolframalpha.com/input/?i=(-2x^2+++6x+-+8)+/+(-2+++6x+-+8x^2)+=+0

Now you see that there are no real solutions. So matphysik's claim that a < -9/8 gives an onto function is not true. Now, there may be SOME value of a < -9/8 that works, but till we hear from matphysik, we have to reject his claim. He noted that with the denominator never zero, the rational function is analytic. That's true. But I don't see how it follows that f must then be onto the reals as a function of the reals.

 

[nickalh]

VKash,
The very first step HallsOfIvy gave on one of the very first responses gets us into trouble.
y= \frac{ax^2+6x-8}{a+6x-8x^2}
then
y(a+ 6x- 8x^2)= ax^2+ 6x- 8
For x-values, which make the denominator zero, the first line is invalid. See II & III on my Tuesday, 9:43am post. However, for the second line, those x-values masquerade as solutions to the original question.
see
False solutions a=2, x=1 and a=14, x=-1


Essentially, it boils down to you're original method may include vertical tangents, which is technically not f:R->R.

using shredharanachrya principle D<0

This appears related to the discriminant of a quadratic function.
However, b^2 -4ac becomes
36-4*(-8)a<=0
36 plus 32a <=0
So I'm unsure what you're getting at.


The main reason, HallsOfIvy's original solution doesn't work

HoI's method is perfectly correct. His method will give that the function f is onto if a\geq 2. However, since the function must be entire, it must also hold that a+6x-8x^2 has no roots. But no a\geq 2 satisfies this. So there are no solutions.

I also think we're beating a dead horse. I'll try to move on.
Bye.

 

[SteveL27]

I also think we're beating a dead horse. I'll try to move on.

Nobody has suggested a value of 'a' that makes f onto. Nobody has proved that no such 'a' exists. This is an open problem. matphysik has suggested a promising approach, but I believe there's a counterexample to his claim that any a < -9/8 works. Perhaps some other value works.

An open problem isn't a dead horse in my opinion.

 

[disregardthat]

It can be proven algebraically that no such a exists as by hall's of ivy's method.

Alternatively using calculus, one can simply look at the limit as x approaches infinity. The function will have horizontal asymptotes for any 'a' which makes surjectivity impossible.

 

[nickalh]

In general, surjectivity, also known as onto, does not require horizontal asymptotes.
See
http://www.wolframalpha.com/input/?i=y+=+(cos+x)/x"

Similar, examples of polynomial rational functions, may exist especially with higher degree polynomials.

 

[disregardthat]

In general, surjectivity, also known as onto, does not require horizontal asymptotes.
See
http://www.wolframalpha.com/input/?i=y+=+(cos+x)/x"

Similar, examples of polynomial rational functions, may exist especially with higher degree polynomials.

Uhm, that function does not have domain R. A continuous function f : R --> R with horizontal asymptotes on both sides will not be surjective.

 

[matphysik]

Here's a Wolfram Alpha link for the case of a = -2. Note that -2 is less than -9/8.

http://www.wolframalpha.com/input/?i=(-2x^2+++6x+-+8)+/+(-2+++6x+-+8x^2)+=+0

Now you see that there are no real solutions. So matphysik's claim that a < -9/8 gives an onto function is not true. Now, there may be SOME value of a < -9/8 that works, but till we hear from matphysik, we have to reject his claim. He noted that with the denominator never zero, the rational function is analytic. That's true. But I don't see how it follows that f must then be onto the reals as a function of the reals.

Hi. If the function is analytic, then it`s continuous.

x↦logx takes (0,∞) onto ℝ. Rewrite, y=f(x)=(ax²+6x-8)/(a+6x-8x²)≡(x-A)(x-B)/(x-C)(x-D). Then, logy=log(Re[(x-A)(x-B)]-log[(x-C)(x-D)]). So then, ONE possibility for f(x) being onto occurs if ALL of {A,B,C,D} are complex numbers. In this way we will not get log(0+) for all x∈ℝ.

 

[vkash]

I think we're beating a dead horse. I'll try to move on.
Bye.

I also think so.
Actually my main motto for this post was to know why method written by HallsOfIvy is wrong and now i got that.
I does not post it(real answer and my question) directly in first post because i want to know is there any mathematical dude here who think so. unfortunately no such dude come here to answer.

 

[matphysik]

CORRECTION:

f (as given) takes ℝ into ℝ so that, y=f(x)=Re{f(x)}. Let z↦logz take ℂ\(-∞,0] onto ℂ. Write, f(x)=(ax²+6x-8)/(a+6x-8x²)≡(x-A)(x-B)/(x-C)(x-D). For `a` to be determined such that f takes ℝ onto ℝ. ONE possibility for f(x) being onto occurs if ALL of {A,B,C,D} are complex numbers.Then,
Re[logf(x)]=Re[log(x-A)+...-log(x-D)] takes ℝ (x) into ℝ\(-∞,0] (f(x)) onto ℝ (Re[logf(x)]).

If z=eʷ and w=u+iv then |z|=e^(Re(w)) and log|z|=Re(w). Let w=logf(x).

 

[SteveL27]

Hi. If the function is analytic, then it`s continuous.

CORRECTION:

f (as given) takes ℝ into ℝ so that, y=f(x)=Re{f(x)}. Let z↦log₁z take ℂ\(-∞,0] onto ℂ. Write, f(x)=(ax²+6x-8)/(a+6x-8x²)≡(x-A)(x-B)/(x-C)(x-D). For `a` to be determined such that f takes ℝ onto ℝ. ONE possibility for f(x) being onto occurs if ALL of {A,B,C,D} are complex numbers.Then,
Re[log₁f(x)]=Re[log₁(x-A)+...-log₁(x-D)] takes ℝ onto ℝ\(-∞,0]. Next, let z↦log₂z take ℂ\[0,∞) onto ℂ. Then, Re[log₂f(x)]=Re[log₂(x-A)+...-log₂(x-D)] takes ℝ onto ℝ\[0,∞). So that, given such `a` f will take ℝ onto ℝ\{0} (since, (-∞,0]∩[0,∞)={0}). Last, since Re{f(0)}=Re{(A)(B)/(C)(D)} we have proved (for this particular case) that f takes ℝ onto ℝ.

I'm sorry, I had a hard time following that. You used the phrase "for this particular case," but I don't see any particular case.

Consider f(z) = z^2. It's onto the complex numbers as a function of complex numbers; but its restriction to the reals is not onto the reals. Why not? Well, f(x) = x^2 does not hit -1, for example. If you take z = i, then f(z) = z^2 does hit -1. But Re(i) = 0 so you haven't really gotten any benefit out of taking the real part of a complex function.

Earlier you said that any a < -9/8 gives an onto function (with f as in the original problem); but I showed that with a = -2, the resulting rational function has two complex roots and no real ones, so as a function from the reals to the reals, it fails to hit 0 and is therefore not onto.

Do you have a specific 'a' that works?

(edit)
I think you're getting in trouble with y=f(x)=Re{f(x)}. It's true that you can take the real part of f(z) to get a function from C to R. But it doesn't follow that you now have a function from R to R that preserves the attributes you care about. You only have a function from C to R. When you restrict the domain to R, you lose surjectivity, as the example of f(z) = z^2 shows.

Another example is f(z) = e^z. We have f(i*pi) = -1, but the real part of i*pi is 0; and the the real part of -1 is -1. You can't freely interchange Re and f.

By the way, what is log₁? What does the subscript '1' do? Do you just mean complex natural log?

 

[matphysik]

"By the way, what is log₁? What does the subscript '1' do? Do you just mean complex natural log?"

They are different branches of the complex logarithm.

"I'm sorry, I had a hard time following that. You used the phrase "for this particular case," but I don't see any particular case."

Find B²-4AC<0 for BOTH quadratics in order to find an interval for `a` for each case, then find the intersection of the two intervals for `a` obtained.

 

[matphysik]

 

[SteveL27]

"By the way, what is log₁? What does the subscript '1' do? Do you just mean complex natural log?"

They are different branches of the complex logarithm.

Oh, I'd never seen that notation before. Thanks.

Find B²-4AC<0 for BOTH quadratics in order to find an interval for `a` for each case, then find the intersection of the two intervals for `a` obtained.

I think you have a clever idea to use complex analysis to attack this problem. It's clear that the discriminants of both the top and bottom polynomials are the same, namely 32a + 36; and that this is negative exactly whenever a < -9/8.

What you've found, assuming the rest of your proof holds up, are functions from the complex numbers that are onto the reals. Unfortunately the original problems specifies f:R -> R.

Now you are making the leap to say that restricting the domain to R preserves surjectivity. But this is demonstrably false, since if you take a = -2 you get a rational function that has two complex roots and no real ones. That means it doesn't hit 0. So you've lost surjectivity when you restrict the domain to the reals.

I already presented this example earlier and you have not commented on it.

I also gave the example of f(z) = z^2, which is surjective when considered as a function from C to C; but is not surjective onto the reals when the domain is restricted to R.

I've just said the exact things as I did earlier. One, there is a clear counterexample to your claim that {x : discriminant < 0} gives a solution to the problem. And two, that your ambitious attempt to use complex analysis is a good idea, and DOES produce a surjection C -> R. But it does not produce a surjection R -> R.

Back atcha

 

[matphysik]

Oh, I'd never seen that notation before. Thanks.



I think you have a clever idea to use complex analysis to attack this problem. It's clear that the discriminants of both the top and bottom polynomials are the same, namely 32a + 36; and that this is negative exactly whenever a < -9/8.

What you've found, assuming the rest of your proof holds up, are functions from the complex numbers that are onto the reals. Unfortunately the original problems specifies f:R -> R.

Now you are making the leap to say that restricting the domain to R preserves surjectivity. But this is demonstrably false, since if you take a = -2 you get a rational function that has two complex roots and no real ones. That means it doesn't hit 0. So you've lost surjectivity when you restrict the domain to the reals.

I already presented this example earlier and you have not commented on it.

I also gave the example of f(z) = z^2, which is surjective when considered as a function from C to C; but is not surjective onto the reals when the domain is restricted to R.

I've just said the exact things as I did earlier. One, there is a clear counterexample to your claim that {x : discriminant < 0} gives a solution to the problem. And two, that your ambitious attempt to use complex analysis is a good idea, and DOES produce a surjection C -> R. But it does not produce a surjection R -> R.




Back atcha

Lol, Yeah. I`ve never really thought deeply about a question like this. Even though at first glance it seems simple. It`s not.

The plan put forth by `HallsofIvy` is NOT good either. For you end up with a quadratic in `a` without any meaningful conditions imposed upon it.

I tried (unsuccessfully) to use the well known properties of the complex log to make some progress indirectly. But what i wrote are just attempts, and not a polished end result. Oh well.

 

[matphysik]

I also think so.
Actually my main motto for this post was to know why method written by HallsOfIvy is wrong and now i got that.
I does not post it(real answer and my question) directly in first post because i want to know is there any mathematical dude here who think so. unfortunately no such dude come here to answer.

NO. What you`re doing is giving up when the going gets tough.

 

[vkash]

NO. What you`re doing is giving up when the going gets tough.

Do you not read micromass 13th post. He explain it.

HoI's method is perfectly correct. His method will give that the function f is onto if a≥2. However, since the function must be entire, it must also hold that a+6x−8x2 has no roots. But no a≥2 satisfies this. So there are no solutions.

You are little bit correct that i am giving up. Mainly because of around nil knowledge of complex numbers.

 

[matphysik]

Oh, I think I know what it is know. Isn't it weird how there is an -8 on top where there is an a in the bottom and vise versa? You simply put in a=-8 and the answer is always 1!

((-8)*x^2+6x-8) / ((-8)+6x-8*x^2)=1

-8 is smaller than -9/8 so the calculus was true.
I don't quite get what the question is though.

NO. Then f will take ℝ onto {1}. We want f to take ℝ onto ℝ. As a matter of fact, for a=-8 we have f(x)≡1.

 

[disregardthat]

Calculus proof of that such an a does not exist:
First of all, the denominator will have to be non-zero for all x. Hence the discriminant Delta = 6^2-4(-8)a = 36+32a will have to negative, i.e. a < -36/32 = -9/8.

For such a the function is well-defined. Now consider \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{ax^2+6x-8}{a+6x-8x^2} = \lim_{x \to \infty} \frac{ax^2+6x-8}{a+6x-8x^2} = \lim_{x \to \infty} \frac{a+6\frac{1}{x}-8\frac{1}{x^2}}{a\frac{1}{x^2}+6\frac{1}{x}-8} = \frac{a}{-8}.

Similarly \lim_{x \to -\infty} f(x) = \frac{a}{-8}.

So there are real numbers A and B such that |f(x)-\frac{a}{-8}| &lt; 1 for all x > A, and |f(x)-\frac{a}{-8}| &lt; 1 for all x < B, and we can assume that A > B.
i
f(x) on [B,A] is continuous and thus bounded, so there is an M such that |f(x)|<M for all x in [A,B]. Hence |f(x)| &lt; \max(1+|\frac{a}{-8}|,M) for all x in (-\infty,B) \cup [B,A] \cup (A,\infty) = \mathbb{R}, and is thus not surjective.

 

[matphysik]

Calculus proof of that such an a does not exist:
First of all, the denominator will have to be non-zero for all x. Hence the discriminant Delta = 6^2-4(-8)a = 36+32a will have to negative, i.e. a < -36/32 = -9/8.

For such a the function is well-defined. Now consider \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{ax^2+6x-8}{a+6x-8x^2} = \lim_{x \to \infty} \frac{ax^2+6x-8}{a+6x-8x^2} = \lim_{x \to \infty} \frac{a+6\frac{1}{x}-8\frac{1}{x^2}}{a\frac{1}{x^2}+6\frac{1}{x}-8} = \frac{a}{-8}.

Similarly \lim_{x \to -\infty} f(x) = \frac{a}{-8}.

So there are real numbers A and B such that |f(x)-\frac{a}{-8}| &lt; 1 for all x > A, and |f(x)-\frac{a}{-8}| &lt; 1 for all x < B, and we can assume that A > B.
i
f(x) on [B,A] is continuous and thus bounded, so there is an M such that |f(x)|<M for all x in [A,B]. Hence |f(x)| &lt; \max(1+|\frac{a}{-8}|,M) for all x in (-\infty,B) \cup [B,A] \cup (A,\infty) = \mathbb{R}, and is thus not surjective.

Your proof looks good!