Rational Connection!
f ( x ) = x 2 − 1 a x 2 + 6 x + b and g ( x ) = x 2 − 1 ( a + b ) x 2 + 6 0 x − 5 4
Let f ( x ) and g ( x ) be defined by the given formulas on the set R − { − 1 , 1 } . If the range of f ( x ) is the set R , then which one of the following is the range of g ( x ) .
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2 solutions
After identifications and thorough checks, a = 1.04 and b = 4 is at least the proximity for
f(x) = x 2 − 1 2 5 2 6 x 2 + 6 x + 4
but neither g(x) = x 2 − 1 2 5 1 2 6 x 2 + 6 0 x − 5 4 nor g(x) = x 2 − 1 2 5 3 5 4 x 2 + 6 0 x − 5 4 .
(0, -4), (-5, 0), (2, 6.72) and (7, 2.02) were proximity of f(x) applied.
Between -1 and 1, both f(x) and g(x) are indeed between + ∞ and − ∞ and there is no discontinuity.
f(-0.769230769230769) = 0 and g(0.840639341755765) = 0 are within continuities respectively.
Discontinuity for g(x) only occur for x <-1 which is not within our range of discussion!
This is a question with an attempt to capture preset intuition by fault. Careful work can get rid of fault.
Specific rather than general. However, it must be true before a general case can be true.
Answer: T h e S e t O f A l l R e a l N u m b e r s .
@Lu Chee Ket , thank you very much for finding this problem interesting. You are right! Assuming that the problem has been set correctly, you can find a selection of values for a and b that justifies your answer. Of course, this is a valid way in this case. If you want to see a general proof that applies for any possible selection of values of a and b you can see my solution above. Thank you again for taking time to try this problem!
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Much appreciate that specific case is not the only situation that can satisfy. Not easy to imagine in the way you described for general cases. You are very strong at seeing "function variables" to state in a way synonym to a topic called complex variables. I am still trying to understand the whole mathematics that you described. To be honest however, I think I am doing a right way for a correct answer. But another question followed this question try to give a lesson that this thinking may not be right. Nevertheless, simplifying complicated general way into specific way is usually a technique to answer questions which are given with certain correct answers, because when a specific case cannot be correct, a general case makes no point to be considered. This way rely onto correctness of question instead of doing an absolute analysis. Yet, it meets the purpose of answering correctly. The question followed is another story. Thanks for setting this question in Brilliant to share with all.
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I agree with you. The question was asked in a way that your solution by taking into consideration a particular set of values is completely right. In my solution, I am just trying to show a more a general approach for Brilliant members that want to know the reasons why this works. I don't think that what I am using here is very advanced. The only result that I use is the fact that a quadratic equation a x 2 + b x + c = 0 with real coefficients has real solutions if and only if its discriminant b 2 − 4 a c ≥ 0 . Another related result that I use is that under similar conditions, if a > 0 then a x 2 + b x + c ≥ 0 for all values of if and only if b 2 − 4 a c ≤ 0 .
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@Arturo Presa – Was b 2 − 4 a c applied in this particular question?
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@Lu Chee Ket – Yes, when I said that the following equivalence is true:
∀ y : ( a − y ) x 2 + 6 x + ( b + y ) = 0 has real solutions with respect to x that are distinct from 1 and -1.
⟺ ∀ y : 3 6 − 4 ( a − y ) ( b + y ) ≥ 0 and a + b = ± 6
I am considering the equation ( a − y ) x 2 + 6 x + b + y = 0 as a quadratic equation with respect to x . This equation has real solutions if and only if its discriminant 6 2 − 4 ( a − y ) ( b + y ) ≥ 0 .
Additionally, we have to consider solutions of that equation distinct from 1 or -1, because 1 and -1 are not solutions of the equation f ( x ) = x 2 − 1 a x 2 + 6 x + b = y , due to the fact that f ( 1 ) and f ( − 1 ) are undefined. The way of excluding those solutions from the equation ( a − y ) x 2 + 6 x + b + y = 0 is by adding the condition a + b = ± 6 . The reason is simple: if you evaluate the expression ( a − y ) x 2 + 6 x + b + y at x = 1 or x = − 1 , you get a − y + 6 ( ± 1 ) + b + y = 0 , or equivalently a + b ± 6 = 0 . The latter condition is equivalent to a + b = ± 6 .
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@Arturo Presa – Let me take some time to read.
@Arturo Presa – Let me have much more time to think.
@Arturo Presa – All right. I can see that you relate g ( x ) to f ( x ) to show that they are actually similar within the range in between x = -1 and x = 1 as they are of the same form. Perhaps you can add in the explanation of how you applied b 2 − 4 a c in the first place. A lot of patience is required to write for a very clear guiding solution. You may like to rectify later.
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@Lu Chee Ket – Thank you @Lu Chee Ket . I really appreciate your comments. If you have any other questions about this problem, feel free to ask me.
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@Arturo Presa – What is the significance about graph f (x) and g (x) as introduced which you may think that they are relevant or quite important to study? Are they feature of any equipment or any thing which require our caution from thinking wrongly about them?
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@Lu Chee Ket – I can't tell you exactly, but the only thing that I noticed about using graphs here is that there are many different possibilities. For example, the numerator function can have a zero, two zeros or no zeros in between -1 and 1, Again, what I am telling applies only to the case when you are trying to get a general proof that if the range of the first function is the set of all real numbers, then the range of the second is also the same set.
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@Arturo Presa – You mean more onto technical applications and cautions onto pure mathematics in a form of creativity and challenge. One more question: What or where is the source of this question or what have made you thought about such an f (x)?
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@Lu Chee Ket – The inspiration has been another problem. You can see the problem here. That problem was my starting point and then I began trying with similar functions until I got these ones. Once you know that the range of the original function f ( x ) is the set of all real numbers if and only if ∣ a + b ∣ < 6 , the idea was to modify the function in a way that you get another one satisfying the same condition. I noticed that the set of pairs K = { ( a , b ) / ∣ a + b ∣ < 6 } is convex. Actually it is the open region bounded by the square with vertices ( − 6 , 0 ) , ( 0 , 6 ) , ( 6 , 0 ) , and ( 0 , − 6 ) . Using the definition of this set, and its convexity, you can get that if ( a , b ) is a point of K many other points that are also in K like for example ( a + b , 0 ) , ( 0 , a + b ) , ( 1 0 1 ( a + b ) , 1 0 9 ( − 6 ) ) . The latter point is in the linear segment joining ( a + b , 0 ) and ( 0 , − 6 ) . Any of these point could be used to generate a function whose range is the set of all real numbers. For example, we might have considered the function g ( x ) = x 2 − 1 ( a + b ) x 2 + 6 x that corresponds to the point ( a + b , 0 ) . Instead, I originally decided to use the function g ( x ) = x 2 − 1 1 0 1 ( a + b ) x 2 + 6 x − 1 0 9 ( − 6 ) that corresponds to the point ( 1 0 1 ( a + b ) , 1 0 9 ( − 6 ) ) . But then I noticed that I could take the factor 1 0 1 out, by multiplying by 10, and this does not change the range in the case that this is the set of all real numbers. That is how I got the second function g ( x ) = x 2 − 1 ( a + b ) x 2 + 6 0 x + ( 9 ) ( − 6 ) = x 2 − 1 ( a + b ) x 2 + 6 0 x − 5 4 .
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@Arturo Presa – Thanks! As a friend, I think I should inform you that there is a word "integer" being written as "integral" at the question you introduced at "here" (blue). The question was not created by you anyway.
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@Lu Chee Ket – I understand, what you are saying. Actually, it is not so common but I have seen the use of "integral" instead of "integer" before.
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@Arturo Presa – Several times in Brilliant but I suddenly thought of mentioning. I think I shall tell the person when I can have an opportunity later.
First we are going to prove that the range of f ( x ) = x 2 − 1 a x 2 + 6 x + b is R if and only if ∣ a + b ∣ < 6 . Indeed, if a + b = ± 6 the solutions of any equation of the form ( a − y ) x 2 + 6 x + b + y = 0 for whatever value of y are always distinct from 1 and -1. Therefore,
The range of f ( x ) is R ⟺
⟺ ∀ y : f ( x ) = y has real solutions that of course are distinct from 1 and -1.
⟺ ∀ y : ( a − y ) x 2 + 6 x + b + y = 0 has real solutions that are distinct from 1 and -1.
⟺ ∀ y : 3 6 − 4 ( a − y ) ( b + y ) ≥ 0 and a + b = ± 6
⟺ ∀ y : 9 − ( a − y ) ( b + y ) ≥ 0 and a + b = ± 6
⟺ ∀ y : y 2 + ( b − a ) y + 9 − a b ≥ 0 and a + b = ± 6
⟺ ( b − 1 ) 2 − 4 ( 9 − a b ) ≤ 0 and a + b = ± 6
⟺ a 2 + 2 a b + b 2 − 3 6 ≤ 0 and a + b = ± 6
⟺ ( a + b ) 2 ≤ 3 6 and a + b = ± 6
⟺ ∣ a + b ∣ ≤ 6 and a + b = ± 6
⟺ ∣ a + b ∣ < 6
Second we are going to prove that the range of the function h ( x ) = 1 0 1 g ( x ) is R , and then this would imply that the range of g ( x ) is also R . Actually, you can notice that h ( x ) = x 2 − 1 1 0 1 ( a + b ) x 2 + 6 x − 1 0 5 4 and ∣ 1 0 1 ( a + b ) − 1 0 5 4 ∣ ≤ 1 0 1 ∣ a + b ∣ + 1 0 5 4 < 1 0 6 + 1 0 5 4 = 6 . According to the first part of our proof, the range of h ( x ) is R , and this would complete our proof. So the range of g ( x ) is R .