A question that give wrong answer?

[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 $|f(x)-\frac{a}{-8}| < 1$ for all x > A, and $|f(x)-\frac{a}{-8}| < 1$ 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 $|f(x)| < \max(1+|\frac{a}{-8}|,M)$ for all x in $(-\infty,B) \cup [B,A] \cup (A,\infty) = \mathbb{R}$, and is thus not surjective.

Given, f(x)= (ax² + 6x - 8)/(a + 6x - 8x²). We are asked to find `a` s.t. f is onto (surjective). Set, (Ax² + Bx + C)(a + 6x - 8x²) = ax² + 6x - 8, for {A,B,C} to be determined. Expanding and collecting terms gives:

Aa + 6B - 8C = a
-6A - 8B = 0
A=0 ⇒ B=0
Ba + 6C = 6 ⇒ C=1
Ca = -8 ⇒ a=-8.

There is, in fact, only one possible value of `a`, and the given f cannot take ℝ onto ℝ.

Thus, f(x) ≡ 1. Viz., f: ℝ→{1}⊂ℝ. The lesson to be learned is that the given problem statement is ambiguous, and misleading.

 

[disregardthat]

mathphysics, what you have proved is that in order for f(x) to be a polynomial a must be -8. But the question is whether f(x) is surjective.

 

[matphysik]

NO. What i did was divide the numerator quadratic by the denominator quadratic directly, in order to reveal the true identity of f(x).

ADDENDUM:

If we should assume the simplest non-polynomial in this case, say:

(Ax² + Bx + C + D/x)(a + 6x - 8x²) = ax² + 6x - 8.

Then expanding, and collecting terms gives:

A=0
-8B + 6A=0 ⇒ B=0
-8C + 6B + Aa=a ⇒ C=a/-8
-8D + 6C + Ba=6
6D + Ca=-8
6Da + Ca²=-8a ⇒ C= -8/a
Da=0 ⇒ D=0
-8Da + 6Ca=6a ⇒ C=1 ⇒ a= -8.

 

[Dschumanji]

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 $|f(x)-\frac{a}{-8}| < 1$ for all x > A, and $|f(x)-\frac{a}{-8}| < 1$ 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 $|f(x)| < \max(1+|\frac{a}{-8}|,M)$ for all x in $(-\infty,B) \cup [B,A] \cup (A,\infty) = \mathbb{R}$, and is thus not surjective.

I love this proof.

 

[SteveL27]

NO. What i did was divide the numerator quadratic by the denominator quadratic directly, in order to reveal the true identity of f(x).

And it turns out to be ... wealthy playboy Bruce Wayne!

 

[RandomMystery]

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 $|f(x)-\frac{a}{-8}| < 1$ for all x > A, and $|f(x)-\frac{a}{-8}| < 1$ 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 $|f(x)| < \max(1+|\frac{a}{-8}|,M)$ for all x in $(-\infty,B) \cup [B,A] \cup (A,\infty) = \mathbb{R}$, and is thus not surjective.

Why can't we make a equal some function?
a = -8x
$\frac{a}{-8}=\frac{-8x}{-8}$
Here the function is globally onto.

I don't understand why you choose y=0 in the discriminant:
Delta = 6^2-4(-8)a
instead of
Delta = (6 - 6y)^2 - 4(a + 8y)(-8 - ay)

I think you can still use the quadratic equation:

We can either replace a for one function that solves the dilemma, or we could make a into a piece wise value (or function).
The question says find the value of a, well the value of a could be different for each x

 

[Dschumanji]

Why can't we make a equal some function?
a = -8x
$\frac{a}{-8}=\frac{-8x}{-8}$
Here the function is globally onto.

I don't think there is anything stopping someone from making a into a function. However, the OP seems to have made it clear later on that a does not depend on x.

I don't understand why you choose y=0 in the discriminant:
Delta = 6^2-4(-8)a
instead of
Delta = (6 - 6y)^2 - 4(a + 8y)(-8 - ay)

The OP states that the domain of f is R. For this to be true, f must not contain points of discontinuity. We can see that f will have such points of discontinuity whenever the denominator is equal to zero. The only way to prevent the denominator from equaling zero is to choose a value of a such that quadratic has no real roots. Such an a must make the discriminant less than zero.

I think your confusion is coming from the fact that DisregardThat is not using the method suggested by HallsOfIvy.