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$$ y=f(x)=\frac{ax^2+bx+c}{dx^2+ex+f} $$ We have to find the conditions for this it takes all real values.

MY solution

One approach is to equate it to y and for a quadratic of x and put discriminant greater than equal to 0..That is very lengthy.Is their any better method..?

Inceptio
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maths lover
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  • You mean the range of $f(x)$ is $\mathbb{R}$, right? – newbie Jul 06 '13 at 13:29
  • ya range is real . – maths lover Jul 06 '13 at 13:31
  • I can't think of a better way to get range of y – evil999man Mar 06 '14 at 07:50
  • I can give an easy example of a function with this property from my experience tutoring, but I can't generalize. Maybe someone else will see how to generalize a function like $f(x)=\frac {x}{x^2-1}$ which should have this property. – A. Thomas Yerger Feb 09 '15 at 05:58
  • Actually on second thought I think I do see why that function works - it is continuous and decreasing between its two infinite discontinuities. At $x=1$ the function is undefined, but approaches positive infinity from the right. It decreases over the interval $(-1,1)$, where it approach negative infinity from the left. So the range includes all real numbers. – A. Thomas Yerger Feb 09 '15 at 06:02
  • Hint: If $\lim_{x\rightarrow-\infty} f(x)=-\infty$ and $\lim_{x\rightarrow\infty} f(x)=\infty$ and $dx^2+ex+f \neq 0$ then the range of $f(x)$ will be $\mathbb{R}$. Also you can change the signs of $\infty$ in the above conditions. Anyways, $d \neq 0$ is a necessary condition. – MasB Jul 22 '15 at 20:34

3 Answers3

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Assuming $a,d\neq0$, Factor out $a$ and $d$, introducing $b',c',e',f'$ so I don't have to write fractions:

$$y=f(x)=\frac{a}{d}\frac{x^2+b'x+c'}{x^2+e'x+f'}=\frac{a}{d}\left(1+\frac{(b'-e')x+(c'-f')}{x^2+e'x+f'}\right)$$

So what you are looking for is the range of $\frac{(b'-e')x+(c'-f')}{x^2+e'x+f'}$ to be $\mathbb{R}$.

This is impossible if the denominator has no real zeros, because in that case the curve has no vertical asymptotes, is continuous over all of $\mathbb{R}$, and has $y=0$ as a horizontal asymptote. So the denominator has zeros, and we may consider $\frac{(b'-e')x+(c'-f')}{(x-r)(x-s)}$ where maybe $r=s$.

Assume $r\neq s$. If the numerator is not identically $0$, then this function definitely has at least one vertical asymptote of degree $1$, and a horizontal asymptote at $y=0$. There is an issue of whether the function achieves the output of $0$. It does if and only if $b'\neq e'$ and the root of the numerator, $\frac{f'-c'}{b'-e'}$, is neither $r$ nor $s$. Now, if the function achieves the output of $0$, it achieves all outputs when you consider the possible arrangements of the two degree-one vertical asymptotes.

If $r=s$, then there are two cases, and both fail

  • The numerator has a root and it is $r$. But then we are looking at the reciprocal of a linear function and such things never have all of $\mathbb{R}$ as their range.
  • Otherwise, there is a lone vertical asymptote of degree $2$ with both adjacent arms either pointing up or pointing down. Since the horizontal asymptote is still $y=0$, either arbitrarily high positive outputs or arbitrarily high negative outputs are missed.

So in summary, all assuming $a,d\neq0$, you must have $$\begin{align} b'&\neq e'&e'^2&>4f'&\frac{f'-c'}{b'-e'}&\neq r,s\\ \iff b'&\neq e'&e'^2&>4f'&\frac{f'-c'}{b'-e'}&\neq \frac{-e'\pm\sqrt{e'^2-4f'}}{2}\\ \iff \frac{b}{a}&\neq \frac{e}{d}&\frac{e^2}{d^2}&>4\frac{f}{d}&\frac{\frac{f}{d}-\frac{c}{a}}{\frac{b}{a}-\frac{e}{d}}&\neq \frac{-\frac{e}{d}\pm\sqrt{\frac{e^2}{d^2}-4\frac{f}{d}}}{2}\\ \iff bd&\neq ae&e^2&>4df&\frac{af-cd}{bd-ae}&\neq \frac{-e\pm\sqrt{e^2-4df}}{2d}\\ \iff bd&\neq ae&e^2&>4df&2d(af-cd)&\neq (bd-ae)\left[-e\pm\sqrt{e^2-4df}\right] \end{align}$$

I believe the last of these three conditions is equivalent to: $$(af-cd)^2\neq (ae-bd)(bf-ce)$$ I found this by isolating the radical, squaring, and condensing like terms and factoring what was left. It's possible I made an error during all that.

2'5 9'2
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$$f\left( x \right) =\frac { a{ x }^{ 2 }+bx+c }{ d{ x }^{ 2 }+ex+f } =\frac { a{ x }^{ 2 }+bx+c }{ d\left( { x }^{ 2 }+\frac { e }{ d } x+\frac { f }{ d } \right) } =\frac { a{ x }^{ 2 }+bx+c }{ d\left( { x }^{ 2 }+\frac { e }{ d } x+\frac { e^{ 2 } }{ 4{ d }^{ 2 } } -\frac { e^{ 2 } }{ 4{ d }^{ 2 } } +\frac { f }{ d } \right) } =\\ =\frac { a{ x }^{ 2 }+bx+c }{ d{ \left( x+\frac { e }{ 2d } \right) }^{ 2 }+\frac { 4fd-{ e }^{ 2 } }{ 4{ d } } } $$ so range is $$R\setminus \left\{ -\frac { e }{ 2d } \right\} \\ $$

haqnatural
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  • I don't follow this. If $f(x)=\frac{x^2-1}{x^2+1}$, you are saying the range is all real numbers except $0$. But this function's range is only $[-1,1)$. – 2'5 9'2 Dec 05 '15 at 05:17
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Find the numbers that make the bottom zero. You can't divide by zero.

Kiro
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