Finding the Inverse of a Quadratic Function?

cyh96 · **Joined:** Wed, 13 Jun 2012 09:29:08 UTC **Posts:** 2

Hi, need some help in the aspect of finding the inverse of a quadratic function. I know how to solve simple ones, but what happens when the quadratic function is in the form of a fraction

f(x) = (x^2 - x + 1) / (x^2 + x + 1)

So far I have only gotten this far:
y = (x^2 - x + 1) / (x^2 + x + 1)
yx^2 + yx + y = x^2 - x + 1
(y-1)x^2 + (y+1)x + (y-1) = 0
(y-1)(x^2 +1) + (y+1)x = 0

Thanks in advance!

aswoods · **Posted:** Wed, 13 Jun 2012 12:13:31 UTC

cyh96 wrote:

Hi, need some help in the aspect of finding the inverse of a quadratic function. I know how to solve simple ones, but what happens when the quadratic function is in the form of a fraction

f(x) = (x^2 - x + 1) / (x^2 + x + 1)

So far I have only gotten this far:
y = (x^2 - x + 1) / (x^2 + x + 1)
yx^2 + yx + y = x^2 - x + 1
(y-1)x^2 + (y+1)x + (y-1) = 0
(y-1)(x^2 +1) + (y+1)x = 0

Thanks in advance!

Notice that because f(x) is not a one-to-one function, there are zero, one, or two solutions for a given y, so there is no single inverse function.

$$ \begin{align*} y&=\frac{x^2 - x + 1}{x^2 + x + 1}\\ &=\frac{(x^2 + x + 1)-2x}{x^2 + x + 1}\\ &=1-\frac{2x}{x^2 + x + 1}\\1-y&=\frac{2x}{x^2 + x + 1}\\ \frac{2x}{1-y}&=x^2+x+1\\ 0 &= x^2+(1-\tfrac{2}{1-y})x+1 \end{align*}$

This is a quadratic equation in x, and now you can apply the quadratic formula.

cyh96 · **Joined:** Wed, 13 Jun 2012 09:29:08 UTC **Posts:** 2

So is it safe to say that there is no inverse function for all quadratic eqns?
Thanks

outermeasure · **Posted:** Wed, 13 Jun 2012 15:42:40 UTC

cyh96 wrote:

So is it safe to say that there is no inverse function for all quadratic eqns?
Thanks

No.

Soroban · **Posted:** Wed, 13 Jun 2012 15:50:22 UTC

Hello, cyh96!

Quote:

$f(x) \:=\:\dfrac{x^2 - x + 1}{x^2 + x + 1}$

So far I have only gotten this far:

. . $y \:=\: \dfrac{x^2 - x + 1}{x^2 + x + 1}$

. . $yx^2 + yx + y \:=\: x^2 - x + 1$

. . $(y-1)x^2 + (y+1)x + (y-1) \:=\: 0$

. . $(y-1)(x^2 +1) + (y+1)x \:=\: 0$ . . . . keep going

. . $(y-1)x^2 + (y-1) + (y+1)x \:=\:0$

. . $(y-1)x^2 + (y+1)x + (y-1) \:=\:0$ . . . a quadratic in

Quadratic Formula: . $x \;=\;\dfrac{-(y+1) \pm\sqrt{(y+1)^2 - 4(y-1)^2}}{2(y-1)}$

. . . . . . . . . . . . . . $x\;=\;\dfrac{-(y+1) \pm \sqrt{-3y^2 + 10y -3}}{2(y-1)}$

. . . . . . . . . . . . . . $x\;=\;\dfrac{-(y+1) \pm \sqrt{(3y-1)(3-y)}}{2(y-1)}$

. . . . . . . . . . . . . . . . $\text{where }\,\frac{1}{3} \le y \le 3\,\text{ and }\,y\,\ne\,1$

Shadow · **Posted:** Wed, 13 Jun 2012 15:57:58 UTC

This function is NOT 1-1, so it does not have an inverse. Take for example the image point 0, what is $f^{-1}(0)$ is it ${1\over 2}+{1\over 2}\sqrt{5}$ or is it ${1\over 2}-{1\over 2}\sqrt{5}$ ?

Shadow · **Posted:** Wed, 13 Jun 2012 16:01:36 UTC

cyh96 wrote:

So is it safe to say that there is no inverse function for all quadratic eqns?
Thanks

What do you mean by "inverse function for an equation"?

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Finding the Inverse of a Quadratic Function?

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