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If the inverse exists, how do I find the inverse to this function:

$$ f(x)= x^2 - 6x + 11 $$

with $x \le 3$

Stuck at the quadtric formula. I think i have got the right answer which is $x = 3 ± \sqrt{y-2}$ ? But it doesnt seem right.

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Have you tried the quadratic formula? –  EuYu Oct 17 '12 at 15:35
    
Why doesn't the answer (which is correct, by the way) seem right? –  Rick Decker Oct 17 '12 at 16:05

2 Answers 2

If the inverse exists, you just write $y=x^2-6x+11$ and use the quadratic formula to get $x$ in terms of $y$. To see if it exists, you need to ensure that for a given $y$ there is only one $x$. The obvious threat is the $\pm$ sign in the quadratic formula.

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I have tried it. But Im stuck at the quadtric formula. I think i have got the right answer which is x = 3 ± squareroot (y-2) ? But it doesnt seem right. –  Karoline Oct 17 '12 at 15:39
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Karoline, your efforts (including thought process and the answer which you got) would be a welcome addition to the body of the question. Feel free to edit your original question and add anything that you feel would help those who are trying to help you. –  The Chaz 2.0 Oct 17 '12 at 15:41
    
@Karoline: you have used the quadratic formula correctly. What doesn't seem right about it? If you plot the graph, the inverse function consists of interchanging the $x$ and $y$ axes. This corresponds to turning the paper over along the diagonal from lower left to upper right and looking through the paper. Then you can apply the vertical line test. –  Ross Millikan Oct 17 '12 at 15:46
    
@TheChaz: to notify a user (at least as I understand it) you need to precede the username with an @ sign. Probably a typo and you get this, but if not it will help. –  Ross Millikan Oct 18 '12 at 5:20
    
I was under the impression that the author of the question is notified whenever there is a comment anywhere on the page. Maybe I'm wrong! Just in case... @Karoline, see above :) –  The Chaz 2.0 Oct 18 '12 at 5:36

You don't need the quadratic formula, just complete the square! You start with $$ y = x^2 - 6x + 11 $$ or in other words $$ x^2 - 6x + 11 - y = 0 $$ Now your goal is to write that as $(x + a)^2 + \ldots = 0$ for some $a$. Observe that by expanding that square you get $x^2 + 2ax + \ldots$. Matching that to your original equation shows that you have to pick $a=-3$. That produces the correct coefficients for $x^2$ and $x$, so all you need to do is correct for the differing constant term. You get $$ (x - 3)^2 + 2 - y = 0 $$ which via simple algebra yields $$ x = 3 \pm \sqrt{y - 2} $$

Note that this always works! If the coefficient of $x^2$ in your equation isn't $1$, just divide the whole equation by the coefficient before you start. Once you've praticed this square completion a few times, you'll be at least as fast as with the formula, and you won't have to remember the formula anymore.

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In fact the quadratic formula just automates completing the square. It is another route to the same answer, which to me takes less thought. It is a matter of taste. –  Ross Millikan Oct 17 '12 at 16:40
    
Thanks! This was very helpful. –  Karoline Oct 18 '12 at 4:56

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