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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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    Notice that $ f(x) = (x - 3)^{2} + 2 $ for all $ x \in \mathbb{R} $. Hence, $ f(2) = f(4) $, so $ f: \mathbb{R} \to \mathbb{R} $ is not $ 1 $-$ 1 $. Also, $ f(x) \geq 2 $ for all $ x \in \mathbb{R} $, so $ f: \mathbb{R} \to \mathbb{R} $ is not onto. If, however, we restrict the domain and co-domain of $ f $ appropriately, then an inverse exists. Consider $ f: [3,\infty) \to [2,\infty) $. Then $ f $ is both $ 1 $-$ 1 $ and onto, which yields the existence of $ f^{-1} $.2014-10-03

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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 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 :)2012-10-18
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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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    Thanks! This was very helpful.2012-10-18
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A trick I use is to switch x with y and y with x. Then isolate y on one side. Then do the math! (Look on the other posts).

Also, some functions (or relations) don't have inverses. They only have an inverse if they pass the vertical line test.