# How to find the inverse to $f(x)= x^2 - 6x + 11$

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
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 $f^{-1}$. – Berrick Caleb Fillmore Oct 3 '14 at 4:30

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
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

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.

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