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My question is ; How can I solve the following inverse of function question?

$y=x^2-6x+4$

Thanks in advance,

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Did you ever learn the discriminant method? Note that $x^2-6x+9=(x-3)^2$ –  Raskolnikov Jan 13 '11 at 20:27
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Also called "completing the square" I think. –  Noldorin Jan 13 '11 at 20:31
    
In light of Isaac's point that your question could be asking for the relation that is the inverse to the relation defined by your function, could you please provide more context indicating precisely what is being asked for? That is, could you please specify whether or not the inverse is supposed to be a function, or provide the source of the question for readers to help determine the context? –  Jonas Meyer Jan 13 '11 at 21:26

5 Answers 5

up vote 7 down vote accepted

This function does not have an inverse, since \begin{align*} y\left(3+\sqrt{5}\right) &= (3+\sqrt{5})^2 - 6(3+\sqrt{5}) + 4\\ &= 9 + 6\sqrt{5} + 5 -18 - 6\sqrt{5} + 4 = 0,\\ y\left(3-\sqrt{5}\right) &= (3-\sqrt{5})^2 - 6(3-\sqrt{5}) + 4\\ &= 9 - 6\sqrt{5} + 5 - 18 + 6\sqrt{5} + 4 = 0. \end{align*} Since the function is not one to one, it does not have an inverse.

You can, on the other hand, restrict the domain; for instance, since $$ y = x^2 -6x + 4 = (x-3)^2 - 5$$ you can restrict yourself to $[3,\infty)$, where the function is one-to-one. In that case, note that $y\geq -5$, and you have \begin{align*} y &= x^2 - 6x + 4\\ x^2 - 6x + (4-y) &= 0 \end{align*} and now you can use the quadratic formula to find an expression for $x$ in terms of $y$. The function will have domain $[-5,\infty)$, and will have range $[3,\infty)$.

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"This function does not have an inverse"—that depends entirely on how one defines "inverse." There are a number of texts that define inverse at the level of relations so that all functions (which are special types of relations) have inverses (which are relations), though it is not guaranteed that the inverse relation of a function is also a function. –  Isaac Jan 13 '11 at 21:10
    
@Isaac: In my experience, such books are not at this level; but even those that do make that definition usually distinguish the two cases by explicitly using "inverse relation" when the inverse is not a function. –  Arturo Magidin Jan 13 '11 at 21:16
    
UCSMP, which is a widely-used k-12 textbook series, uses "inverse" of a function to refer to the relation and specifies "inverse function" when discussing whether or not the inverse relation is a function. I'm pretty sure it's not the only pre-college-level book to do so, but I don't have my full library at hand to check. –  Isaac Jan 13 '11 at 21:19
    
@Isaac: That is a good point, but I also took it to mean "function that is an inverse", given that the question is asking for the inverse of a function (which is pretty close to asking for an inverse function) without saying anything about relations. I think that more context of the problem would be needed to be 100% sure about what is being asked. In the precalculus level book I have at hand, this language is not used to mean inverse relation of a function, but on the other hand the authors are careful to refer to finding "the inverse function", not just the inverse of a function. –  Jonas Meyer Jan 13 '11 at 21:20
    
@Isaac: I confess to being wholly unfamiliar with K-12 textbooks; that said, this alone would not recommend UCSMP to me at that level. Smacks too much of new math for my taste (I say that as a "survivor" of new math). –  Arturo Magidin Jan 13 '11 at 21:29

First you have to be careful, because without restricting your domain, the function is not invertible. For example $2^2-6\cdot 2+4 = -4 = 4^2 - 6\cdot 4 +4$, which shows that $x\mapsto x^2-6x+4$ is not one-to-one. If you restrict the domain to $x\geq3$ or $x\leq3$, then the resulting function will be one-to-one. You can see this by finding the vertex of the parabola $y=x^2-6x+4=(x-3)^2-5$, and noticing that the function is strictly monotone (increasing or decreasing) on either side of the vertex.

If you define $f(x)=x^2-6x+4$ on $[3,\infty)$, then you can find the inverse function as indicated in the other answers. The domain will be the range of the previous function $[-5,\infty)$, and for each $x\in[-5,\infty)$, $f^{-1}(x)$ is the unique solution $y$ to $x=y^2-6y+4$ in the interval $[3,\infty)$, which can be found by completing the square or using the quadratic formula. Something similar can be said for the function $g(x)=x^2-6x+4$ on $(-\infty,3]$, where $g^{-1}$ will now have a different range than $f^{-1}$.

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My answer overlaps with Arturo Magidin's, which was posted 27 seconds earlier. Hopefully it will be useful (rather than overly redundant) to leave my complementary answer. –  Jonas Meyer Jan 13 '11 at 20:39

If your function is $f(x)=x^2-6x+4$, then its inverse (which is not a function) can be found by using the quadratic formula to find the solutions to the equation $0=x^2-6x+4-f(x)$ (which will give a formula for $x$, which becomes $f^{-1}(x)$, in terms of $f(x)$, which becomes $x$).

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Write $x = y^2-6y+4$ and solve for $y$. You need to restrict the domain and codomain so that the function is bijective (i.e. the inverse exists).

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Swapping the variables to start is slightly confusing, because you're not maintaining consistency. –  Noldorin Jan 13 '11 at 20:37

Simply rearrange the equation to solve for $x$ in terms of $y$. This is precisely the definition of the inverse.

Hint: factorise the quadratic expression.

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