# For $f:\mathbb R^{<0}\to\mathbb R$, $f(x)=2x^2-3$, find the values of a for which $f(a)=f^{-1}(a)$

Okay, i've got the answer for this with some luck I guess, however i'm still left wondering specifically what this part of the question means: "find the values of a for which $f(a)=f^{-1}(a)$"

My understanding of this is, that the question is asking me to find a value of a where the output and input given by the function $f$ are equal?

Could someone do a better job of explaining this to me please, thank you.

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And $f(f(a))= a.$ –  Will Jagy Oct 7 '13 at 22:28
The inverse function of $f$ is not well defined, because $f$ is not one-to-one. –  egreg Oct 7 '13 at 22:29
This question is a little bit silly, since $f^{-1}(a)$ is a set rather than a value (e.g. $f^{-1}(-1)=\{-1,1\})$. The best you can do is what everyone's suggesting. –  Ian Coley Oct 7 '13 at 22:29
Ah sorry, the domain is x<0 for all reals –  seeker Oct 7 '13 at 22:29
@Assad $f(a)=f^{-1}(a)$ is equivalent, since $f$ is one-to-one, to $f(f(a))=f(f^{-1}(a)$, assuming $f(a)$ belongs to the domain of $f$. But $f(f^{-1}(a))=a$ by definition (assuming $a$ belongs to the set of values of $f$). –  egreg Oct 8 '13 at 15:10

The values for which $f(x)=f^{-1}(x)$ must lie on the line $y=x$, since by the sheer definition of inverse function this is an axis of symmetry. This means that if a function and its inverse intersect, the points of intersection must lie on that line. Hence, you have to solve for which values of $x$ we have $f(x)=x$. So $2x^2-3=x$ and this leads to $x=-1$ or $x=1\frac{1}{2}$. Since your domain is the negative reals, you are left with $x=-1$.

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This is only a sufficient condition; also $x=-(1+\sqrt{21})/4$ is a solution to the OP's problem. –  egreg Oct 8 '13 at 8:07
I don't think so ... –  Nicky Hekster Oct 8 '13 at 8:36
@NickyHekster, your answer is the right one. I dont understand what the others are telling me to do, why does $f(f(a))=a$? Surely it's ${f^{ - 1}}(f(a)) = a$? –  seeker Oct 8 '13 at 12:29
Assad if you deem my answer the right one please tick it as such. Thanks! –  Nicky Hekster Oct 8 '13 at 16:09
BTW: it is correct that if $f^{-1}(a)=f(a)$, then $f(f^{-1}(a))=f(f(a))$ and hence $id(a)=a=f^2(a)$ (where the 2 does not denote squaring, but abbreviates $f \circ f$). But this way of solving the problem introduces more "roots" in the fourth degree equation you worked out. –  Nicky Hekster Oct 8 '13 at 16:21

Hint: $$f(a) = f^{-1}(a) \Longrightarrow f(f(a))=a.$$

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Could you explain why this is? –  seeker Oct 7 '13 at 22:57
@Assad By definition of inverse function. –  egreg Oct 7 '13 at 22:57
I dont get it... –  seeker Oct 7 '13 at 23:02
If I do f(f(a))=a, I get this: \eqalign{ & f(x) = 2{x^2} - 3 \cr & f(a) = 2{a^2} - 3 \cr & f(f(a)) = a: \cr & a = 2{(2{a^2} - 3)^2} - 3 \cr & a = 2(4{a^4} - 12{a^2} + 9) - 3 \cr & a = 8{a^4} - 24{a^2} + 15 \cr & 8{a^4} - 24{a^2} - a + 15 = 0 \cr} How on earth do I solve that? –  seeker Oct 7 '13 at 23:31
@Assad You can apply rational root theorem to decrease the degree of the degree of the polynomial, otherwise this is quartic equation and can be "easily" solved using Ferrari's method. –  Stefan4024 Oct 8 '13 at 1:09

You are looking for periodic points of period 2.

First find $f(f(a))$. Then set it equal to $a$ and solve for $a$.

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