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Suppose '$f$' is a continuous function from $\mathbb{R}$ to $\mathbb{R}$ and $f(f(a))=a$ for some $a \in \mathbb{R}$ then find the number of solutions of the equation $f(x)=x$.

Options given:

(a) no solution

(b) exactly one solution

(c) at most one solution

(d) atleast three solutions

I tried to solve it like this

$$f(f(a))=a$$ $$f(a)=f^{-1}(a)$$

So the value of function is equal to its inverse at $x=a$.

I tried thinking about that function but it will become a special case.

Is there a mathematically rigorous and simple way to tackle this problem?

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  • $\begingroup$ Is $f$ given in your situation? $\endgroup$ – wythagoras Jun 7 '15 at 11:42
  • $\begingroup$ No, it's not given. @Wythagoras $\endgroup$ – me_ravi_ Jun 7 '15 at 11:44
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You can say there is at least one solution to $f(x)=x$. There may be more, but there is at least one.

Here is a proof. If $f(a)=a$ then $a$ is a solution and we are done. Assume for now that $f(a)<a$.

Consider the function $g(x)=f(x)-x$. Note that $g$ is continuous since $f$ is continuous. Then

$$g(a)=f(a)-a<0$$

and

$$g(f(a))=f(f(a))-f(a)=a-f(a)>0$$

Hence, by the Intermediate Value Theorem for continuous functions, $g(x)$ has a zero between $f(a)$ and $a$. Let's call that zero $c$. Then

$$0=g(c)=f(c)-c$$

so $f(c)=c$ and we have found a solution.

You can repeat this argument if $f(a)>a$. That covers all possibilities regarding $f(a)$ and $a$, so there is at least one solution to $f(x)=x$.


In the edited version of your question, which gives four multiple choice possible answers, none of those choices is correct. An example where there is exactly one solution is $f(x)=-x$, which disproves choices (a) and (d). An example where there are infinitely many solutions is $f(x)=x$. This disproves choices (b) and (c).

Are you sure there was no choice "at least one solution"?

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  • $\begingroup$ The answer given in my book is (d) atleast three solutions. @RoryDaulton $\endgroup$ – me_ravi_ Jun 7 '15 at 11:57
  • $\begingroup$ @Integrator: Then your book is wrong, as the example $f(x)=-x$ shows. $\endgroup$ – Rory Daulton Jun 7 '15 at 11:59
  • $\begingroup$ Thanks @RoryDaulton I think the author overlooked the fact that given question ( with given options ) does not suffice for $f(x)=-x$. This example is great. $\endgroup$ – me_ravi_ Jun 7 '15 at 12:42

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