We are required to determine all continuous real valued functions $f$ such that $$f(f(x))=-x$$  I’ve determined that if such a function exists, it must be bijective. But I don’t know if such a function exists, let alone find all such functions. Any hints and suggestions will be appreciated.
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I've been perusing the internet looking for interesting problems to solve. I found the following problem and have been going at it for the past 30 minutes with no success:     Find a function $f: \mathbb{R} \to \mathbb{R}$ satisfying $f(f(x)) = -x$ for all $x \in \mathbb{R}$.   I am also wondering, can we find $f$ so that is continuous?  I was thinking of letting $f$ be a periodic function, and adding half the period to x each time. I had no success with this, and am now thinking that such a function does not exist.  Source:
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exercise: Suppose that $f:(0,∞) → R$  satisfies $f(x) - f(y) = f(x/y)$  for all $x,y$ in $(0, ∞)$ and $f(1) = 0$.   Prove f is continuous on $(0, ∞)$ if and only if f is continuous at $1$.    (proof→:) Suppose that $f:(0,∞) → R$ satisfies $f(x) - f(y) = f(x/y)$ for all $x,y$ in $(0, ∞)$ and $f(1) = 0$. Suppose f is continuous on (0,∞), then by definition f is said to be continuous at a point $a$ if for every $ε &gt; 0$  there is a $δ &gt; 0$ such that $|x-a| &lt; δ$ implies $|f(x) - f(a)| &lt; ε$.   Then suppose f is continuous at $1$, then by our defined function  f(x) - f(y) = f(x/y) , we get $|x-1| &lt; δ$ implies $|f(x) - f(1)| = |f(x) - 0| = |f(x)| =  &lt; ε$.   (proof ←) :Suppose that $f$ is continuous at $1$, then we need to show $f$ is continuous on every point $a$ on $(0, ∞)$. Thus, by definition, f is said to be continuous at a point $a$ if for every $ε &gt; 0$ there is a $δ &gt; 0$ such that $|x-a| &lt; δ$ implies $|f(x) - f(a)| &lt; ε$. Thus, $|x-a| &lt; δ → |f(x) - f(a)| = |f(x/a)| &lt; ε$. Then $f$ is continuous on on $(0, ∞)$.  Please can anyone please help me verify this makes sense. If not, can someone please give any feedback/hints. Thank you in advance.
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Does $f(f(f(x)))=x \implies  f(x)=x$? Is it necessary for it to follow? How do we prove this? Do we have to substitute some special $x$? Or is it some other consideration of its properties?  Edit: does it follow in the special case when f is surjective? Or would we wind up with the same thing?
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I want to find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all $x,y\in\mathbb{R}$: $$f(x^2+y+f(y))=2y+f(x)^2.$$  Suppose we have given a solution $f$. Then $f$ is surjective (set $x=0$). Moreover, I have shown that $f(-x)^2=f(x)^2$ for all $x\in\mathbb{R}$ as well as $f(x)=0\Rightarrow x=0$. But I could not make any progress.  I tried for example $y:=-f(x)^2/2$. Then we get $x^2-f(x)^2/2+f(-f(x)^2/2))=0$. Can someone tell me how to proceed?  Best wishes
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Prove that $[0,1]$ isn't homeomorphic to $\mathbb{R}$  My first thought is that there can not be a continuous bijection, $f$, from $[0,1]$ to $\mathbb{R}$ because a continuous function that maps $[0,1]\rightarrow \mathbb{R}$ must be bounded so there can't be a surjection. So that would then be the proof.   Though I believe I am incorrect in my thinking because a hint for the assignment says to use the intermediate value theorem. That is "Suppose $f:[a,b]\rightarrow\mathbb{R}$ is continuous. If $f(a)&lt;\delta&lt;f(b)$ or $f(b)&lt;\delta&lt;f(a)$ then $\delta=f(c)$ for some $c\in[a,b]$".   Why is my first thought wrong and how is the IVT useful?
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Prove that the function : $f(x) = x$ when $x$ is rational and $f(x) = 1 - x$ when $x$ is irrational is continuous at $x = \frac{1}{2}$  To prove the continuity it is necessary to prove that for every $\epsilon$ there exists a $\delta$ such that $|f(x) - f(\frac{1}{2})| &lt; \epsilon$ whenever $|x -c| &lt; \delta$ $- (1)$ Now if we choose $a  € (\frac{1}{2} - \delta, \frac{1}{2} + \delta)$ and also $a $ is rational and choose $\epsilon &lt; |a - \frac{1}{2}|$ then condition in $(1)$ is not satisfied. How to move from here?
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Is there a function $f\colon \mathbb{R} \to\mathbb{R} $ such that $ f(f(x)) = -x$ ?
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One of my friend gave the following problem to me but I am unable to solve it even after trying it for nearly two days.     Problem. Let $f:[0,1]\to\mathbb{R}$ be a continuous function such that $f(0)=f(1)=0$. If $f$ is differentiable on $(0,1)$ then prove that there exists $c\in(0,1)$ such that $f'(c)=f(c)$.   Can anyone help me solving this problem?
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Let $f$ be a real valued and continuous function over $[a,b]$ with $a &lt;b$ such that $f(a)=f(b)$ , My question here is :     Question (claim): How to show that : there  exists  $c \in [a, \frac{a+b}{2} ]$ such that : $f(c)=f(c+\frac{b-a}{2})$.   Note: I have tried intermediate values but I didn't succeed.
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Let $f: A \rightarrow B$ be a function. How can we show that for all subsets $S$ of $A$, $S \subseteq f^{-1}(f(S))$? I think this is a pretty simple problem but I'm new to this so I'm confused.  Also, how can we show that $S = f^{-1}(f(S))$ for all subsets $S$ iff $f$ is injective?
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So I had an exam today and one of the questions were: Give an example of a function $f$ which is nowhere continuous but $|f|$ should be continuous at all points. At first I had no idea how to do it then I came up with this, even though I know it's wrong:  $$f(x)= \begin {cases} \ \sqrt{x}&amp;\text{if }x &lt;0\\ \sqrt{-x}\ &amp;\text{if }x &gt; 0\\ \end {cases}$$  Like I said I know it's wrong..but can someone give me an example of this? Thanks.
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Prove that if $f:X\to Y$ is a surjective function between sets, then there must exist a function $g:Y\rightarrow X$ such that $f\circ g=1_Y$.   I know that the identity function is onto, and if $f$ has right inverse, then $f$ must be onto; although, I haven't seen a proof of this. A good hint would be nice.
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Determine all $a\in[0,1]$ such that for ${\it every}$ continuous function $f:[0,1]\to \mathbb{R}$ with $f(0)=f(1)$ there exists at least one $x$ where $f(x) = f(x+a)$.   First of all, $a=0,1/2,1$ are obviously good values. Secondly, no $1/2&lt;a&lt;1$ works (look at $f = \sin(2\pi x)$).  I remember trying to solve this several years ago. I think I started by dividing $[0,1]$ into regions where $f\geq 0$ and $f&lt;0$. Then studied each interval and all pairs of intervals with the same sign. However, this seems like a problem that should have a simple solution and I would love to see it.
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Find all (if that's possible) functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$f(f(x))=\sin x.$$  This is a problem a friend gave me. He said he saw it on a book a long time ago, but he couldn't remember what book. I literally have no idea how to solve this so I can't post my attempt here.
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If $f:\mathbb{R} \to \mathbb{R}$ is a continuous function such that $f(x)=x$ has no real solution, then show that $f(f(x))=x$ has no real solution either.   Is the proof trivial as it seems or does it need an analytical approach?
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I am wondering whether or not there is a reasonable characterization of differentiable functions $f: \mathbb{R}\to \mathbb{R}$ such that $f'(f(x))=f(f'(x))$ for each $x\in\mathbb{R}$. (Or, if you like the composition sign, $f'\circ f=f\circ f'$).   I could only come up with trivial examples of such functions: $f(x)=0$ and $f(x)=e^{x}$.   This reminds me of a recent Putnam problem (2010), which asked whether or not there exists a strictly increasing function $f:\mathbb{R}\to\mathbb{R}$ satisfying $f'(x)=f(f(x))$. (The answer is: No).    Note: I see that a question of similar type has been asked .
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Question is to prove :  $f : S^1 \to \mathbb R$ is continuous then $f(x)=f(-x)$ for some $x\in S^1$  I guess it would be helpful to use intermediate value theorem  Assuming $f(x)\neq f(-x)$ then given any $p\in (f(-x),f(x))$ (assuming $f(-x)&lt;f(x)$) there exists $y\in (-x, x)$ such that $f(y)=p$  I am not very sure of how to use this..   It would be helpful if someone can give some hint which would help me to solve this..  Thank you.