Be a continuous function $f: \mathbb{R} \to \mathbb{R}$ such that $f(x)f(f(x)) = 1$ and $f(2020) = 2019$. What is the value of $f(2018)$?  I am having a problem with this question. I already tried, through various attempts, to find the function explicitly which satisfy this condition and $f(0)$ different from $0$. But it leads me to nothing. I am really thinking there is no way to find an explicit function, so I would appreciate any solution to this problem (normally I would prefer a tip, but my ideas are exhausted).
123240
If $f$ is continuous real function such that $f(11) = 10$ and $\forall x, f(x)f(f(x)) = 1$, then $f(9) =$ ?
58579
Let  $f : \mathbb{R}^2\rightarrow \mathbb{R}$ be a continuous map such that $f(x)=0$ only for finitely many values of $x$. Which of the following is true?   Either $f(x)\leq 0$ for all $x$ or $f(x)\geq 0$ for all $x$. the map $f$ is onto. the map $f$ is one one. None of the above.   What I have done so far is :   I would take polynomial in two variables.. This need not be like $f(x)\geq 0$ for all $x$ or $f(x)\geq 0$ for all $x$.So,first option is eliminated. The map is not one one assuming that $f$ has more than one zero. So, third option is wrong.   I could not think of an example in which it is not onto..  Only examples i am getting in my mind are polynomials and they are onto..  So, I am having trouble with surjectiveness of the function.  Please help me to clear this.  Thank you..
52466
Let $f:[0,1]\to\mathbb{R}$ be continuous and $f(0)=f(1)$. Show that for any integer $n\geqslant 2$, there is $\xi\in(0,1)$ s.t.$f(\xi)=f\left(\xi+\frac{1}{n}\right).$  I think this requires the intermediate value theorem somewhere.
28546
Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous. Is it possible for $f$ to assume each value in its range an even number of times?  To clarify, some values might be taken 0 times, some 2, some 4, etc., but always an even (and therefore finite) number. I don't require that there be a value that is assumed any particular number of times.  For example, the function might be surjective, or never take on any value exactly twice.  This question is possibly the same as . That question may instead have meant, "If $n$ is even, is it possible for $f$ to assume every value in its range exactly $n$ times?"  In any event, it has not been answered.
101332
Suppose $f:[0,\infty)$ $\rightarrow$ R and has continuous $f'(x)$ and $f''(x) $ .  $f(x) $$\rightarrow $0 as $x$ $\rightarrow$ $\infty$. Given $f'(x)\rightarrow b$ as $x \rightarrow \infty$  show that $b=0$  Get stuck in the first place, but I tried to use Mean Value Theorem and then connected it to the limits.  Any hint?
21472
Let $f:[0,1] \to \mathbb{R}$ be a continuous function such that $f(x)=f(x^2)$ for all $x \in [0,1]$.  Any hint/idea for proving that $f$ has to be constant?
97011
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function with $f(1) = 5$ and $f(3) = 11$. If $g(x) = \int_1^3 f(x+t)dt$ then find $g'(0).$   Please explain your answer.
88187
Let $$f\colon [0,1]\rightarrow [0, \infty )$$  be  continuous. Suppose that  $$\int_{0}^{x} f(t)\,\mathrm dt \ge {f(x)} \quad\text{for  all }x\in[0,1].$$  Then   $A.$ No  such  function  exists.  $B.$ There  are  infinitely  many  such  functions.  $C.$ There is  only  one  such  function  $D.$ There  are  exactly  two  functions.  Now  I  was  thinking,  since $\int_{0}^{x} f(t) dt$  is  the  area  under  the  graph  of  $f(t)$  from $0$  to  $x$ so it  can  be  easily  made  greater  than  the  value  of  $f(t)$  at  one  point, namely  $x$.  So  there   will  be  infinitely  many  such  functions  satisfying  this  condition.   Is  my  reasoning  correct or  not?  Thanks.
27606
If $f$ is continuous in [0,1] then what will be the value of $$\lim_{n\to\infty}\sum_{j=0}^{\left[\frac{n}{2}\right]}\frac{1}{n}f\left(\frac{j}{n}\right)?$$  We know that,$$\lim_{n\to\infty}\sum_{j=0}^{{n}}\frac{1}{n}f\left(\frac{j}{n}\right)=\int_0^1f(x)dx$$But what will be the value if we have $\left[\frac{n}{2}\right]$ instead of $n?$
30995
If $f:[0,1]\to\mathbb R$ is continuous and $\int_0^x f(t)dt = ∫_x^1f(t)dt$ then prove that $f(x)=0$.  I suppose that $f(x)$ isn't zero, set $g(x)= \int_y^xf(t)dt$ with $y\in[0,1]$ and use Bolzano but I don't know if it's right.
87970
Given a continuous function $f:\mathbb{Q}\to\mathbb{Q}$ ,does there exist a continuous function  $g:\mathbb{R}\to\mathbb{R}$ such that   $g|_{\Bbb Q} = f$?  What I have no Idea about how to attempt this Question! Any suggestion will be very helpful.
59440
Let $f:[0,1] \rightarrow \mathbb{R}$ be a continuous function. If $f(0)=f(1)$ then exists $x\in [0,1]$ s.t. $f(x) = f(x+\frac{1}{2})$   We have that $f\left(\frac{1}{2}\right)=f(0)$ or $f\left(\frac{1}{2}\right)&gt;f(0)$ or $f\left(\frac{1}{2}\right)&lt;f(0)$.  If $f\left(\frac{1}{2}\right)=f(0)$ then $x = 0$ Solves the problem.  if $f\left(\frac{1}{2}\right)&lt;f(0)$ then by the intermediate value theorem  $$\exists k \in \left[0,\frac{1}{2}\right], f(0)&gt;f(k)&gt;f\left(\frac{1}{2}\right) \implies$$  $$\implies f(1)&gt;f(k)&gt;f\left(\frac{1}{2}\right)\implies \exists t \in \left[\frac{1}{2},1\right] s.t. f(t)=f(k).$$   Now how can I show that exists at least one value of k such that $t = k+\frac{1}{2}$? (The solution is similar if $f(\frac{1}{2})&gt;f(0)$)
92142
We have the function $f: \mathbb{R} \to \mathbb{R}$.     Question: Show that if the sets $\{x:f(x) &gt; a\}$ and $\{x:f(x) &lt; a\}$ are open for every $a\in\mathbb{R}$, then $f$ is continuous.   My attempt:$f$ is continuous if $A \subset \mathbb{R}$ is open, then $f^{-1}(A) \subset \mathbb{R}$ is open. I can show that the image of $\{x:f(x) &gt;a\}$ is open. Namely $\{f(x) &gt; a\} = (a,\infty)$ is open in $\mathbb{R}$. But this doesn't prove that for any open set $A \subset \mathbb{R}$, we have that $f^{-1}(A)\subset\mathbb{R}$ open as well.    Thanks in advance!
114512
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?
47071
Let $f:\mathbb{R}\to\mathbb{R}$ be a map sending closed intervals to closed intervals. Prove that $f$ is continuous or find a counter example.  WLOG we just have to prove continuity at $0$ and we can also assume that $f(0)=0$ and $f(1)=1$ (by considering $x\mapsto (f(x)-f(0))/(f(1)-f(0))$ ).  I build up this question while thinking of the question .
30496
Problem: Given $f:[0,1] \rightarrow \mathbb{R}$ ($f$ continuous ) and $f(x^2) = f(x)$ $\forall x \in [0,1]$. Show that function $f$ is a const.
4826
I need to show that the following function is constant: $f:(0,\infty)\to\mathbb{R}$, $f(2^x)=f(3^x)$ and $f$ is continuous. I need to find the value of $f(x)$.    So I did something but I do not know how to continue. Let $x\to \log_2 x$ and we get that $f(x)=f(3^{\log_2 x)}$. Now, let $x\to 3^{\log_2 x}$ and we get that $f(3^{\log_2 x)}=f(3^{\log_2 3^{\log_2 x})}$. And so on. I think I need to find a rule. Do you have any idea?