I am not sure how to prove this.  Although could a proof of even-degree polynomials with positive leading coefficient be helpful in this case?     Let $f:\mathbb R\to \mathbb R$ be continuous and such that $f(x)\to+\infty$ as $x\to\pm\infty$. Show that $f$ has a minimum, i.e. that there exists $x_0\in \mathbb R$ such that $f(x)\geq f(x_0)$ for all $x\in \mathbb R.$
51384
Let $f:\mathbb R\to\mathbb R$ continuous and $$\lim_{x\to-\infty }f(x)=\lim_{x\to\infty }f(x)=+\infty$$ then $f$ takes its minimum. This is a homework. My solution looks to simple, that's why I would like to have confirmation if it's correct or not.  My attempts  Let $a,b\in\mathbb R$, $a&lt;b$. Let denote $M$ the maximum of $f$ on $[a,b]$. By definition of the limits, there is $N_1&lt;a$ and $N_2&gt;b$ such that $f(x)\geq M$ when $x&lt;N_1$ and $x&gt;N_2$. If $f(x)\geq M$ for all $x\in [N_1,N_2]$, then $M$ is the minimum. Otherwise, $f$ takes its minimum on $[N_1,N_2]$ (since $f$ is continuous on the compact $[N_1,N_2]$), and it's a global minimum.   Do you think it's correct ?
56332
Let $f\colon\mathbb{R}\to\mathbb{R}$ be an infinitely differentiable function, and suppose that for each $x\in\mathbb{R},$ $\exists\  n=n(x)\in\mathbb{N}$ such that $f^{(n)}(x)=0$. For each fixed $n\in\mathbb{N}$, consider the sets $F_n=\{x\in\mathbb{R} : f^{(n)}(x)=0\}$ and let $I=[a,b]$ a finite closed interval.  Show that $$I=\bigcup_{n=0}^\infty(I\cap F_n).$$ Show that there is a non empty open set $A\subset I$ where $f$ is a polynomial.   My attempt:  If $x\in I$ then by hypotesis there is $n$ such that $f^{(n)}(x)=0$ then $x\in F_n$ so then $\bigcup F_n\supset I$ then $$I=I\cap\bigcup_{n=0}^\infty F_n=\bigcup_{n=0}^\infty(I\cap F_n).$$  Let $$B_k=\bigcap_{n=k}^\infty F_n.$$ I want to show that $I\cap B_k$ has non-empty interior for some $k\in\mathbb{N}$. Then there is $A\subset B_k$ and by Taylor's Theorem $f$ will be a polynomial. But I didn't know how to prove this. I tried using the following bigger set: $$B=I\cap\bigcup_{k=0}^\infty B_k=I\cap\bigcup_{k=0}^\infty \bigcap_{n=k}^\infty F_n.$$ I think that it suffices to show that $B^o\neq\emptyset$ maybe using the Baire's Theorem proving that $B$ is a not meagre set.   More general Problem  Let $f$ be an infinitely differentiable function on $[0,1]$ and suppose that for each $x \in [0,1]$ there is an integer $n \in \mathbb{N}$ such that $f^{(n)}(x)=0$. Then does $f$ coincide on $[0,1]$ with some polynomial? If yes then how. I thought of using Weierstrass approximation theorem, but couldn't succeed. Answer: The proof is by contradiction. Assume $f$ is not a polynomial. Consider the following closed sets: $$ S_n = \{x: f^{(n)}(x) = 0\} $$ and $$ X = \{x: \forall (a,b)\ni x: f\restriction_{(a,b)}\text{ is not a polynomial} \}. $$ It is clear that $X$ is a non-empty closed set without isolated points. Applying Baire category theorem to the covering $\{X\cap S_n\}$ of $X$ we get that there exists an interval $(a,b)$ such that $(a,b)\cap X$ is non-empty and $$ (a,b)\cap X\subset S_n $$ for some $n$. Since every $x\in (a,b)\cap X$ is an accumulation point we also have that $x\in S_m$ for all $m\ge n$ and $x\in (a,b)\cap X$. Now consider any maximal interval $(c,e)\subset ((a,b)-X)$. Recall that $f$ is a polynomial of some degree $d$ on $(c,e)$. Therefore $f^{(d)}=\mathrm{const}\neq 0$ on $[c,e]$. Hence $d&lt; n$. (Since either $c$ or $e$ is in $X$.) So we get that $f^{(n)}=0$ on $(a,b)$ which is in contradiction with $(a,b)\cap X$ being non-empty.
37755
Assume that $f: [0,1] \rightarrow \mathbb{R}$ has a continuous $f''$ and $f$ is positive on the interval $(0,1)$ and $0$ at the endpoints.    I want to prove that $$\int_0^1 \Big| \frac{f''(x)}{f(x)} \Big| \,dx &gt;4. $$  I could not make any progress. Since the unit interval is compact, we know that the continuity of $f, f'$ and $f''$ are uniform and they attain their maximum and minimum values. Moreover, I tried to apply the Mean Value Theorem but I failed.
65877
Let $f: \mathbb R\to \mathbb R$ be a continuous function such that $f$ attains neither its minimum nor its maximum at any open interval $I \subseteq  \mathbb R$ , then how to prove that $f$ is monotone ?
36912
Theorem. Let $f\colon I \to J$, where $I$ is an interval and $J$ is the image $f(I)$, be a function such that:   $f$ is strictly increasing on $I$. $f$ is continuous on $I$.   Then $J$ is an interval, and $f$ has an inverse function $f^{-1}\colon J\to I$ such that:   $f^{-1}$ is strictly increasing on $J$. $f^{-1}$ is continuous on $J$.   I need to prove only that $f^{-1}$ is continuous on $J$.  Proof. Let $y\in J$, and (for simplicity) assume that $y$ is not an end-point of $J$. Then $y=f(x)$, for some $x\in I$, and we want to prove that $$y_n\to y\implies f^{-1}(y_n)\to f^{-1}(y)=x.$$ Thus we want to deduce that for each $\varepsilon&gt;0$, there is some number $X$ such that $$x-\varepsilon&lt;f^{-1}(y_n)&lt;x+\varepsilon\qquad \forall n&gt;X\qquad(1).$$ Since $f$ is strictly increasing, we know that $$f(x-\varepsilon)&lt;f(x)&lt;f(x+\varepsilon).$$ also, since $y_n\to y=f(x)$, there is some number $X$ such that $$f(x-\varepsilon)&lt;y_n&lt;f(x+\varepsilon), \qquad\forall n&gt;X.$$ Then, applying the strictly increasing function $f^{-1}$ to these inequalities, we obtain (1).  Edit  I don't understand why, since $y_n\to y=f(x)$, there is some number $X$ such that  $$f(x-\varepsilon)&lt;y_n&lt;f(x+\varepsilon), \qquad\forall n&gt;X.$$ I would have written the last line as follows: $$f(x)-\varepsilon&lt;y_n&lt;f(x)+\varepsilon, \qquad\forall n&gt;X.$$ Applying the $f^{-1}$ to my last line, which I think it's correct, I obtain $$f^{-1}[f(x)-\varepsilon]&lt;f^{-1}[y_n]&lt;f^{-1}[f(x)+\varepsilon], \qquad\forall n&gt;X.$$ which is not (1).  What am I misunderstanding? Why the author of this proof wrote that last line and not my version?  Thank you.
97541
I'm having troble with this problem, I don´t know what to do, is for a final exam tomorrow.  Let $f:R^2\to R$ given by  $(y-x^2)(y-2x^2)$  a) Draw the sets $A=${$(x,y)$|$f(x,y)&gt;0$}$\ and\ $$B=${$(x,y)$|$f(x,y)&lt;0$}  b) Shows that for all $\vec u$=$(u_{1},u_{2})$ with ||$\vec u$||=$1$, the function: $g(t)=f(t\vec u$) has a local minimum in $t=0$  c) Shows that notwithstanding the foregoing, $\vec 0$ is not a local minimum of $f$
99999
The IVT says that, given a closed interval of real numbers, $[a, b]$, and a real-valued function $f$ continuous on $[a,b]$, for any point $z\in [f(a), f(b)]$ (WLOG, $f(a) &lt; f(b)$), there exist a point $c\in [a,b]$ such that $f(c) = z$.  Now, I'm wondering, why is the theorem only applicable to $z\in[f(a), f(b)]$ and not to $z\in [\min\limits_{x\in [a,b]}\{f(x)\},\max\limits_{x\in [a,b]}\{f(x)\}]$?
77549
Let $f:[0,1]\to\mathbb R$ is a continuous function, and for all $x\in [0,1]$, $f(x)$ is either a local maximum or a local minimum. Then prove that $f$ is a constant.  Here is what I have tried, I don't know whether it is correct:  Assume $f$ is not a constant, by continuity, there exist $x_1$, $x_2$ such that $f(x_1)=m$ is a global minimum and $f(x_2)=M$ is a global maximum. Now if $m\ne M$, choose any $c\in(m,M)$ and define  $$x_0=\sup (x\in[x_1,x_2]:f(x)&lt;c)$$  Then we have $f(x_0)$ NOT a local minimum nor a local maximum (since according to the definition of $x_0$, there are always some points $&lt;c$ in the left neighborhood of $x_0$ and the points in right neighborhood are always $&gt;c$). Contradiction arises and so $f$ must be a constant.
107389
What is the minimum of $$f(a,b,c):=\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}$$ where $a,b,c$ are positive real numbers?  When $a=b=c$, we have $f(a,b,c)=\dfrac{3}{\sqrt{2}}\approx 2.12$  When $a=1,b=c\rightarrow\infty$, we have $f(a,b,c)\rightarrow 2$. So the minimum is at most $2$.
119842
Let $n$ be a positive integer, $x_0, x_1,\ldots,x_n$ be $(n+1)$ pairwise distinct real numbers, and $y_0, y_1,\ldots,y_n$ be $(n+1)$ real numbers. We have know that there exists a unique polynomial $P(x)$ where its degree is not greater than $n$.  What is the necessary and sufficient condition guaranteeing $P(x)$ has a degree of $n$.  Thank you for all comments and helping.
21449
I can't prove this statement, can anybody show me how to prove it?    $$f:\mathbb{C}\rightarrow \mathbb{C} \in \mathcal{O}(\mathbb{C}), \exists n\in \mathbb{N}, R &gt;0 , M&gt;0 : |f(z)| \le M|z|^{n} \ \ \forall |z|&gt;R \Rightarrow \deg(f)\le n $$  To show is that if there exists such an $M$, that then $f$ is a polynomial of max degree $n$. I started like this:   $$f(z) = \sum_{n=0}^{\infty} a_n (z-z_0)^n$$ So if I put this into the inequality: $$|f(z)| = \left| \sum_{n=0}^{\infty} a_n (z-z_0)^n \right| \le M |z|^n .$$
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.
42672
This is from Putnam:     If a function $f: \mathbb{R} \to [-1, 1]$ is such that $f''(x)$ exists for all $x \in \mathbb{R}$ and $\{f(0)\}^{2} + \{f'(0)\}^{2} = 4$ then prove that there is a point $c$ such that $f(c) + f''(c) = 0$.   Now the condition concerning values of $f(0), f'(0)$ (and the fact that range of $f$ is subset of $[-1, 1]$) implies that $f'(0) \neq 0$ so that function is not a constant. If we take $g(x) = \{f(x)\}^{2} + \{f'(x)\}^{2}$ then we can see that $g' = 2f'(f + f'')$. One hope of solving the problem is to show that $g$ attains local extrema at some point different from an extrema of $f$. This would ensure that $g'$ vanishes without making $f'$ vanish and that will lead to $f + f'' = 0$. However the constraint on $f(0), f'(0)$ does not help in analyzing the extrema of $f$ or $g$.  Another line of thought which could be helpful here is to consider $F(x) = f'(x)\sin x - f(x)\cos x$ so that $F'(x) = \sin x\{f(x) + f''(x)\}$ but in this case I am not able to see how to use the condition on values $f(0), f'(0)$ to impose some constraint on $F(x)$.  Perhaps both the approaches which came to my mind are not in right direction (or may be they are, but I am unable to see). Please provide any hints.  Update: Based on the answer by Sandeep Thilakan, I have dropped the requirement for continuity of $f''(x)$.
109759
It's a really basic question,in these days, I've been thinking why a polynomial $p(x)\in F[x]$ ($F$ a field) with degree $n$ can have at most n roots. It seems easy to prove, but I've been trying to prove this since yesterday, maybe I forgot some important details necessary to prove, I don't know. I'm solving some questions about Field Theory and I noticed that almost every question I should use this theorem, I need help.  Thanks
81123
Let $f : \mathbb{R} \longrightarrow \mathbb{R}$ be a differentiable function such as $\forall x \in \mathbb{R}, f'(x) &gt; 2$. What could I do to show that $\exists c \in \mathbb{R} : f(c) = 0$ ? At first I thought I could use the mean value theorem but it did not help me. In fact, I don't really understand why $f(x)$ couldn't be strictly positive $\forall x \in \mathbb{R}$. Would you know how to proceed ? Thanks for your help.
36536
This is problem 10 from the International Mathematical Competition for University Students of 2015, from day 2, in Bulgaria. I think it is an interesting problem!     Let $n$ be a positive integer, and $p(x)$ be a polynomial of degree $n$ with integer coefficients. Prove that   $$ \max_{x\in [0,1]}\left|p(x) \right| &gt; \frac{1}{e^{n}}. $$   Proposed by Géza Kós, Eötvös University, Budapest.
62291
If $f(x)$ is a quadratic expression such that $f(x)&gt;0\;\forall x\in\mathbb{R},$ and if $g(x)=f(x)+f'(x)+f''(x),$  Then prove that $g(x)&gt;0\; \forall \; x\in \mathbb{R}$.  $\bf{My\; Trial \; Solution::}$ If $f(x)&gt;0\;\forall x\in \mathbb{R}$. Then function $f(x)$ has Minimum value .  Let Minimum occur at $x=x_{0}$. Then $f(x)_{Min} = f(x_{0})&gt;0$  Now Given $g(x) = f(x)+f'(x)+f''(x)$. Then $g'(x) = f'(x)+f''(x)+f'''(x).$  Now at $x=x_{0}\;\;,$ Value of $g(x_{0}) = f'(x_{0})+f''(x_{0})+f'''(x_{0}) = 0+f''(x_{0})+f'''(x_{0})$  Now I did not understand How can i solve after that,   Help me   Thanks