If $f(x)$ is a quadratic expression such that $f(x) &gt; 0,\ x\in\mathbb{R}$ and if $g(x)= f(x) + f'(x) + f''(x)$, then prove that $g(x) &gt; 0, \ x\in\mathbb{R}$.
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
63149
Let $f$ be a twice differentiable function on the open interval $(-1,1) $such that $f(0)=1$. Suppose $f$ also satisfies $f(x) \ge 0, f'(x) \le 0 $and $f''(x) \le f(x)$, for all $ x\ge 0$. Show that $f'(0) \ge -\sqrt2.$  My attempt $ \rightarrow$  From Taylor's we see that $0 \leq f(x)=f(0)+f'(0)x+\frac{f''(\zeta)x^2}{2} $ for some $\zeta \in (0,x) .$  $f'(x)\leq 0$ and $  f'(x) \le 0 \implies$  $  f''(\zeta) \leq f(\zeta)\leq 1$  Thus $1+f'(0)x+\frac{x^2}{2} \geq 0 $.  I'm stuck upto this,I can see that if the discriminant of the quadratic is less than $ 0$ we're done but how do I conclude that the discriminant would be $0$ ?
106543
I don't know how to prove this example. Let $y(t):\mathbb{R}^+\to \mathbb{R}$ be a nonincreasing function. Assume that there is a constant $A&gt;0$ such that \begin{equation} \int_{s}^{+\infty}y(t)dt \leq Ay(s), \quad 0\leq s&lt;+\infty \end{equation} Then $y(t)\leq y(0)e^{1-\frac{t}{A}}$, for all $t&gt;0$.
69535
Let $f(x) = |x|$ for $x \in [-1,1]$ and extend $f$ to all of $\mathbb R$ by requiring $f(x+2)=f(x)$. Prove that $|f(x)-f(y)|\le|x-y|$ for all $x$ and $y$.   This is easy to observe on a graph but my proof is long and a bit tedious. Just curious if there is a quick/elegant way?
94331
Problem: Hi, I'm trying to solve the following proof:  If $a &gt; b \geq 0$, Prove that $a^2 &gt; b^2$.  But I'm a bit new with proofs and I don't manage to write a proper one. I would be very grateful if someone could explain me which are the steps required to demonstrate the proof. Thank you.
31188
Suppose that $f\colon\mathbb{R}\to\mathbb{R}$ without any further restriction. If $f(f(x))=x^2-x+1$, how can one find $f(0)$?   Thanks in advance.
115044
Let $f(x)=ax^2+bx+c$ where $a,b,c$ are real numbers. Suppose $f(-1),f(0),f(1) \in [-1,1]$. Prove that $|f(x)|\le \frac{3}{2}$ for all $x \in [-1,1]$.    I made quite a few attempts but could not explicitly derive the required condition. I can understand that it happens, but cannot derive that. Please help.
11179
Let $f:R\rightarrow R$ be a function such that all its successive derivatives exist in all $R$ and also $f(x)f''(x)\leq 0$ everywhere.  If $\alpha$ and $\beta$ be two successive roots of $f(x)=0$.   Then prove that $f'''(x)=0$ for atleast one $\gamma \in(\alpha,\beta)$.  My Attempt:  I began by taking an example $f(x)=(x-1)(2-x)$ and let $x=1$ and $x=2$ be its two successive roots.The statement is trivially true.  Then I took the function $f(x)=e^{-x}(x-1)(2-x)$.The statement is true here also so on and so forth.  But what would be the exact proof I wonder. It appears to be a question of Rolle's Theorem.
43964
Determine all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for all $x, y \in \mathbb{R}$, $$f((x + 1) f(y)) = y (f(x) + 1).$$    Source: Vuong Lam Huy, on a Facebook group.
111724
The Problem:   Let $f : \mathbb R \to \mathbb R$ be a real-valued function defined on the set of real numbers that satisfies $$f(x + y) \leq yf(x) + f(f(x))$$ for all real numbers $x$ and $y$.  Prove that $f(x) = 0$ for all $x ≤ 0$.  (IMO $2011$ , Pr: $3$)  Here is the solution:  Before we prove that $f (x) = 0$ for all $x ≤ 0$, we will prove that $f (0) = 0.$ $\color{black}{\large\text{Case} \thinspace  1:}$ $f(0)&gt;0$ Let, $\varepsilon &gt;\dfrac{f(f(0))}{f(0)}$ and $x&lt;- \dfrac{\varepsilon+f(f(0))}{f(0)}$  we have, from $f(x)\leq xf(0)+f(f(0))$,  $f(x)&lt;-\varepsilon.$ Then from, $f(f(x))\leq f(x)f(0)+f(f(0))$ we have $f(f(x))&lt;0$. Finally,  since $0&lt;f(0)\leq f(f(0))$ ,we get $f(0)\leq-xf(x)+f(f(x))&lt;0$ which imply $f(0)&lt;0$, which gives a contradiction. $\color{black}{\large\text{Case} \thinspace  2:}$ $-\alpha \leq f(0)&lt;0$, where $\alpha\in\mathbb{R^+}.$ From $f(x + y) \leq yf(x) + f(f(x))$ we have: $\begin{cases} f(x)\leq f(f(x)) \\ f(x) \leq xf(0)+ f(f(0)) \end{cases} \Longrightarrow f(x)\leq f(x)f(0)+f(f(0)) \Longrightarrow f(x)(1-f(0))\leq f(f(0))$. Then applying $x=f(0)$, we get $f(f(0))\leq 0$, which imply $f(x)\leq 0$, which gives $f(f(x))\leq 0$. In this case, we have $f(x)&lt;0.$ Because, if $f(x)=0$ at least for some $x$, from $f(x)\leq f(f(x))$, we get $f(0)\geq 0$, which gives a contradiction. Then let, $x&lt;-\sqrt {\alpha}$,  from  $f(x + y) \leq yf(x) + f(f(x))$  we have: $$\begin{align} f(z)\leq(z-x)f(x)+f(f(x)) \Longrightarrow f(f(x)) \leq (f(x)-x)f(x)+f(f(x))\Longrightarrow f(x)(f(x)-x)\geq 0 \Longrightarrow f(x) \leq x &lt;-\sqrt {\alpha} \Longrightarrow f(x)&lt; -\sqrt{\alpha}.\end{align}$$ Finally, from $f(0)\leq-xf(x)+f(f(x))$ we get $f(0)&lt;-\alpha$, which gives a contradiction. So, we deduce that $f(0)=0$. Then, applying $f(0)=0$, from  $f(0)\leq-xf(x)+f(f(x))$ and $f(x) \leq xf(0)+ f(f(0))$ we have, $$\begin{cases} xf(x)-f(f(x)) \leq 0 \\ f(x) \leq 0 \\ f(f(x)) \leq 0 \end{cases} \Longrightarrow  \begin{cases} xf(x)\leq0 \\ f(x)\leq 0 \end{cases} \Longrightarrow \begin{cases} xf(x)\leq0, x\in\mathbb {R} \\ xf(x)\geq 0, x\leq 0 \end{cases} \Longrightarrow xf(x)=0, x\leq 0 \Longrightarrow \color{blue} {\boxed{ f(x)=0, \text{for all} \thinspace x\leq 0.}}$$ End of the Proof.  Can you verify this solution? Are there any mistakes, gaps, etc.? Thank you very much!
38171
I have to do the following exercise:     Let $f$ and $g$ two differentiable functions such that $f(0)=g(0)$ and $f'(x)\leq g'(x)$ for all $x$ in $\mathbb{R}$. Prove that $f(x)\leq g(x)$ for any $x\geq0$.   Now, I know this is true because the first derivative of a function is the angular coefficient of the function in a point $x$. So, $f'(x)\leq g'(x)$ means, in other words, that the function $g(x)$ grows faster than $f(x)$. I think this is the base for a more formal proof, could someone help me to figure out a more formal proof?
103459
Suppose $f:\mathbb{R}\to\mathbb{R}$ has period 1, and for some $q\in(0,1)$:  $$|f(x)-f(y)|\leq q|x-y|\quad \forall x,y$$  Let $g(x)=x+f(x)$, for any $a\in\mathbb{R}$,define the following sequence:  $$a_1=a,\quad a_2=g(a_1),\quad a_3=g(a_2),\quad \dots,\quad a_{n+1}=g(a_n)$$  Show that  $$\lim_{n\to\infty}\frac{a_1+a_2+\dots+a_n}{n^2}$$  exists and does not depend on the choice of $a$.  I don't even know what is the possible approach, any hints? Thanks!
44911
Given a function $g$ which has derivative $g'$ for all x $\in {R}$ and satisfying $g'(0)=2$ and $g(x+y)=e^yg(x)+e^xg(y)$ for all $x,y\in {R}$ Show that $g'(x)+g(x)-2e^x=0$  $\dfrac{g(x+y)}{e^{x+y}}=\dfrac{g(x)}{e^{x}}+\dfrac{g(y)}{e^{y}}$ Putting $x=0$, $g'(y)=2e^y+g(x)$ Also putting $y=0$,we get, $0=e^xg(0)\implies g(0)=0$ I had a strong hunch that $g(x)=e^x-e^{-x}$ but it does not satisfy $g(x+y)=e^yg(x)+e^xg(y)$ Please help.
111496
Show:   if $f^{(n)}(x_0),g^{(n)}(x_0)$ exist and   $$\lim_{x\to x_0} \dfrac{f(x)-g(x)}{(x-x_0)^n}=0$$   then $f^{(r)}(x_0)=g^{(r)}(x_0 )$ for all $0\leq r\leq n.$   My attempt:  let n = 1,  since $f'(x_0),g'(x_0)$ exists then The taylor polynomial exists: $$T_{f,1}(x)=f(x_0)+f'(x_0)(x-x_0), \lim_{x\to x_0} \dfrac{f(x)-T_{f,1}(x)}{(x-x_0)}=0$$  $$T_{g,1}(x)=g(x_0)+g'(x_0)(x-x_0), \lim_{x\to x_0} \dfrac{g(x)-T_{g,1}(x)}{(x-x_0)}=0$$  Then, \begin{align} \lim_{x\to x_0} \dfrac{f(x)-g(x)}{(x-x_0)^n} &amp; =\lim_{x\to x_0} \dfrac{f(x)-g(x)+T_{f,1}(x)-T_{f,1}(x)+T_{g,1}(x)-T_{g,1}(x)}{(x-x_0)^n}\\ &amp;=\lim_{x\to x_0} \dfrac{f(x)-g(x)+T_{f,1}(x)-T_{f,1}(x)+T_{g,1}(x)-T_{g,1}(x)}{(x-x_0)^n}\\ &amp;=\lim_{x\to x_0} \dfrac{f(x)-T_{f,1}(x)}{x-x_0}-\lim_{x\to x_0} \dfrac{g(x)-T_{g,1}(x)}{x-x_0}+\lim_{x\to x_0} \dfrac{T_{f,1}(x)-T_{g,1}(x)}{x-x_0}\\ &amp;=\lim_{x\to x_0}\dfrac{T_{f,1}(x)-T_{g,1}(x)}{x-x_0}\\ &amp;=0 \end{align}  Then, \begin{align} \lim_{x\to x_0}\dfrac{T_{f,1}(x)-T_{g,1}(x)}{x-x_0} &amp;=\lim_{x\to x_0}\dfrac{f(x_0)+f'(x_0)(x-x_0)-g(x_0)-g'(x_0)(x-x_0)}{x-x_0}\\ &amp;=\left(f(x_0)-g(x_0)\right)\lim_{x\to x_0}\dfrac{1}{x-x_0}+f'(x_0)-g'(x_0) \\&amp;=0 \end{align} I am confused because  $\lim_{x\to x_0}\dfrac{1}{x-x_0}$ does not exist.
84717
Prove that $$\forall x,y \in \mathbb{R}:\vert\cos x- \cos y\vert\leq\vert x-y\vert.$$  My try : $ f(x) =\cos x$ and use the mean value theorem.
12493
Assume $f$ and $g$ are differentiable at $x$. Prove that $(fg)^{(n)} = \sum_{k=0}^n \binom{n}{k}f^{(k)}(x)g^{(n-k)}(x)$   I am assuming here $fg = f(x) g(x)$. Then we can prove this via induction. If $n = 0$ we have $1 = 1$ which is true. Now assume it is true for some $m$ we need to show it is possible for $m+1$. We have $(fg)^m = \sum_{k=0}^m f^{(k)}(x)g^{m-k}(x)$. How do we show that?
113885
Let $f$ be a $C^2$ function on $[0,1]$ such that $f(0)=f(1)=f'(0)=0,f'(1)=1.$ Prove that $$\int_0^1[f''(x)]^2dx\ge4.$$ Find all $f$ for equality to occur.