I am facing the following problem.      Let $f$ be a continuous and derivable function from $[0, 1]$ to $R$,   such that $f(0) = 0$. Show that if $f \neq 0$ then there exist $x_0$   such that $|f'(x_0)| &gt; |f(x_0)|$,   I have been thinking for a while on this problem but I have no idea how to demonstrate it. Can someone provide a hint?
121002
Let $f:[-1,1]\rightarrow \mathbb{R}$ be continuous with $f(0)=0$ and continuously differentiable in $]-1,1[$. Suppose that  $$|f'(x)|\leq |f(x)|,\;\mbox{ for all }x\in]-1,1[.$$  Prove that $f(x)=0$ for all $x\in[-1,1]$.  My idea is take one point (arbitrary) $x_0&gt;0$ in $]-1,1[$, then using mean value theorem, exist $y_0\in[0,x_0]$ such that   $$|f(y_0)|\geq|f'(y_0)|=\frac{|f(x_0)-f(0)|}{|x_0-0|} \mbox{ then }|f(x_0)|\leq x_0|f(y_0)|,$$  So if $f(y_0)=0$ implies $f(x_0)=0$ and we ended. But is $f(y_0)\neq0$ then I can follow and obtain a decreasing sequence but I not sure that converge to zero.  Thank you by some hints.
53635
$f(0)=0$ and $\lim_{x \to \infty}f(x)=0$ and $f$ is differentiable in $\mathbb{R}$  How can I show there's $x_0$ such that $f'(x_0)=0$.  I know it's true since the values of f(x) for large enough x are close to 0 as much as I want. So, that means that the same value would have to appear at least twice, and then it can be concluded. However, I don't know how to show it formally.
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The statement is: $f$ is a real fucntion on $\mathbb R$. Then if $f'(x)=f(x)$ and $f(0)=1$, then $f(x)\neq 0.$
75223
Suppose $f(x)$ and $g(x)$ are continuous functions on $[a,b]$ with $f$ monotone increasing. Assume there exists a sequence $x_n \in [a, b]$ such that for all $n \in N$ , $g(x_n) = f(x_{n+1})$. Show that there exists $x_0 \in [a,b]$ such that $g(x_0) = f(x_0)$.  Can someone provide an example of functions that fulfills this condition?
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I'm trying to make a problem for my advanced calculus students.  I was thinking, if we have a differentiable function $f:\mathbb{R}\to\mathbb{R}$ such that $f'(q)=0$ for all $q\in\mathbb{Q}$, can we say that $f$ is constant?
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Prove that if $f:(0,\infty)→\mathbb{R}$ satisfying $f(xy)=f(x)+f(y)$, and if $f$ is continuous at $x=1$, then $f$ is continuous for $x&gt;0$.   I let $x=1$ and I find that $f(x)=f(x)+f(1)$ which implies that $f(1)=0$. So, $\lim_{x\to1}f(x)=0$, but how can I use this to prove continuity of $f$ for every $x \in \mathbb R$?   Any help would appreciated. Thanks
39321
Let $f:[0,1]\to \mathbb{R}$ is continuous, such that $f(0)=f(1)$. For each $n\in\mathbb{N}$, prove that $\exists x\in [0,1]$ such that $x+\dfrac{1}{n}\in[0,1]$ and $f\left(x+\dfrac{1}{n}\right)=f(x)$.   My approach: The interval $[0,1]$ is connected, then if $f$ is continuous, we have $f([0,1])\subset\mathbb{R}$ is a interval. And I suppose that, I can define a closed path such that $f(0)=f(1)$. Any hint pls,
113748
Possible Duplicates:      I found a nice problem I would like to share. Problem: If $f$ is continuous on $[0,1]$, and if $$\int_0^1 f(x)x^n \ dx =0$$ for every positive integer $n$, prove that $f(x)=0$ on $[0,1]$. Source: W. Rudin, Principles of Mathematical Analysis, Chapter 7, Exercise 20. I have posted a proposed solution in the answers.
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Let $f$ be differentiable on $\Bbb{R}$ and let $\lim_\limits{x\to \infty}f(x)=\lim_\limits{x\to -\infty}f(x)=0$. Prove there exists $a\in \Bbb{R}$ such that $f'(a)=0$.  Attempt: If $f$ is constant we are done. Otherwise, $f$ is not constant. Suppose there is no such $a$. Then, in particular, $f$ has no extremum $\implies$ $f$ has no point where is goes from increasing to decreasing or from decreasing to increasing. $f$ is continuous since it is differentiable. Hence, since $\lim_\limits{x\to \infty}f(x)=\lim_\limits{x\to -\infty}f(x)=0$, $f$ must be constant. A contradiction. Therefore either $f$ is constant and $f'(x)=0$ everywhere, or $f$ has an extremum at $x=a$ and $f'(a)=0$.   I feel like my attempt is not formal enough, any correction?  I would appreciate your help.
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This is an exercise from Abbott's second edition of Understanding Analysis.     Let $f$ be differentiable on an interval $A$. Show that if $f'(x) \neq 0$ on $A$, show that $f$ is one-to-one on $A$. Provide an example to show that the converse statement need not be true.   Is my solution (below) correct?   Let $x_{1},x_{2} \in A $ such that $x_1 \neq x_2$. Since the function satisfies all the conditions of the mean value theorem, there exists $c \in (x_1,x_2)$ [without loss of generality we consider $x_1 &lt; x_2$] such that $f(x_2) - f(x_1) = (x_2 -x_1) f'(c) \neq 0$ as it is given that $f'(x)$ is nonzero on $A$ and $x_1 \neq x_2$ is our assumption. Therefore $x_1 \neq x_2$ implies $f(x_1) \neq  f(x_2)$ for every $x_1,x_2 \in A$. Thus $f$ is one-to one on $A$.  The converse may not be true. Consider $A=[-1,1)$ and $f(x)= x^3$. Therefore $f$ is injective on $A$. Again $f'(x)= 2x^2$. Therefore $f'(0)=0, 0 \in A$. So the converse need not be true.
10490
Let $f:\mathbb R \to \mathbb R$ be a twice continuously differentiable function, with $f(0)=f(1)=f'(0) = 0$. Then   $f^{"}$ is the zero function. $f^{"}(0)$ is zero. $f^{"}(x)=0$ for some $x \in $ (0,1). $f^{"}$ never vanishes.   I don't know where to start, tried rolles theorem but cant catch the point..please help me to start..
51915
Prove that there is no continuous functions $f:[0,1]\to R$, such that   $$ f(x)+f(x^2)=x. $$   My try. Assume that there is a continuous function with this property. Thus, for any $n\ge 1$ and all $x\in [0,1]$, \begin{align*} f(x)&amp;=x-f(x^2)=x-\big(x^2-f(x^4)\big)=x-x^2+\big(x^4-f(x^8)\big)=\cdots\\ &amp;=x-x^2+x^4-\cdots+(-1)^n\left(x^{2^n}-f\big(x^{2^{n+1}}\big)\right) \end{align*}  since $f(0)=0$ and $\displaystyle\lim_{n\to \infty}x^{2^{n+1}}=0$ for any $x\in(0,1)$,it follows by the continuity of $f$ that $\displaystyle\lim_{n\to \infty}f\big(x^{2^{n+1}}\big)=0$, hence $$f(x)=x-x^2+x^4-x^8+\cdots+(-1)^nx^{2^n}+\cdots$$ for any $x\in (0,1)$  Why do I have prove that there is exists such a continuous functions $f$? Maybe  my example is wrong? Why?  if my   method is wrong,then How prove this problem ? Thank you.
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 $f''(x)$.
117332
$f(x)$ is continuous and differentiable function defined in $x \geq 0$. If $f(0)=1$ and $f'(x) &gt;3 f(x)$, then prove that $$f(x) \geq  e^{3x}, \forall x \geq 0$$  Could someone give me little hint as how to proceed in this question? I am not able to initiate.
89754
Let $f(x)$ be continuous on $(0,1]$. Show that $f$ is uniformly continuous IFF $\displaystyle \lim_{x\to0^+} f(x)$ exists.    Thoughts:  Backward Proof:  Let another function $\overline f(x)$ be continuous on $[0,1]$ which is equal to $f(x)$ plus the limit point. Thus the limit exists and it is uniformly continuous. So, as $f(x)$.  Forward Proof: I Don't really have any idea...??  Please help guys
60477
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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Suppose $f$ is continuous, $f(x) \ge 0$ on $[a,b]$ and $\int_a^b f(x) dx=0$. Show that $f(x)=0$ on $[a,b]$.   I know that if $f$ in continuous at $x_0$ and $f(x_0) &gt; 0$, then there exists $\delta &gt; 0$ s.t. $f(x) \ge \frac{1}{2} f(x_0)$ for $|x-x_0| &lt; \delta$ and I think it should help me but I'm stuck at this point.