Find all continuous real valued function such that $$(f(x))^2+C=\int\limits_0^xf(t)dt$$   for some $C\in\mathbb{R}$   If I set $F(x)=\int\limits_0^xf(t)dt$ then $F$ is differentiable and $F'(x)=f(x)$,  so $(f(x))^2+C$ is diferentible as it does $\sqrt{F(x)-C}=|f(x)|$,  and then $F'(x)=(|f(x)|^2+C)'=2|f(x)||f(x)|'=f(x)$.  and then I don't know how to continue. I just see the solution $f(x)=0$ whith $C=0$.
114591
Find all continuous functions $f$ satisfying   $$\int_0^xf=(f(x))^2+C$$ for some constant $C \neq 0$, assuming that $f$ has at most one $0$.  I have a question about the solution to this problem. It says that   Clearly $f^2$ is differentiable everywhere, its derivative at $x$ is $f(x)$ (I'm assuming that this is because the left hand side is differentiable, the right hand side must be differentiable as well). So $f$ is differentiable at $x$ whenever $f(x) \neq 0$, and  $$f(x)=2f(x)f'(x),$$ so $f'(x) = \frac{1}{2}$ at such points. Thus, $f= \frac{1}{2}x+b$ for some $b$ on any interval where it is non-zero; if $f$ has a zero, this gives two possible solutions, with two possible values of $b$, but since $f$ is assumed continuous, they must be the same. So we need $$\int_0^x(\frac{1}{2}t+b)dt=(\frac{1}{2}x+b)^2+C$$  so we must have $b=\sqrt{-C}$ or $-\sqrt{-C}$, leading to two solutions for $C \lt 0$.  I don't understand the bolded statements. First, how does the differentiability of $f^2$ guarantee that $f$ is differentiable, and why only at $x$ where $f(x) \neq 0$?  Next, why are there two possible solutions if $f$ has a zero, but since $f$ is continuous they must be the same?  Finally, does the last statement mean that if $C \gt 0$, then there is no solution, since we're restricted to real numbers here?  I would appreciate it if anyone answers my questions.
11457
Theorem: If $f$ is finitely differentiable at $c$, then $f$ is also continuous at $c$.  But according to the definition, for a function to be differentiable at $c$, it need not even be defined at $c$, just that it should have a finite value in the vicinity of $c$.    My question: If function is not defined at $c$, how can it be continuous at $c$?  For example, say we have a straight horizontal line function which is broken at $x=1$, the function is defined at all the points except $x=1$, it has a finite value at all the points except $x=1$, since it is having a value in the nearby region of $x=1$ we say that it is differentiable at $x=1$ but can we say that it is continuous at $x=1$?
89554
Suppose that $f(x)$ is continuous on $(0, +\infty)$ such that for all $x &gt; 0$,$f(x^2) = f(x)$. Prove that $f$ is a constant function. My attempt is to show that for any point $a \neq b$ , we have $f(a)=f(b)$. But I have no idea on how to get this. Anyone can help?
77630
Consider $f\in C^2$ so that $$f''(x)+f(x)=-x\,g(x)f'(x), \ \forall x\ge0 $$ where $g(x)\ge 0$. Then ($\forall x\ge 0$)         (A) $f(x)^2+f'(x)^2$ is non-increasing,   (B) $f(x)^2&lt;3f(0)^2+(2f'(0))^2$,   (C) $|f(x)|\le\alpha$, where $\alpha$ is a fixed real constant.   (D) $\lim_{x\to\infty} f(x)\sin\left(\dfrac{1}{x}\right)$ exists.        Answer Given: (A),(B),(C),(D)  What I have tried : I differentiated option (A) and then with the help of the given equation in the question I was able to show that  it is less than zero . Hence I was able to infer that option (A) is correct .   Now coming to option (B).  I assumed the function in option (A) to be $h(x)$.   $$ h(x) = f(x)^2 + f'(x)^2 $$  Since $h'(x)&lt;0$ ,therefore $h(x)$ is a decreasing function   hence for all $x\ge 0$  $h(x)&lt; h(0)$  which implies $(f(x))^2 + (f'(x))^2&lt;(f(0))^2 + (f'(0))^2$  From the above expression we can get the second option .   Is my method correct ?. If not , please show the right method .  I am unable to get the the 3rd and 4th option .
62941
Find all functions $f$ that assign a real number $f(x)$ to every real number $x$ such that $$(x+y)f(x)+f(y^2)=(x+y)f(y)+f(x^2)$$  I've tried subbing in heaps of values but I keep getting things like $f(0)=f(0)$ and other such useless results.  Any help would be hugely appreciated.
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.
61825
Suppose $f$ is continuous on $[0,1]$ and   $$ \int\limits_0^1 f(x) x^n d x = 0 $$   for all $n=0,1,2,...$. Then, $f(x) = 0$  I was trying to use by parts  $$ 0 = f(x) \frac{x^{n+1}}{n+1} \bigg|_0^1 - \int\limits_0^1 \frac{x^{n+1}}{n+1} f(x) dx $$  Thus,  $$ \frac{f(1)}{n+1} = \int\limits_0^1 \frac{ x^{n+1} f(x) dx }{n+1} \implies f(1) = \int\limits_0^1 x^{n+1} f(x) dx $$  but here I am stuck. Maybe my idea is not the best one? How would you approach this problem?
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.
51565
Find all continuous functions $f:\mathbb R^{+}\to\mathbb R^{+}$ such that $f(x)^2=f(x^2)$ for all positive reals $x$.  If $f$ is a constant function, then $f(x)=1$. If $f$ is non-constant, I'm suspecting the only solutions are of the form $f(x)=x^k$, where $k$ is a constant, but I have no idea how to prove this. Thanks in advance.
65966
Regarding: Let $f:(0, \infty) \to \mathbb{R}$ be a differentiable function for which $f'(x)f'(\frac{1}{x})&lt;0$ for every $0&lt;x \ne 1$. Then $f'(1)=0$. -- So I am trying to prove this, and I said: We know that $f$ is differentiable at $(0, \infty)$, so there exists a $0&lt; \delta &lt; 1$ such that $f$ is differentiable at $[1- \delta, 1+\delta]$ . So there exists a $x_{0} \in [1- \delta, 1+\delta]$ such that $f'(x_{0})f'(\frac{1}{x_{0}})&lt;0$ So by Darboux Theorem, there exists a $x_{0}&lt;c&lt; \frac{1}{x_{0}}$ such that $f'(c)=0$. But now, I do not know how to link $c$ with $1$... Any hints? Thanks! :)
88860
Suppose that a function $f:[0,1] \to \mathbb R$ is continuous in $[0,1]$. Prove that there is a point $c\in [0,1]$ such that $\int_0^1 x^2 f(x)\,dx = \frac{f(c)}{3}$.   Do I somehow use the mean value theorem or FTC?
101432
I believe that the correct answer to this question will involve applying both the Mean Value Theorem and the Intermediate Value Theorem, perhaps several times.  However, I can't see how to proceed.  I went down what I thought were the typical paths and just reached dead ends.  Does someone maybe have a hint?  I think that with the appropriate nudge I should be able to solve this question.  If not, I'll edit and ask if someone knows how to solve the whole thing.  Here's the question:  Let $f: \mathbf R \to \mathbf R$ be such that $f$, $f'$, $f''$, and $f'''$ exist and are continuous on $\mathbf R$ and satisfying $f(-3)=-1$, $f(0)=0=f'(0)$, and $f(3)=8$.  Prove that there exists $\xi \in (-3, 3)$ such that $f'''(\xi) \ge 1$.
90256
Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a continuously differentiable function such that $f'(0)=0$.  Define for all $x,y\in \mathbb{R}$,  $$g(x,y)=f(\sqrt{x^2+y^2})$$ Pick out the true statements :   $g$ is differentiable on $\mathbb{R}^2$ $g$ is differentiable on $\mathbb{R}^2$ if and only if $f(0)=0$ $g$ is differentiable on $\mathbb{R}^2- \{ (0,0)\}$   I do not have much idea of differentiation of multivariable calculus.  I just tried out some examples to get some idea..  I have to take some function whose derivative at $0$ has to be zero.  for simplicity I would take $f(x)=x^2$ Then I would get :     $g(x,y)=x^2+y^2$ which is differentiable on $\mathbb{R}^2$   This would suggest me that second option has more chances as $f(0)=0$ in this case.  I would now take $f(x)=x^2+1$ then :     $g(x,y)=x^2+y^2+1$ which is differentiable on $\mathbb{R}^2$ though $f(0)\neq 0$   So, I would now eliminate second option and third option also...  So, It is now got confirmed that i have to work out to "Prove" that first option is correct.  Now, I have to prove that $g(x,y)=f(\sqrt{x^2+y^2})$ is differentiable.  I have to look for partial derivatives first :  $$\frac{\partial g}{\partial x}= f'(\sqrt{x^2+y^2}).\frac{2x}{2\sqrt{x^2+y^2}}=f'(\sqrt{x^2+y^2}).\frac{x}{\sqrt{x^2+y^2}}$$  $$\frac{\partial g}{\partial y}= f'(\sqrt{x^2+y^2}).\frac{2y}{2\sqrt{x^2+y^2}}=f'(\sqrt{x^2+y^2}).\frac{y}{\sqrt{x^2+y^2}}$$  I have used here that $f$ is differentiable on $\mathbb{R}$  As $f$ is continuously differentiable, we have $f'$ is continuous and I know that :  $\frac{x}{\sqrt{x^2+y^2}}$ and $\frac{y}{\sqrt{x^2+y^2}}$ are also continuous.     As $f'$ and $\frac{x}{\sqrt{x^2+y^2}}$ are continuous So is the product $\frac{\partial g}{\partial x}=f'(\sqrt{x^2+y^2}).\frac{x}{\sqrt{x^2+y^2}}$      As $f'$ and $\frac{y}{\sqrt{x^2+y^2}}$ are continuous So is the product $\frac{\partial g}{\partial y}=f'(\sqrt{x^2+y^2}).\frac{y}{\sqrt{x^2+y^2}}$   So we have continuous partial derivatives for $g(x,y)$..  So, I am willing to conclude that:     $g(x,y)$ is  differentiable on $\mathbb{R}^2$   I would be thankful if some one can confirm if the answer correct.  I would be much more thankful if some one can let me know if there are any gaps in my argument.  This is just a proof verification Question..  Thank you.
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.
99902
Let $f$ be a continously differentiable function on $\mathbb R$.  Suppose that   $$L=\lim_{x\to\infty}(f(x)+f'(x))$$ exists.  If $0&lt;L&lt;\infty$, then which of the following statements is\are correct?               If $\lim_{x\to\infty} f'(x)$ exists, then it is $0$.     If $\lim_{x\to\infty} f(x)$ exists, then it is $L$.     If $\lim_{x\to\infty} f'(x)$ exists, then $\lim_{x\to\infty}f(x)=0$.     If $\lim_{x\to\infty} f(x)$ exists, then $\lim_{x\to\infty}f'(x)=0$.           My Guess  I could not conclude the answer and prove that properly. But, I guess that it must be 1 and 2. help me.
46625
$f:\mathbb{R} \to  \mathbb{R}$ be defined as  $f(x)$= $x^2 , x \in \mathbb{Q} \\ 0 \ \ , x \in \mathbb{Q^c}\\ $  How to show that $f$ is differentiable at $x=0.$  I want to try this approach.  here  $f'(x)=$ $2x  \in \mathbb{Q}  \\  0  \in \mathbb{Q^c} $  since $lt \\ x \to 0+$$f(x)$ = $lt \\ x \to 0-$$f(x)= 0$   so $f$ is differentiable at $x=0$
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.