Let f be a continuously differentiable real-valued function on [0,1] such that  $\int_{1/3}^{2/3} f(x)dx = 0$. Find the minimum value of  $$I = \frac{\int_{0}^{1}f′(x)^2 dx} {(\int_{0}^{1}f(x)dx)^2}$$
35672
Let $f$ be a continuously differentiable real valued function on $[0,1]$.  It is given that $\displaystyle \int_{\frac{1}{3}}^{\frac{2}{3}}f(x) dx=0$     Find the minimum value of $\dfrac{\int_{0}^{1} (f'(x))^2 dx}{\left( \int_{0}^{1} f(x) dx \right)^2}$   I tried to use Cauchy-Schwarz to show that  $$\frac{\int_{0}^{1} (f'(x))^2 dx}{\left( \int_0^1 f(x) dx \right)^2} \ge \frac{\left( \int_0^1 \bigl| f(x)f'(x) \bigr| dx \right)^2}{ \left( \int_0^1 f(x) dx \right)^2} \ge \frac{ f(1)^2 - f(0)^2}{2 \left( \int_0^1 f^2 (x) dx \right)^2}$$ But I can't proceed from here.  Also, I don't know how to use the condition  $\int_{1/3}^{2/3}f(x) dx=0$
90543
Let $f:[0;1]\to\mathbb R$ be continiously differentiable  and $ f(0)=0$.   Prove that $\int\limits_0^1 |f(x)|^2\, dx\leq \dfrac{1}{2}\int\limits_0^1 |f'(x)|^2\, dx.$
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.
101507
I am trying to find a function such that   $\dfrac{d}{dx}f(x)=f(x-1)$   Is there such function other than $0$ ?
63411
Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f(0)=0$ and $\forall x \in \mathbb{R}$, we have $f'(x)=[f(x)]^2$. Show that $f(x)=0, \forall x \in \mathbb{R}$.
73846
I've been stuck on this one for a while. Comes from an analysis qual question.  Let f be a function that is continuous on $\left[0,1\right]$ and differentiable on $(0,1)$. Show that if $f(0)=0$ and $|f'(x)| \leq |f(x)|$ for all $x \in (0,1)$, then $f(x)=0$ for all $x \in \left[0,1\right]$.  What I've tried doing so far is see if there was anything I could do with MVT. I didn't really see anything to do with definitions either..to which I have a feeling I'll be playing around with them. Drawing a picture was a little difficult with these conditions as well  Any hints/suggestions?
17745
I found this problem on an old FB post and have just been humbled by it.  Let $f(x)$ be continuous on $[0,n]$ such that $n &gt; 0$ and $f(x) + f(n - x)$ does not vanish on $[0,n]$ then evaluate  $$ \int_0^n \frac{f(x)}{f(x) + f(n-x)}dx $$  Any clues/hints would be appreciated.
46266
Let $f\in C[a,b]$. Assume that $\min_{[a,b]}f=m&gt;0$ and $M=\max_{[a,b]}f$. Which one is true?  a. $$\frac{1}{M}\int_a^bf(x)dx+m\int_a^b\frac{1}{f(x)}dx\geq 2\sqrt{\frac{m}{M}}(b-a)$$  b. $$\int_a^bf(x)dx\int\frac{1}{f(x)}dx\geq (b-a)^2$$  c. $$\int f(x)dx\int_a^b\frac{1}{f(x)}dx\leq (b-a)^2$$  For the first one  $$\frac{1}{M}\int _a^bf(x)+m\int _a^b\frac{1}{f(x)}dx\geq {\frac{m}{M}}(b-a)+{\frac{m}{M}}(b-a)$$  and the last two I am not getting. How to find these?
48826
Suppose that $f$ is continous on $[0,1]$ , differentiable on $(0,1)$ , and $f(0)=0$ and  $f(1)=1$.For every integer $n$ show  that there must exist $n$ distinct points $\alpha_1,\alpha_2\cdots,\alpha_n$ in that interval so that    $\sum \limits_{k=1}^n \frac{1}{f'(\alpha_k)} = n $   Is it theorem or i can show it with basic differentiation knowledge .Please give me hint .
80158
Let $f(x)=\int_0^1|t-x|t~dt$ for all real $x$. Sketch the graph of $f(x)$, what is the minimum value of $f(x)$    I could not in any way understand how to approach this problem. I think I will be able to plot the graph if I know how the function looks like, so one can wish to avoid that part.   To find the minimum value I thought of differentiating the function. But, there is $x$ on one side and $t$ along with $x$ on the other side. I got really confuse how should I do that. Please help.
110437
Let $f$ be a function defined on the interval $(-1,1)$ such that for all $x,y\in(-1,1)$,  $f(x+y)=\frac{f(x)+f(y)}{1-f(x)f(y)}$.  Suppose that $f$ is differentiable at $x=0$.  Show that $f$ is differentiable on $(-1,1)$.  For me I can see that $\tan(x)$ is obviously a case of $f$, but I cannot find a general proof for this $f$. Could any kind soul help?
75005
If $f$ is real-valued and continuously differentiable on $\mathbb{R}$, prove that $$  \left(\int|f|^2dx\right)^2\le 4\left(\int|xf(x)|^2dx\right)\left(\int|f'|^2dx\right) $$  Attempt: I tried the Cauchy-Schwarz inequality as well as the Plancherel theorem, but none of them seems to work.
41002
I was trying to solve a question of an entrance exam. I am completely stuck in the problem. I am not able to find idea how to proceed. Please help me.     Let $A$ be the set of twice continuously differentiable functions on the interval $[0, 1]$ and let $B = \{f \in A : f(0) = f(1) = 0, f'(0) = 2\}$. Find the value of $$\text{min}_{f\in B} \displaystyle \int_0^1 (f''(x))^2dx.$$    I am really sorry for not showing some effort from my side but I can not find ant way to proceed.  Please help me. Thnx in advance.
103888
I am working on proving the below inequality, but I am stuck.  Let $g$ be a differentiable function such that $g(0)=0$ and $0&lt;g'(x)\leq 1$ for all $x$. For all $x\geq 0$, prove that $$\int_{0}^{x}(g(t))^{3}dt\leq \left (\int_{0}^{x}g(t)dt  \right )^{2}$$
54831
This question was asked in an entrance test for an undergraduate mathematics program today, held all over India.   Question: $f$ is a differentiable function in $[0,1]$ such that $f(f(x))=x$ and $f(0)=1$.  Find the value of $\int_{0}^{1} (x-f(x))^{2016} dx$.  I tried to solve it by substituting $f(f(x))$ in place of $x$, but could not proceed much further. Any hints or solutions will be appreciated.
21849
Let $f(x,y)=x(y\ln y-y)-y\ln x.$ Find $\max_{1/2\le x\le 2}(\min_{1/3\le y\le1}f(x,y))$.   This problem is quite easy and it is from Spivak; it is the part $c)$ of the general exercise 2-41 page 43 Calculus on manifolds; here it is:     Let $f:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ be twice continuously differentiable. For each $x\in \mathbb{R}$ define $g_x(y)=f(x,y)$. Suppose that for each $x$ there is a unique $y$ with $g'_x(y)=0$; let $c(x)$ be this $y$.      $a)$: If $D_{2,2}f(x,y)\ne0$ for all $(x,y)$ show that $c$ is differentiable and $c'(x)=-\frac{D_{2,1}f(x,c(x))}{D_{2,2}f(x,c(x))}$      $b)$: Show that if $c'(x)=0$, then for some $y$ we have $D_{2,1}f(x,y)=0$, $D_2f(x,y)=0$.   I cannot visualize how part c) relates to the previous ones. Can you give me a hint?
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 $2$.