I wanna prove this problem. I tried it with Mean Value Theorem but cannot proceed to any plausible result. So could I have some hints?
31460
Let $f:\mathbb{R} \to \mathbb{R}$ be a twice differentiable function. And let  $$\eqalign{   &amp; M_0  = \sup \left|f(x)\right|  \cr                &amp; M_1  = \sup \left|\frac{d}{dx} f(x) \right|  \cr                &amp; M_2  = \sup \left|\frac{d^2}{dx^2} f(x) \right| \cr }$$ Prove that  $$M_1 ^2  \leqslant 4M_0 M_2$$ I can not think how I can relate these values ​​in some inequality )=
86705
Let $f:[0,1]\to\mathbb{R}$ be continuous with $f(0)=0$, $f(1)=1$ and $f$ is differentiable on $(0,1)$. Show that there are distinct $\xi,\eta\in(0,1)$ so that $f'(\xi)f'(\eta)=1$.  I think this requires mean value theorem. But this does not help since if apply the theorem to $f$ we can only get $f'(c)=1$ for some $c$.
113749
This is an exercise from Kelley's book. Could someone help to show me a proof?   It seems very natural, and it is easy to prove by utilizing the arctan function in $\mathbb{R}^1$.  Thanks a lot.
30673
I am asked to prove the second mean value theorem:  Let $f$ and $g$ be defined on $[a,b]$ with $g$ continuous, $f\ge 0$, and $f$ integrable. Then there is a point $x_0 \in (a,b)$ such that $$ \int_{a}^{b}f(x)g(x)dx = g(x_0) \int_{a}^{b}f(x)dx. $$  My (incomplete) proof goes as follows:  Since $ [a,b] $ is compact and $ g $ is continuous, we know there exist $ x_1, x_2 \in [a,b] $ such that $ g(x_1) \le g(x) \le g(x_2) $ for any $ x \in [a,b] $. Thus, we have the following string of inequalities $$ f(x)g(x_1) \le f(x)g(x) \le f(x)g(x_2), \, \, \forall x \in [a,b], $$ $$ \Rightarrow g(x_1)\int_{a}^{b} f(x)dx \,\le \, \int_{a}^{b} f(x)g(x) dx \, \le  \, g(x_2)\int_{a}^{b}f(x)dx. $$  My problem now is to show the final equality. It seems to me the intermediate value theorem is somehow involved, but I cannot see how to use it. Furthermore, how do I make use of the $f\ge 0$ condition in the hypothesis?
29377
Let $f$ be differentiable and let $\lim_{x \to \infty}f'(x)=+\infty$ prove that: 1) $\lim_{x \to \infty}(f(x)-f(x-1))=+\infty$ and 2) $\lim_{x \to \infty}f(x)=+\infty$.  1) I'll prove by contradiction. Let $\lim_{x \to \infty}(f(x)-f(x-1))=a$, where $a$ is from and $\Bbb R$. $\lim_{x \to \infty}(f(x)-f(x-1))=\lim_{x \to \infty}\frac{f(x)-f(x-1)}{x-(x-1)}=\lim_{x \to \infty}f'(c)=a$, where c is from $(x-1,x)$ (mean value theorem). As $x$ goes to infinity so does $c$ i.e. we have that $\lim_{x \to \infty}f'(c)=\lim_{c \to \infty}f'(c)=a$. Which is a contradictition since we have that  $\lim_{x \to \infty}f'(x)=+\infty$ .   2) I will also prove this by contradiction. We have that $\lim_{x \to \infty}f(x)=a$ This is equivalent to $\lim_{x \to \infty}f(x)-f(x-1)=0$. I use mean value theorem and get $\lim_{x \to \infty}(f(x)-f(x-1))=\lim_{x \to \infty}\frac{f(x)-f(x-1)}{x-(x-1)}=\lim_{x \to \infty}f'(c)=0$, where c is from $(x-1,x)$. As $x$ goes to infinity so does $c$ i.e. we have that $\lim_{x \to \infty}f'(c)=\lim_{c \to \infty}f'(c)=0$. Which is a contractition since we have that  $\lim_{x \to \infty}f'(x)=+\infty$ .
46536
I need to show that $\dfrac{2}{\pi}&lt;\dfrac{\sin(x)}{x}&lt;1$ for $0&lt;x&lt;\dfrac{\pi}{2}$.  I know I need to use the mean value theorem, would I just say that since $f$ is continuous in the interval (call it I).  We know $\exists c\in I$ such that $f'(c) = \dfrac{f\left(\dfrac{\pi}{2}\right)-f(0)}{\dfrac{\pi}{2}}$. I don't know where this is going though.  Also, would I need to show $\lim_{x \to 0}\dfrac{\sin(x)}{x} = 1$? Is there a way to do this without L'H Rule?
67714
Show that  $$\left | \sin a - \sin b \right | \leq \left | a-b \right |$$  I saw a proof using The Mean Value Theorem, but I could not grasp it well. Are there any other proofs or can someone clarify it by using The Mean Value Theorem?
19779
Show that if $u$ is harmonic in $\mathbb R^n$ and $u=o(|x|)$, then $u$ is a constant.(Hint: use the solid version of the mean value property $u(x)=\frac{1}{\omega R^n} \int _{B_{R(x)}} u(y)dy$, and estimate $|u(x)-u(x')|$.)  Can anyone tell me how to use this hint?
5503
I'm currently trying to prove the following:    And this is what I've come up with so far:    And now I'm stuck. Perhaps the next step might be to eliminate for all x?  I'm sorry for using links, but it seems like I'm not allowed to do it otherwise due to needed reputation point.  Help would be appreciated.
104441
I'm interested in the different ways of proving this, any proof is welcome. I understand one way is the cosine sum/difference formula, another is using a right angled triangle. Are there any others? Thanks.
117421
Use the mean value theorem to prove that if $0 \leq x \leq 1$, then $$x\sqrt{1-x^2} \leq \sin x \leq x$$ The theorem guarantees the existence of a point, but not an inequality, so I don't know how to begin.
79153
I want to prove or disprove this problem: If there exist $\lim\limits_{x\rightarrow \infty} (f'(x)+f(x))=L&lt;\infty$ then $\lim\limits_{x\rightarrow\infty} f(x) =L$.  When I assume problem below:  If there exist $\lim\limits_{x\rightarrow\infty} (f'(x)+f(x)) =L&lt;\infty$, There exists $\lim\limits_{x\rightarrow\infty} f(x)$?  I can use mean-value theorem to show that.  So my question is:     If $\lim\limits_{x\rightarrow\infty} (f'(x)+f(x))=L&lt;\infty$, does $\lim\limits_{x\rightarrow\infty} f(x)$ exist?
56137
There is a probabilistic method to solve it. But I am not familiar with probability. I am trying to compute it by analytic method, such as using L Hospital's rule or Stolz formula, but they are not working.
105003
If $f \colon [a,b]\rightarrow \mathbb{R}$, then show that there exists $c \in [a,b]$ such that $c \in [a,b]$ such that $\int_a^c f = \int_c^b f$.  I think that I have to proceed with the Mean Value Theorem of Integrals...
70462
I want to show that $$\frac{x}{1+x}&lt;\log(1+x)&lt;x$$ for all $x&gt;0$ using the mean value theorem. I tried to prove the two inequalities separately.  $$\frac{x}{1+x}&lt;\log(1+x) \Leftrightarrow \frac{x}{1+x} -\log(1+x) &lt;0$$  Let $$f(x) = \frac{x}{1+x} -\log(1+x).$$ Since $$f(0)=0$$ and $$f'(x)= \frac{1}{(1+x)^2}-\frac{1}{1+x}&lt;0$$ for all $x &gt; 0$,  $f(x)&lt;0$ for all $x&gt;0$.  Is this correct so far?  I go on with the second part: Let $f(x) = \log(x+1)$. Choose $a=0$ and $x&gt;0$ so that there is, according to the mean value theorem, an $x_0$ between $a$ and $x$ with   $f'(x_0)=\frac{f(x)-f(a)}{x-a} \Leftrightarrow \frac{1}{x_0+1}=\frac{ \log(x+1)}{x}$.  Since $$x_0&gt;0 \Rightarrow  \frac{1}{x_0+1}&lt;1.$$ $$\Rightarrow 1 &gt; \frac{1}{x_0+1}= \frac{ \log(x+1)}{x} \Rightarrow x&gt; \log(x+1)$$
100760
So I've spent a few hours trying to find a counter example to this statement, and then gave up and spent additional few hours trying to prove it and Ive finally became frustrated with this problem. :) I seem to get stuck in the scenario where $f\left(\frac{1}{x}\right)$ &lt; 0, I'm just unable to find an epsilon that will satisfy all the conditions.. plus I'm not even sure if the statement is true or false anymore..
121277
This statement seems so simple yet I don't quite know how to start with this proof in a substantial way.  Can anybody help me here?