Let $f$ be a differentiable function with $|f'(x)|\leq1$ and $f(-3)=-3, f(3)=3$. Then find $f(0)$.   I think this is mean value theorem problem. But I can't solve... help me please.
122493
Given $f(x)$, which is differentiable at every point such that: $f'(x) \ge -5$ for every $ x \in R$   $f(2) = -13$, $f(9) = -48$  Prove that:$ f(3) = -18$  Now it's quite obvious that $ f(3) = -18$ since that if we know that the lowest value of the derivative is -5, the difference between $f(2)$ and $f(9)$ is $-48-(-13)=-35$  that means that that in the area $[2,9]$, $f'(x)=-5$ right? that's why I told that $f(3) = -18$, but I feel that my proof is insufficient, probably because I'm missing some theorems or such. can someone help me structure this proof into something RIGHT :)?  Thanks a lot!
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?
94971
let $f(\frac{x}{3})+f(\frac{2}{x})=\frac{4}{x^2}+\frac{x^2}{9}-2$ then find $f(x)$     My Try :  $$f(\frac{x}{3})+f(\frac{2}{x})=(\frac{2}{x})^2-1+(\frac{x}{3})^2-1$$  So we have :  $$f(x)=x^2-1$$  it is right ?Is there another answer?
43089
I have no idea how to do this. Any help would be appreciated. The chapter I'm on is about differentiation and the mean value theorem.  Prove $a^\alpha b^{1-\alpha} \le \alpha a + (1 - \alpha)b, \; a,b &gt; 0,\; 0 &lt; \alpha &lt; 1$
30364
I know and understand the mean value theorem. But at the moment I don't have the intuition to understand the generalized mean value theorem     If $f$ and $g$ are continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then there exists a point $c\in(a,b)$ where$$[f(b)-f(a)]g'(c)=[g(b)-g(a)]f'(c).$$If $g'$ is never zero on $(a,b)$, then the conclusion can be stated as$$\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}.$$   What is the intuition? And how can you prove this (generally) ?
93339
Let $f$ be differentiable in $[a,b]$ and $f(a)=0$. If $\exists M \in R$ such that $\vert f'(x) \vert \leq M \vert f(x)\vert \ $,  $\forall x \in [a,b]$, then $f(x)=0, \ \forall x \in [a,b]$.  $\\$ I have an idea, but I can´t see it clear. I have arrived to the inequality $\vert f(x)\vert \leq  K\vert f(y)\vert $ for $y \in (a,x)$, supposing $k&gt;0$ and repeating the process $\vert f(x)\vert \leq  K^n\vert f(z)\vert$ for $a&lt;z&lt;y&lt;x$. If $K&lt;1$, then if $n \rightarrow \infty \ $, $f(x)=0$. That´s a sort of intuitive idea.
105572
Let's assume the functions in question are continuously differentiable. If $f(g(x)) = f(x)g(x)$, then: $$f'(g(x))g'(x) = f'(x)g(x)+f(x)g'(x)$$ I attempted to find a particular solution by letting $g(x) = x$. Then, $$f'(x) = xf'(x) + f(x)$$ $$df = xdf + fdx$$ $$\frac{df}{f} = \frac{dx}{1-x}$$ $$\ln(f(x)) = -\ln(1-x)+C$$ $$f(x) = \frac {A}{1-x}$$ Where $A$ is a constant. This solution was fairly unsatisfying to me, and upon checking it, I see it doesn't actually work, as letting $g(x) = x$ leads to $f(x) = xf(x)$. I was wondering if there were better answers, or at least a less trivial way to go about the problem.
108439
Let $f:(-1,1) \rightarrow \mathbb{R}$ be infinitely differentiable on $(-1,1)$, $f(0)=1$ and   $|f^{(n)}(x)| \le n!$ for every $x\in(-1,1)$ and for every $n\in \mathbb{N} $  $f'(\frac{1}{m+1})=0$ for every $m \in \mathbb{N}$  Determine the value of $f(x)$ for every $x\in(-1,1)$.  I guess $f(x)=1$ for every $x\in(-1,1)$.
40532
$f(x)$ is a differentiable function and $g(x)$ is a double differentiable function such that $|f(x)|\leqslant 1$ and $f'(x)=g(x)$. If $$f(0)^2+g(0)^2=9$$ then prove that there exists some $c\in(-3,3)$ such that$ \ \ g(c) \cdot g''(c)&lt;0$.  Attempt: Let us define a function $h(x) = g(x) g~'(x)$. Then $$h'(x) =  g(x)g''(x) + \left( g'(x) \right)^2 \tag 1$$ If we prove that for some $c \in (-3,3), h~'(c) &lt; 0,$ then $$g(c)g''(c) &lt;0 \tag 2$$ Also, $$\left|f(0)\right| &lt; 1 \implies f'(0) \in (-3,-2\sqrt 2 ) \cup (2\sqrt 2,3) $$ Could someone please advise me how do I move forward from here. Thank you very much for your help in this regard.
59481
Let $f(x)$ be twice-differentiable and $f(a)=f(b)=0$ with $a&lt;b$  How to prove that  $$\max|f(x)| \le \frac 18(b-a)^2 \max|f''(x)|$$  (With $\max|f(x)|$ beeing maximum (or minimum) between a and b.)  How would I use Taylor-Expansion and Mean Value Theorem here?
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! :)
110245
Problem.  Show that $$e^x &gt; 1 + x , \ \ x \neq 0$$  My attempt.  Using Mean Value Theorem: $$f'(c) = \frac{f(b) - f(a)}{b-a}$$  $$\Rightarrow e^c = \frac{e^x - 1}{x}$$ $$\Rightarrow xe^c = e^x -1$$ $$\Rightarrow x+1 = \frac{e^x}{e^c}$$  I'm not feeling very comfortable with these calculations...Am I on the right track? Can I even choose $e^o$ in my initial calculation as it says that $x \neq 0 $ ?
88381
Let $\,f$ be three times differentiable on the interval $[a, b]$. Show that there is a $c$ such that: $$f(b) - f(a) = \frac{b-a}{2}(f'(a)+f'(b)) - \frac{(b-a)^3}{12}f'''(c).$$ Answered in the comments.
101332
Suppose $f:[0,\infty)$ $\rightarrow$ R and has continuous $f'(x)$ and $f''(x) $ .  $f(x) $$\rightarrow $0 as $x$ $\rightarrow$ $\infty$. Given $f'(x)\rightarrow b$ as $x \rightarrow \infty$  show that $b=0$  Get stuck in the first place, but I tried to use Mean Value Theorem and then connected it to the limits.  Any hint?
72025
Let $f(x)$ be a function which is differentiable on $[0,1]$ with $f(0)=0$ and $f(1)=1$. Show that for every $n\in \Bbb N$ there exists numbers $x_1,x_2,\ldots,x_n\in [0,1]$ such as $$  \sum_{k = 1}^n \frac{1}{f' (x_k)} = n $$ I think the mean value theorem should be applied. So there exists $x_1$ in $[0,1]$  such that $f ' (x_1) = \frac{f(1) - f(0)}{1-0} =1$  and there exists $x_2$ in $[0,x_1]$ such that $f ' (x_2) = \frac{f(x1) - f(0)}{x1-0} = \frac{f(x1)}{x1}$, so on and so forth for $x_3 ,x_4, \ldots,x_n$ and we have the sum $$1+\frac{x_1}{f(x1)} + \frac{x_2}{f(x2)} +\cdots+\frac{x_{n-1}}{f(x_{n-1})}$$  and from here I have no idea what to do . I was wondering if anyone could be so kind to help ?
76314
This is an exercise from Abbott's second edition of Understanding Analysis.     Let $f$ be differentiable on an interval $A$. Show that if $f'(x) \neq 0$ on $A$, show that $f$ is one-to-one on $A$. Provide an example to show that the converse statement need not be true.   Is my solution (below) correct?   Let $x_{1},x_{2} \in A $ such that $x_1 \neq x_2$. Since the function satisfies all the conditions of the mean value theorem, there exists $c \in (x_1,x_2)$ [without loss of generality we consider $x_1 &lt; x_2$] such that $f(x_2) - f(x_1) = (x_2 -x_1) f'(c) \neq 0$ as it is given that $f'(x)$ is nonzero on $A$ and $x_1 \neq x_2$ is our assumption. Therefore $x_1 \neq x_2$ implies $f(x_1) \neq  f(x_2)$ for every $x_1,x_2 \in A$. Thus $f$ is one-to one on $A$.  The converse may not be true. Consider $A=[-1,1)$ and $f(x)= x^3$. Therefore $f$ is injective on $A$. Again $f'(x)= 2x^2$. Therefore $f'(0)=0, 0 \in A$. So the converse need not be true.
63879
Find the maximum value of $f(x)$ when $0 \leq x \leq 1$ and $$ f(x) = \int^{x}_{0} \sqrt{(x^2-x)^2+y^4}~dy$$   I'm guessing that the Fundamental Theorem of Calculus can't be used because the integrand is a function of $x$.  Any help would be greatly appreciated.