Let $f\in C^1[a,b]$, $f(a)=0$ and suppose that $\lambda\in\mathbb R$, $\lambda &gt;0$, is such that   $$|f'(x)|\leq \lambda |f(x)|$$for all $x\in [a,b]$. Is it true that $f(x)=0$ for all $x\in [a,b]$?   Solution: Say there exits an interval $[c,d]$ where funtion $f(x)$ is non zero so $|f(x)|$ is differentiable in $[c,d]$ with  $f(c)=0$. Now consider the function $g(x)=e^{-\lambda x}|f(x)|$ and see that $g'(x)\leq 0$. So $g $ is decreasing. So, $x\geq a \implies g (x)\leq g(a)=0$. Hence, $|f (x)|\leq 0\implies f (x)=0$  Is this solution correct? basically i am asking is the differentiablity of $|f(x)|$ makes sense
123647
$f$ is differentiable on $[a,b]$, $f&#39;(x) \leq A|f(x)|$ where $A$ is a non-negative constant.  If $f(a)=0$ show $f(x)=0, \forall x\in [a,b]$  I imagine the proof uses the Mean Value Theorem but I have not been able to get it to work.  I know $|f(x)|=|f&#39;(c)|(x-a)$ where $c \in [a,x]$, so $|f(x)| \leq A\  |f(c)|(x-a)$ where $c\leq x$ And I guess I could sort of iterate this to keep getting a smaller $c$ but I don't see why it must go all the way to zero.
79431
Let $f: \mathbb{R} \to \mathbb{R}$ be differentiable on $(a,b)$. Suppose $f' &gt; 0$ on $(a,b)$ except at a point $c \in (a,b)$ (that is, $f'(c) \leq 0$).  Is $f$ increasing on $(a,b)$? Must $f'(c)$ be zero, or can it be negative?  Clearly $f$ is increasing on $(a,c) \cup (c,b)$ but I'm not sure about how the value at $c$ compares with the values at other points. And I think $f'(c)$ must be zero: If $f'(c) &lt; 0$ then for small positive $h$ we have $\frac{f(c+h) - f(c)}{h}$ is also negative (by definition of the derivative as a limit of this ratio), so $f(c+h) - f(c) &lt; 0$. Since $f'$ is positive on $(c,c+h)$, the Mean Value Theorem implies that $f(c+h) - f(c) = f'(d)h$ for some $d \in (c, c+h)$, and $f'(d)h$ is a product of two positive numbers, hence positive. So $f(c+h) - f(c) &gt; 0$, a contradiction.
43481
PROBLEM. Let $f:[a,b]\to\mathbb R$ be a continuous function, such that $f'(x)\ge 0$, for all $x\in [a,b]\setminus A$, where $A\subset [a,b]$ is a countable set. Show that $f$ is increasing.  Attention. In this problem, we DO NOT assume that $f$ is differentiable in the whole $[a,b]$.  Notes. (1) If we assume that $f$ is differentiable in the whole interval, then we can easily show that $f'(x)\ge 0$, everywhere. For otherwise, if $f'(x_0)=c&lt;0$, for some $x_0\in [a,b]$, then by virtue of , $(c,0)\subset f'([a,b])$, and hence, $f'(x)&lt;0$, for uncountably many $x$'s.   (2) The conclusion of the problem does not hold if we replace the assumption $A$ is countable with $A$ is a set of measure zero. Take for example the , with a negative sign in front.  (3) If the hypothesis $f'(x)\ge 0$, is replaced by $f'(x)=0$, then the conclusion becomes f is constant.
81832
Please help me to solve this question:  Suppose $f:[a,b] \to \Bbb R$ satisfies:  $f''(x)+f(x)&gt;0$ and $f(x)&gt;0$ for all $x\in(a ,b)$; $f(a)=f(b)=0$.  Prove that $b-a&gt;\pi$.  Thanks in advance.
28293
I need to prove that  $$\int^1_0 f(x)dx=f(0)+\frac{1}{2}f'(a)$$ where $f$ is continuous and differentiable in $[0,1]$ and $a\in(0,1)$.    I know that as $f$ is continuous in $[0,1]$ there exists $c\in (0,1)$ such that $\int^1_0 f(x)dx=f(c)$. Now maybe I should show that $f(c)=f(0)+\frac{1}{2}f'(a)$ ? If I should I don't know how to do it. The other approach that I had, was going by the definition of $\int^1_0 f(x)dx$ to see if I could somehow get a similar result to the right side of the equation above, it didn't seem to lead anywhere. Do you have any idea how to approach it?
2389
The following result is well known:     If $f$ is continuous on $[a, b]$, differentiable on $(a, b)$ and $f'$ is non-zero on $(a, b)$ then $f$ is strictly monotone on $[a, b]$.   However if the derivative vanishes at a finite number of points in $(a, b)$ and apart from these derivative maintains a constant sign in $(a, b)$ then also the function is strictly monotone on $[a, b]$ (just split the interval into finite number of intervals using these points where derivative vanishes and $f$ is strictly monotone in same direction in each of these intervals).     Let's suppose now that $f$ is strictly monotone and continuous in $[a, b]$ and differentiable in $(a, b)$. What can we say about set of points $$A = \{x \mid x \in (a, b), f'(x) = 0\}$$ Can it be infinite? Can it be uncountable? How large the set $A$ can be?
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.
31576
$f:I \to \mathbb R$ is a convex function on an open interval $I$. I know that if $f$ is differentiable on $I$, then $\forall x_0,x \in I , f(x_0)+f'(x_0)(x-x_0) \le f(x)$  I read somewhere that something similar true even if $f$ is not differentiable, i.e. $\exists a \in \mathbb R: \forall x_0, x \in I, f(x) \ge a(x-x_0)+f(x_0)$.  Does anybody know a proof for that? It would e.g. help to show that $\forall x_0 \in I$ there is a convex differentiable function $g:I \to \mathbb R, g(x_0)=f(x_0)$ and  $\forall x \in I,  g(x) \le f(x)$
94771
I need to find a pair of functions $f$, $g$ such that    $f$ is not differentiable at  $x = 0$ $g$ is not differentiable at $f(0)$ $g \circ f$ is differentiable at $x = 0$   I've tried a lot of functions but just can't seem to find ones that work. Any help would be appreciated, thank you.
85008
Let f be a function continuous on [0,1] and differentiable on (0,1)  My attempt:  I have tried solving using Rolle's theorem for f'(c) = 0  I let c&lt;x then by Mean Value Theorem, f'(c) = (f(x)-f(0))/(x-0) = f(x)/x since f'(c) = 0, f(x) = 0 Hence f'(c) = f(x)  Similarly, I let c>x and by Mean Value Theorem again, f'(c) = (f(1)-f(x))/(1-x)       = f(x)/(1-x) since f'(c) = 0, f(x) = 0 Hence f'(c) = f(x)  I feel like this proof is wrong because I don't know how to continue with the case where c=x
111680
Suppose $f'(x)=0$ for all $x\in (a,b)$. Prove that $f$ is constant on $(a,b)$.   This seems painfully obvious, but I can't prove it rigorously.  $f'(x)=0$ for all $x\in (a,b)$ means that for any $c\in(a,b)$, we have $$\lim_{x\rightarrow c}\frac{f(x)-f(c)}{x-c}=0$$ So for any $c\in (a,b)$, for any $\epsilon$ there exists $\delta$ such that if $|x-c|&lt;\delta$ then $$\left|\frac{f(x)-f(c)}{x-c}\right|&lt;\epsilon$$ which also means $$|f(x)-f(c)|&lt;\epsilon\delta$$ How to continue from here?
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.
77549
Let $f:[0,1]\to\mathbb R$ is a continuous function, and for all $x\in [0,1]$, $f(x)$ is either a local maximum or a local minimum. Then prove that $f$ is a constant.  Here is what I have tried, I don't know whether it is correct:  Assume $f$ is not a constant, by continuity, there exist $x_1$, $x_2$ such that $f(x_1)=m$ is a global minimum and $f(x_2)=M$ is a global maximum. Now if $m\ne M$, choose any $c\in(m,M)$ and define  $$x_0=\sup (x\in[x_1,x_2]:f(x)&lt;c)$$  Then we have $f(x_0)$ NOT a local minimum nor a local maximum (since according to the definition of $x_0$, there are always some points $&lt;c$ in the left neighborhood of $x_0$ and the points in right neighborhood are always $&gt;c$). Contradiction arises and so $f$ must be a constant.
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.
16518
I have a question about the corollary to theorem 5.12 in Rudin's Principles of Mathematical Analysis (page 108):  Suppose $f$ is a real differentiable function on $[a,b]$ and suppose $f'(a)&lt; \lambda &lt; f'(b)$ then there is a point $x \in (a,b)$ such that $f'(x) = \lambda$ Corollary: If $f$ is differentiable on $[a,b]$ then $f'$ cannot have any simple discontinuities on $[a,b]$.  Can someone help me to show how he uses the result in the &quot;main theorem&quot; in the corollary?  (There are two cases of simple discontinuities  $f(x+) = f(x-) \neq f(x)$ and $f(x +) \neq f(x-)$
79487
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.
86039
We know that a function $f: [a,b] \to \mathbb{R}$ continuous on $[a,b]$ and differentiable on $(a,b)$, and if $f'&gt;0 \mbox{ on} (a,b)$ , f is strictly increasing on $[a,b]$. Is there any counterexample that shows the converse fails?   I have been trying to come up with simple examples but they all involve functions that are discontinuous or has derivative $f'=0$ which does not agree with the hypothesis hmmm