I have the following task in my homework:  Let $a,b \in \mathbb{R}, a &lt; b $ and $f:[a,b] \to \mathbb{R} $ be a continuous function that is differentiable on $(a,b)$. Show that if $f'(x)=f(x)$ for all $x \in (a,b)$, then there is a $c \in \mathbb{R}$ with $f(x)=ce^x$ for all $x \in [a,b]$. Hint: Consider the function $x\mapsto f(x)e^{-x} $  I know that the two solutions here would be $c=1$ and $c=0$, but apparently, that has to be shown with the mean value theorem, and that hint just confuses me more. Any help would be appreciated!
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I was wondering on the following and I probably know the answer already: NO. Is there another number with similar properties as $e$? So that the derivative of $\exp(x)$ is the same as the function itself. I can guess that it's probably not, because otherwise $e$ wouldn't be that special, but is there any proof of it?
101313
Let $f\colon\mathbb{R}\to\mathbb{R}$ be a differentiable function. $x\in\mathbb{R}$ is a fixed point of $f$ if $f(x)=x$. Show that if $f'(t)\neq 1\;\forall\;t\in\mathbb{R}$, then $f$ has at most one fixed point.  My biggest problem  with this is that it doesn't seem to be true. For example, consider $f(x)=x^2$. Then certainly $f(0)=0$ and $f(1)=1 \Rightarrow 0$ and $1$ are fixed points. But $f'(x)=2x\neq 1 \;\forall\;x\in\mathbb{R}$. Is there some sort of formulation that makes this statement correct? Am I missing something obvious? This is a problem from an old exam, so I'm assuming that maybe there's some sort of typo or missing condition.
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Question is to prove :  $f : S^1 \to \mathbb R$ is continuous then $f(x)=f(-x)$ for some $x\in S^1$  I guess it would be helpful to use intermediate value theorem  Assuming $f(x)\neq f(-x)$ then given any $p\in (f(-x),f(x))$ (assuming $f(-x)&lt;f(x)$) there exists $y\in (-x, x)$ such that $f(y)=p$  I am not very sure of how to use this..   It would be helpful if someone can give some hint which would help me to solve this..  Thank you.
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If $ f: \mathbb{R} \to \mathbb{R} $ is continuous, and $\forall x \in \mathbb{R} :\;(f \circ f \circ f)(x) = x  $, show that $ f(x) = x $.  The condition that $f$ is continuous on $\mathbb{R}$ is crucial for the proof. I can find functions such as $\displaystyle\frac{x-3}{x+1}$ that satisfies  $ (f \circ f \circ f)(x) = x $. I have tried to negate the conclusion to see if there's a contradiction, but got stuck.
32030
If $f\colon \mathbb{R} \to \mathbb{R}$ is such that $f(x+y)=f(x)f(y)$   and $f'(0)$ exists, then $f$ is differentiable on $\mathbb R$.   I know how to show it is continuous but no clue how to show differentiable.
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Let $f(x)$ be differentiable at $(a,b)$ and continuous at $[a,b]$ prove that there exist $ c \in (a,b)$ such that: $$ \frac{af(a)-bf(b)}{a-b} = f(c) + cf'(c) $$    I started with: let $c \in (a,b) $  $$ f'(c) = \frac{f(b)-f(a)}{b-a} $$  But I have no clue how to continue from here, how can I transform it to this form? :S  Any suggestions?
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I had an exam today, and I was asked about the inverse function theorem, and the exact conditions and statement (as stated in Mathematical Analysis by VA Zorich): Let $X, Y \subset \mathbb{R}$ be open sets and let the functions $f: X \to Y$ and $f^{-1}: Y \to X$ be mutually inverse and continuous at points $x_{0} \in X$ and $y_{0}=f(x_{0})$, respectively. If $f$ is differentiable at $x_{0}$ and $f'(x_{0}) \neq 0$, then $f^{-1}$ is also differentiable at $y_{0}$  and its derivative is $$(f^{-1})'(y_{0})=(f'(x_{0}))^{-1}=\frac{1}{f'(x_{0})}.$$  Then I was asked to come up with an example which would show that the condition that $f^{-1}$ be continuous at $y_{0}$ is not redundant, i.e. that there exists a differentiable bijection $f: \mathbb{R} \to \mathbb{R}$ such that $f'(x) \neq 0$ for all $x \in \mathbb{R}$, but whose inverse $f^{-1}$ is non-differentiable at some point $y_{0} \in \mathbb{R}$ (my professor stated that both the domain and range are $\mathbb{R}$, but I'd be open to any example whose domain/range are open subsets of $\mathbb{R}$). I couldn't come up with an answer on the spot, so I was left with a homework assignment, which I'm shamelessly asking for help with here.  Here are some examples that almost, but don't quite fit the bill:   $f(x)=x^3, f:\mathbb{R} \to \mathbb{R}$, whose inverse $f^{-1}(y)=\sqrt[3]y$ is not differentiable at $0$, but the problem is that $f'(0)=0$. If $f'(x) \neq 0$ were removed as a conditions, many counterexamples could be easily found, because any bijection $f$ whose derivative is zero at $x$ implies that $f^{-1}$ is not differentiable at $f(x)$. An example which is possibly closer to what I'm looking for is the one found in the answer of , which fits the bill completely except for the domain/range, because $f^{-1}$ is not continuous at 1, let alone differentiable, but its domain is not an open or connected set, so it wouldn't be easy to extend its domain to $\mathbb{R}$ and still keep all of its properties.   I'm aware that the inverse $f^{-1}$ of a continuous bijection $f$ defined on an interval ($\mathbb{R}$ in this case) is also continuous, so the "pathological inverse" that I'm looking for is continuous (unlike example 2), but not differentiable (i.e. "spiky") at a point, even though $f$ isn't "spiky" anywhere.
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I ran across this problem in my Analysis class and can't seem to come up with a good solution. Here's the question: Show that if $\lim_{x\to a}f'(x) = A$ then $f'(a)$ exists and equals $A$. $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$.  I tried applying the Mean Value theorem like this: There exists a $c \in (a, x)$ st $f(x)-f(a)=f'(c)(x-a) \implies \frac{f(x)-f(a)}{x-a}=f'(c) \implies f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{c\to a}f'(c)=A$  However, I think this is wrong. I am pretty sure the Mean Value theorem is needed, but my application of it seems to be incorrect. Would somebody be able to point me in the right direction?
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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.
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I'm annoyed by quite a simple problem in calculus (I apologize in advance if I'm not using adequate terms in English, I don't take the course in English nor am I a native speaker):  Let $f:\mathbb R\longrightarrow \mathbb R$ be a function which is continuous at $0$ and satisfies $f(x)=f(2x)\,\forall x\in\mathbb R$.  My question is how do I formally prove that for every $x&lt;0$ $f(x)=c_1$, and for every $x&gt;0$ $f(x)=c_2$, where $c_1,c_2\in \mathbb R$? After proving those two statements (or at least one of them WLOG) it would be quite easy to prove that $f(0)=c_1=c_2$ by the definition of continuity of $f(0)$ and thus $f(x)$ is constant. Those two statements are just so obvious to see that I can't think of any formal way of proving so without "cheating". Thanks in advance!
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I know that this can not be true due to counter-examples but I don't know where the error in my reasoning is.  Assumption: If $f(x)$ is differentiable in $\mathbb{R}$ then the derivative $f'(x)$ is continuous in  $\mathbb{R}$.  Faulty Proof: For every $c \in \mathbb{R}$, using the mean value theorem for $f(x),$ on the interval $x \in [c, c + h] $ where $h$ is positive.  $$ \frac{f(c + h) - f(c)}{h} = f'(\xi(h)) $$  Where $\xi(h) \in (c,c+h)$. Because this equation holds for every $h&gt;0$. It must hold in the limit as $h \rightarrow 0^+$.   $$ \lim_{h\to 0^+}\frac{f(c + h) - f(c)}{h} = \lim_{h\to 0^+}f'(\xi(h))  $$  But the left side of the equation is the right one sided derivative.  $$ f'_{+}(c) = \lim_{h\to 0^+}\frac{f(c + h) - f(c)}{h} = \lim_{h\to 0^+}f'(\xi(h))  $$  The same can be done for $h$ being negative, but because of differentiability at every point the left and right derivatives must be equal.    $$ f'(c) = f'_{+}(c) = f'_{-}(c) = \lim_{h\to 0^-}\frac{f(c + h) - f(c)}{h} = \lim_{h\to 0^-}f'(\xi(h))  $$  As $h \rightarrow 0^+$, $\xi(h) \rightarrow c$. So because the limit $\lim_{h\to 0^+}f'(\xi(h))$ exists and $\xi(h) \neq c$, it is equal to $\lim_{x\to c^+}f'(x)$  It follows that $\lim_{x\to c^+}f'(x) = \lim_{x\to c^-}f'(x) = f'(c)$ so the function $f'(x)$ is continuous.
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Assume that $f: \mathbb{R}^2-\{0\} \rightarrow \mathbb{R}$ is continuously differentiable and $f(\alpha x) = \alpha^2f(x)$ for all $x\neq 0$ and $\alpha &gt; 0$. Then I want to prove that $x \cdot \nabla f(x) = 2 f(x)$ when $x$ is nonzero.  I applied Mean value theorem in several variables and tried some linearity tricks but I could not prove it.
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$f:\Bbb R \to\Bbb R $ is differentiable, $f(0)=0$ and $|f'(x)|\le|f(x)|$ for all $x$ then prove $f$ is identically zero.  I tried to use mean value theorem and end up in $|f(x)|\le |x||f(c)|$ for some constant $c$ which depends on $x$. Can you help me little bit...
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mathematicians! I want to ask to all wise people about a problem I met at the quiz to obtain some ideas. The problem is following. It may not be accurate since the problem is dependent on my memory   Let $f : [0, \infty) \rightarrow \mathbb{R}$ and $f$ is differentiable. $|f'(x) |\; \leq\  |f(x)|$ for all $x \in [0, \infty)$. $f(0) = 0$. Then prove that $f(x) = 0$ for all $x \in [0,\infty)$.  At the quiz, I applied the Mean Value Theorem and the Cauchy-Schwarz inequality.   For any $x$, $|f(x)| = |\{f(x) - f(0)\}(x-0)| = |f'(c_{1})(x)|$ for some $c_{1} \in (0,x)$, and then  $$|f'(c_{1})(x)| \leq |f'(c_{1})||x| \leq |f(c_{1})||x|$$ by Cauchy-Schwarz and given assumption. By doing so consecutively, we can obtain the inequality that  $$ |f(x)| \leq |f(c_{n})||x||c_{1}|\ldots |c_{n-1}| $$ for any $x \in [0,\infty)$. As $\{c_{i}\}, \; i\in \mathbb{N}$ is decreasing sequence and $f$ is continuous, $|f(c_{n})|$ can be arbitrary close to $0$ when $c_{n} \rightarrow 0$. Thus the inequality above can be bounded by $\epsilon$.  I think my answer is right, but a bit uncertain. Could you give me a certainty? Thank you very much.
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Here's a statement in Zygmund's Measure and Integral on page 17:     If $f$ has a continuous derivative on $[a,b]$, then (by the mean-value theorem) $f$ satisfies a Lipschitz condition on $[a,b]$.   This does not seem obvious to me. How can I show it?  Also, what does a continuous derivative imply? Can we conclude the function is differentiable? If so, how can I prove it?
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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) ?
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Assume that $f: \mathbb{R} \rightarrow \mathbb{R}$  is differentiable.  If there exists an $L&lt;1$ such that for each $x \in \mathbb{R}$, $f'(x)&lt;L$, prove that there exists a unique point $x$ such that $f(x)=x$.  So far I have that where I fixed $a$ to be zero and came up with three cases to show existence for  $f(0) = 0$, $f(0) &lt; 0$, and $f(0) &gt; 0$  $f(0)=0$ was fairly trivial following from the mean value theorem but the other two cases I was only able to eliminate half of the possible fixed points by contradiction and I need help showing it explicitly.  Edit as to avoid confusion here is the work I have garnished so far:  From the mean value theorem it follows for some arbitrary $a,b \in \mathbb{R}$ and $a\neq b$ that there exists a point $c$ such that $a&lt;c&lt;b$ or $b&lt;c&lt;a$ then implies that $\frac{f(b)-f(a)}{b-a}=f'(c)$ and this is because it is differentiable on $\mathbb{R}$ and, by convention, is also continuous on $\mathbb{R}$ as well.  Then the result of $\dfrac{f(b)-f(a)}{b-a}=f'(c)$ then follows for $L&lt;1$ such that $\forall x \in \mathbb{R}$ means $f'(x)&lt;1$ from our hypothesis. Which then follows  (*) $\dfrac{f(b)-f(a)}{b-a}=f'(c)&lt;L&lt;1 \Rightarrow f(b)-f(a)&lt;L(b-a)&lt;b-a$  Now without loss of generality choose $a=0$ and let $b$ be some arbitrary value in $\mathbb{R}$ then there are three cases that need to be considered for the value of $f(a)$:  (i) $f(0) = 0$, (ii) $f(0) &lt; 0$, (iii) $f(0) &gt; 0$  (i) For $f(0)=0$   Then $0$ is a fixed point of point of $f(x)$ and it follows from (*) that if $f(0)=a=0$ then $f(b)-f(0)=f(b)&lt;Lb&lt;b$ which shows that $f(b)&lt;b$ thus shows that for any arbitrary value $b\in \mathbb{R}$ that no $f(b)=b$ thus shows that $0$ is the only fixed point if $f(0)=0$  (ii) For $f(0)&lt;0$  Suppose there exists a fixed point $d&gt;0$ then:  $\frac{f(d)-f(0)}{d}&lt;L&lt;1 \Rightarrow d-f(0)&lt; dL&lt;d \Rightarrow -f(0)&lt;d(L-1)&lt;0\Rightarrow 0&lt;-d(L-1)&lt;f(0)$ but this contradicts that $f(0)&lt;0$  thus no other fixed point exists if $d&gt;0$ when $f(0)&lt;0$  (iii) For $f(0)&gt;0$