Let $f(x)$ be defined and continuous on the interval $[a,b]$ and differentiable on $(a,b)$. Prove that there is at least one number $c$ such that  $$\frac{af(a)-bf(b)}{a-b}=f(c)+cf'(c).$$
121892
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?
22355
From Pugh's analysis book, prelim problem 57 from Chapter 4:     Let $f$ and $f_n$ be functions from $\Bbb R$ to $\Bbb R$. Assume that $f_n(x_n)\to f(x)$ as $n\to\infty$ whenever $x_n\to x$. Prove that $f$ is continuous. (Note: the functions $f_n$ are not assumed to be continuous.)   here's my attempt: assume $x_n \to x$. we want to show that $f(x_n) \to f(x)$. so $|f(x_n) - f(x)| \leq |f(x_n)-f_n(x_n)| + |f_n(x_n)-f(x)|$. The second term can be made to be less than any $\varepsilon &gt; 0$ for $n$ sufficiently large. i'm having trouble with the first term. can anyone help? thank you!
29517
$$f\in C[a,b] $$If given that for every $x\in[a,b]\text{ there exists } y\in[a,b]$ such that: $$|f(y)| \le \frac{1}{2}|f(x)| $$ Prove that there exists $ c\in[a,b]$ such that :   $$f(c)=0$$  What I've done so far : We will build a sequence in this way: $$\text{take } x_1\in[a,b]$$ for this $x_1$ there exists $y_1$ such that $ |f(y_1)|\le\frac{1}{2}|f(x_1)|$ $$\text{now $y_1$ will be our $x_2$, } y_1=x_2 $$ we will go on by doing this: $$\text{take } x_n\in[a,b]$$ for this $x_n$ there exists $y_n$ such that $ |f(y_n)|\le\frac{1}{2}|f(x_n)|\le \frac{1}{2^{n}}|f(x_1)|$  I know that this is the only way to prove it but can't seem to get on from here.. any help?
59735
Let $f(x)$ be a continuous function, whose first and second derivatives are continuous on $[0,2π]$ and $f&#39;&#39;(x)≥0\:\:∀\:x∈[0,2π]$. Show that$$ \int _0^{2\pi} f(x) \cos x dx \ge 0$$
70712
Suppose that $f$ is continuous on $[0, 2\pi]$ and that $f(0) = f(2\pi)$.  Prove that there exists a $c$ in $[0, 2\pi]$ for which $f(c) = f(c+\pi)$.  Here is what i did: if $f(c) = f(c+\pi)$, then $f(c) = f(c-\pi)$.  $f(c+\pi) = f(c-\pi)$ for which $\lvert{\sin(x)}\rvert$ is the function that comes to my mind, and I'm stuck here. I must use the mean value theorem and maybe Rolle's theorem since it's a question from that section, but $\lvert{\sin(x)}\rvert$ is not differentiable. Yet the question itself doesn't mention anything about differentiability of the function. Could anyone solve this, please?
117332
$f(x)$ is continuous and differentiable function defined in $x \geq 0$. If $f(0)=1$ and $f'(x) &gt;3 f(x)$, then prove that $$f(x) \geq  e^{3x}, \forall x \geq 0$$  Could someone give me little hint as how to proceed in this question? I am not able to initiate.
100837
Problem $f(x)$ is defined over $[a,b]$ and differentiable over $(a,b)$, where $b-a\geq 4.$Prove that there exists $\xi \in (a,b)$ such that $f'(\xi)&lt;1+f^2(\xi)$. My Proof Since $b-a \geq 4$，we can obtain $$\exists x_1,x_2 \in (a,b):x_2-x_1&gt;\pi.$$ Denote$$F(x):=\arctan f(x).$$ Obviously, $F(x)$ is continuous over $[x_1,x_2]$ and differentiable over $(x_1,x_2)$. Thus, By Lagrange's Mean Value Theorem，$$\exists \xi \in (x_1,x_2) \subset (a,b):F(x_2)-F(x_1)=F'(\xi)(x_2-x_1).$$ Further, $$\frac{f'(\xi)}{1+f^2(\xi)}=F'(\xi)=\frac{F(x_2)-F(x_1)}{x_2-x_1}\leq \frac{|F(x_2)|+|F(x_1)|}{x_2-x_1}&lt;\frac{\frac{\pi}{2}+\frac{\pi}{2}}{\pi}=1,$$ Which implies $$f'(\xi)&lt;1+f^2(\xi).$$ AM I RIGHT? HOPE TO SEE OTHER PROOFS. THX.
53589
If $f$ is continous in $[a,b] \subset \mathbb{R}$ and exist real numbers $\alpha \not= \beta$ such that $$\alpha\int_{a}^{c} f(x) dx + \beta\int_{c}^{b} f(x) dx  =0$$ for all $c \in [a,b]$ show that $f (x)=0$ for all $x \in [a,b].$
52466
Let $f:[0,1]\to\mathbb{R}$ be continuous and $f(0)=f(1)$. Show that for any integer $n\geqslant 2$, there is $\xi\in(0,1)$ s.t.$f(\xi)=f\left(\xi+\frac{1}{n}\right).$  I think this requires the intermediate value theorem somewhere.
40332
Let $f(x) = x^2 \sin{\frac{1}{x}}$ for $x\neq 0$ and $f(0) =0$. (a) Use the basic properties of the derivative, and the Chain Rule to show that $f$ is differentiable at each $a\neq 0$ and calculate $f'(a)$. You may use without proof that $\sin$ is differentiable and that $\sin' =\cos$.  Not even sure what this is asking.  (b) Show that $f$ is differentiable at $0$ and that $f'(0) =0$.  $\frac {f(x)-f(0)}{x-0} \to \lim_{x \to 0} x \sin(1/x)$. $x \sin(1/x) \leq |x|$ and $\lim_{x \to 0} |x|=0$. Thus $f(x)$ is differentiable at $0$; moreover $f^{'}(0)=0$.  (c) Show that $f'$ is not continuous at $0$.  $f{'}(x)=x^{2} \cos(1/x) (-x^{-2}) + 2x \sin (1/x)$. In pieces: $\lim_{x \to 0} \cos (1/x)$. $f^{'}(0-)$ nor $f{'}(0+)$ exists as $x \to 0$ $f^{'}(x)$ oscillates infinity between $-1$ and $1$ with ever increase frequency as $x \rightarrow 0$ for any $p&gt;0$ $[-p,0]$, $[-p,p]$ or $[0,p]$ $f$ is not continuous. Question: How to show more rigorously?
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.
123909
Let $f:\Bbb R \to \Bbb R$ be differentiable on $\Bbb R$ and $|f'(x)| \leq \frac {1}2$ for all $x\in \Bbb R$. Let $a \in \Bbb R$ and define $x_1 = a, x_{n+1} = f(x_n)$ for all $n \in \Bbb N$. Show that $\{x_n\}$ is a Cauchy sequence. And show that $\exists x \in \Bbb R$ such that $f(x) = x$.
10069
Let $f$ be a real-valued function continuous on $[a,b]$ and differentiable on $(a,b)$. Suppose that $\lim_{x\rightarrow a}f'(x)$ exists. Then, prove that $f$ is differentiable at $a$ and $f'(a)=\lim_{x\rightarrow a}f'(x)$.    It seems like an easy example, but a little bit tricky. I'm not sure which theorems should be used in here.    ==============================================================  Using @David Mitra's advice and @Pete L. Clark's notes I tried to solve this proof. I want to know my proof is correct or not.  By MVT, for $h&gt;0$ and $c_h \in (a,a+h)$ $$\frac{f(a+h)-f(a)}{h}=f'(c_h)$$ and $\lim_{h \rightarrow 0^+}c_h=a$.    Then $$\lim_{h \rightarrow 0^+}\frac{f(a+h)-f(a)}{h}=\lim_{h \rightarrow 0^+}f'(c_h)=\lim_{h \rightarrow 0^+}f'(a)$$   But that's enough? I think I should show something more, but don't know what it is.
60427
Prove that if the real-valued function $f$ on the interval $[a,b]$ is bounded and is continuous except at a finite number of points then $\int^1_0f(x)dx$ exists.   I know that I can break up the interval $[a,b]$ into subintervals where each subinterval has one discontinuity point. Then if $f$ is integrable on each subinterval then it is integrable on $[a,b]$. So if I let $c$ be my point of discontinuity in the subinterval $[a_1,b_1]\subset [a,b]$, where $c\in[a_1,b_1]$. Then how can I show this subinterval is integrable?
1790
Let $f$ be a real valued and continuous function over $[a,b]$ with $a &lt;b$ such that $f(a)=f(b)$ , My question here is :     Question (claim): How to show that : there  exists  $c \in [a, \frac{a+b}{2} ]$ such that : $f(c)=f(c+\frac{b-a}{2})$.   Note: I have tried intermediate values but I didn't succeed.
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.
66818
Given $g$ is continuous on $[0,1]$ and differentiable on  $(0,1)$,   $g(0)=0$, and $g(1)=1$.        1) Prove that there exist $0&lt;c_1&lt;c_2&lt;1$ such that   $$\frac{1}{g'(c_1)}+\frac{1}{g'(c_2)}=2.$$       2) Prove that there exist $0&lt;c_1&lt;c_2&lt;1$ such that    $$g'(c_1)g'(c_2)=1.$$   I am thinking if Mean Value Theorem would help, but Mean Value theorem could only prove that in interval (0,1) there exists $g'(c)=1$.