Given $f$ is an even function of interval $(-a, a)$, $a&gt;0$ and $0&lt;c &lt; a$. Prove that if $L'f(c)$ exists, then $Rf'(-c)$ exists and $Lf'(c)=-Rf'(-c)$. Deduce that if $f$ is differentiable on $(-a,a)$, then $f'$ is an odd function on $(-a,a)$.   $f$ is even then $f(-x)=f(x)$, then $f'(-x)=-f'(x)$ i.e $f'$ is odd function. But what exactly the question want? Please help with apropriate answer.
65425
This is the question: "Show that the derivative of an even function is odd and that the derivative of an odd function is even. (Write the equation that says f is even, and differentiate both sides, using the chain rule.)"  I already read numerous solutions online.  was the official solution but I didn't quite understand it (particularly, I'm not convinced why exactly $dz/dx=-1$; even though $z=-x$).  Thanks in advance =]
33976
Prove that if a function $f$ has a jump at an interior point of the interval $[a,b]$ then it cannot be the derivative of any function.  I know that for $f$ is differentiable in $(a,b)$ and that it has one-sided derivative $f_+' (a)≠f_-' (b)$ at the endpoints. If $C$ is a real number between $f_+' (a)$  and $f_-' (b)$, then there exists $c∈(a,b)$ such that $f' (c)=C $. How can I use this to prove the above?
110296
The hint is: Let F has as elements $a_1,...,a_n$. Note that if $$f_i(x)=c(x-a_1)...(x-a_{i-1})(x-a_{i+1})...(x-a_n)$$, then $f_i(a_j)=0$ for $i\neq j$ and the value $f_i(a_j)$ can be controlled by the choice of $c\in F$. Use this to show that every function on F is a polynomial function.  Using Fraleigh’s seventh edition of A First Course in Abstract Algebra. Need help with problem 22.31c. I’m completely lost. Thank you!
49415
I've been thinking about this problem: Let $f: (a, +\infty) \to \mathbb{R}$ be a differentiable function such that $\lim\limits_{x \to +\infty} f(x) = L &lt; \infty$. Then must it be the case that $\lim\limits_{x\to +\infty}f'(x) = 0$?  It looks like it's true, but I haven't managed to work out a proof. I came up with this, but it's pretty sketchy:  $$ \begin{align} \lim_{x \to +\infty} f'(x) &amp;= \lim_{x \to +\infty} \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\ &amp;= \lim_{h \to 0} \lim_{x \to +\infty} \frac{f(x+h)-f(x)}{h} \\ &amp;= \lim_{h \to 0} \frac1{h} \lim_{x \to +\infty}[f(x+h)-f(x)] \\ &amp;= \lim_{h \to 0} \frac1{h}(L-L) \\ &amp;= \lim_{h \to 0} \frac{0}{h} \\ &amp;= 0 \end{align} $$  In particular, I don't think I can swap the order of the limits just like that. Is this correct, and if it isn't, how can we prove the statement? I know there is  already, but I think this is different in two aspects. First, that question assumes that $\lim\limits_{x \to +\infty}f'(x)$ exists, which I don't. Second, I also wanted to know if interchanging limits is a valid operation in this case.
51766
If $f'(c)$ exists then $\lim_{x\to c} f(x)=f(c)$   My attempt is below. Did I do wrong? Since I don't really understand it. I thought $f'(x) = f'(c)$ so $f'(x)-f'(c)=0$. Am I on the right track? Many thanks!  \begin{align} f(x)-f(c)&amp;=\frac{f(x)-f(c)}{x-c}\cdot(x-c) \\ \lim_{x\to c}[f(x)-f(c)]&amp;=\lim_{x\to c}\left[\frac{f(x)-f(c)}{x-c}\cdot(x-c)\right] \\ \lim_{x\to c}[f(x)-f(c)]&amp;=0 \\ \lim_{x\to c} f(x)&amp;=f(c) \end{align}
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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! :)
67634
Let $f:X\to Y$ be a function and $E\subseteq X$ and $F\subseteq X$.   Show that in general $f(E − F)\nsubseteq f(E) − f(F)$.   I have no idea about how to prove this; and could anyone please explain the basic theory of functions(by relating to this question. Thank you.
81925
Question: Let $f:[0,1]\to\mathbb{R}$ be a continuous function such that $f(0)=0$ and $$\int_0^1f(x)dx=0.$$ Prove that $\exists a, b\in(0,1)$ such that $$\int_0^{a} xf(x)dx=0\text{ and }\int_0^bxf(x)dx=\frac{b^2f(b)}{2}.$$  My approach: Let $g:[0,1]\to\mathbb{R}$ be such that $$g(x)=x\int_0^xf(t)dt-\int_0^xtf(t)dt, \forall x\in[0,1].$$ By the first fundamental theorem of calculus we can conclude that $g$ is differentiable on $[0,1]$ and $$g'(x)=\int_0^xf(t)dt, \forall x\in[0,1].$$ Also, observe that $g(0)=0$ and $g(1)=-\int_0^1tf(t)dt$. Thus, by applying MVT to the function $g$ on the interval $[0,1]$, we can conclude that, $\exists c\in(0,1)$ such that $$g'(c)=\int_0^cf(t)dt=-\int_0^1tf(t)dt.$$ Observe that clearly three cases are possible, i.e, either$$\int_0^cf(t)dt&lt;0\text{ or }\int_0^cf(t)dt=0\text{ or }\int_0^cf(t)dt&gt;0.$$ Now let $h:[0,1]\to\mathbb{R}$ be such that $$h(x)=\int_0^xtf(t)dt, \forall x\in[0,1].$$ Please note that the part highlighted below is wrong, but still I have included it to just demonstrate my thinking process, as it might be of some help to others trying out this problem.  Observe that if $\int_0^cf(t)dt&lt;0$, then, $h(1)&gt;0$. This also implies that $\exists$ an open interval $(d,e)\in[0,c]$, such that $f(t)&lt;0, \forall t\in(d,e)$. Now select any point $c_1\in(d,e)$. Applying MVT to the function $h$ on the interval $[0,c_1]$, we can conclude that $\exists c_2\in(0,c_1)$ such that $$h'(c_2).c_1=f(c_2).c_2.c_1=h(c_1)-h(0)=h(c_1)&lt;0.$$ Now $h(c_1)&lt;0$ and $h(1)&gt;0$. Thus, by IVT we can conclude that $\exists a\in(c_1,1)\subseteq(0,1)$, such that $$h(a)=\int_0^af(t)dt=0.$$ A similar reasoning for the case when $\int_0^cf(t)dt&gt;0,$ shows that $\exists a\in(0,1)$, such that $$h(a)=\int_0^af(t)dt=0.$$ Now finally if $\int_0^cf(t)dt=0$, then we will have $h(1)=0$. Now if $f$ is identically equal to $0$ on $[0,c]$, then clearly $tf(t)=0, \forall t\in[0,c]\implies h(x)=0, \forall x\in[0,c].$ Thus choosing any point $x\in(0,c]$ and setting it as $a$, we will have $h(a)=0$ and we will be done in that case. Now if $f$ acquires both positive and negative values on $[0,c]$, then we can conclude that $\exists c_1,c_2\in(0,c)$, such that $f(c_1)&gt;0$ and $f(c_2)&lt;0$. Also, let us assume WLOG that $c_2&gt;c_1$.  I have not been able to make any significant approach other than this. Can someone help me out with this problem? Please note that a solution using integration by parts might not be possible, since $f$ is not a differentiable function.
35060
Can I get help to prove or Disprove that -   $\int_{-a} ^a F_1(x)F_2(x)dx = 0 \ for \ a &gt; 0$  when I know that $F_2$  is even function and $F_1$  is odd.  I would like to get the idea how to delete and extract Minus in integrals.  thanks!
89554
Suppose that $f(x)$ is continuous on $(0, +\infty)$ such that for all $x &gt; 0$,$f(x^2) = f(x)$. Prove that $f$ is a constant function. My attempt is to show that for any point $a \neq b$ , we have $f(a)=f(b)$. But I have no idea on how to get this. Anyone can help?
80847
Let $f$ be a continuous function on a part $I$ of $\mathbb{R}$. Is they exist always a function $F$ differentiable on an interval $J$ containing I such that $F'=f$ on $I$?  If $I$ is an interval , it's ok  if $I$ is an open set , it's ok  But if $I$ is only a part of $\mathbb{R}$  ?
89465
I am struggling to construct a simple proof of the following.  Suppose $f(x)$ is differentiable for any real value $x$ and that $f'(x) \le 4$ for all real values of x. Prove there is at most one real value $c&gt;2$ such that $f(c) =c^2$.  I am completely lost on where to start on this proof and any help would be greatly appreciated!
69480
If $f(x)$ is a continuous function on $(-\infty, +\infty)$ and $\int_{-\infty}^{+\infty} f(x) \, dx$ exists. How can I prove that $$\int_{-\infty}^{+\infty} f(x) \, dx = \int_{-\infty}^{+\infty} f\left( x - \frac{1}{x} \right) \, dx\text{ ?}$$
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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.
22884
Assuming $a&lt;b$, function f(x) is continous at [a,b], and we have  $\int_a^bf(x)dx=\int_a^bxf(x)dx=\int_a^bx^2f(x)dx=0$  Prove that $\exists \ x_1,x_2,x_3\text {(different from each other)}  \in(a,b)$ satisfying $f(x_1)=f(x_2)=f(x_3)=0$.  I have proved the existence of one zero point by differential mean value theorem, but have no idea about going on.  Thanks for any solution or hint.
78312
Is there a counterexample?  Proposition: Let $f:\mathbb{R} \longrightarrow \mathbb{R}$ be a differentiable function  such that $f':\mathbb{R} \longrightarrow \mathbb{R}$ is differentiable at $a\in\mathbb{R}$ (possibly differentiable at a single point). Then $f':\mathbb{R} \longrightarrow \mathbb{R}$ is continuous at an interval $(a-\delta,a+\delta)$ for some $\delta&gt;0$.  Any hints would be appreciated.
13545
Problem:   If $f(x)$ is continous at $x=0$, and $\lim\limits_{x\to 0} \dfrac{f(ax)-f(x)}{x}=b$, $a, b$ are constants and $|a|&gt;1$, prove that $f'(0)$ exists and $f'(0)=\dfrac{b}{a-1}$.   This approach is definitely wrong:    \begin{align} b&amp;=\lim_{x\to 0} \frac{f(ax)-f(x)}{x}\\ &amp;=\lim_{x\to 0} \frac{f(ax)-f(0)-(f(x)-f(0))}{x}\\ &amp;=af'(0)-f'(0)\\ &amp;=(a-1)f'(0) \end{align}  I will show you a case why this approach is wrong:     \[f(x)= \begin{cases} 1,&amp;x\neq0\\ 0,&amp;x=0 \end{cases}\]   $\lim_{x\to0}\dfrac{f(3x)-f(x)}{x}=\lim_{x\to0} \dfrac{1-1}{x}=0$   but $\lim_{x\to0}\dfrac{f(3x)}{x}=\infty$,$\lim_{x\to0}\dfrac{f(x)}{x}=\infty$   Does anyone know how to prove it? Thanks in advance!