Prove that if $f$ is an odd function, then $f ′ (x) = f ′ (- x)$.
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 =]
99523
The statement is: $f$ is a real fucntion on $\mathbb R$. Then if $f'(x)=f(x)$ and $f(0)=1$, then $f(x)\neq 0.$
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!
106435
Let $f$ be continuous and differentiable on $(0,1)$ such that $f(0)=1$ and $f(1)=0$. There is a $c$ in $(0,1)$ such that $f(c) = kc$, where $k$ is a positive constant.  Show that there exists $a$ and $b$ in $(0,1)$ such that $cf'(a) +(1-c)f'(b)=-1$.
80819
What function satisfies $ f(x)+f\left(\frac{1-x}x\right)=x$ ?
102631
Let $f$ a entire function: $f(R)\subset R.\;$ Prove that $f(\bar{z})=\overline{f(z)}, \forall z\in C$
60186
What is the logic/thinking process behind deriving an expression for even and odd functions in terms of $f(x)$ and $f(-x)$?  I've been pondering about it for a few hours now, and I'm still not sure how one proceeds from the properties of even and odd functions to derive: $$\begin{align*} E(x) &amp;= \frac{f(x) + f(-x)}{2}\\ O(x) &amp;= \frac{f(x) - f(-x)}{2} \end{align*}$$ What is the logic and thought process from using the respective even and odd properties, $$\begin{align*} f(-x) &amp;= f(x)\\    f(-x) &amp;= -f(x) \end{align*}$$   to derive $E(x)$ and $O(x)$?    The best I get to is: For even: $f(x)-f(-x)=0$ and for odd: $f(x)+f(-x)=0$  Given the definition of $E(x)$ and $O(x)$, it makes a lot of sense (hindsight usually is) but starting from just the properties. Wow, I feel I'm missing something crucial.
111707
Suppose $f$ is a function such that $f:\mathbb C \to \mathbb C$ such that $f(z)f(iz)=z^2$, then we have to show that $\forall z\in \mathbb C f(z)+f(-z)=0 $.   Actually I have solved this problem and have no doubt about it but actually this question came in MADHAVA-2020. So I posted it on this site.   I am not answering the question and inviting all users to answer it.  It is a nice problem.   I will answer it within a week, but I am letting everyone try.
32828
I'm stumped at the last part of this problem. Could anyone advise, please? Thank you.  Here is  my  proof.   Define  $f: \mathbb{R} \to \mathbb{R}$ by $$ f(x) = \left\{\begin{aligned} &amp;\frac{\sin nx}{nx}&amp;&amp;, x \neq0 \\ &amp;1 &amp;&amp;, x=0 \end{aligned} \right.$$   Hence, $$f(x) = \sum^{\infty}_{k=0}(-1)^k\frac{(nx)^{2k}}{(2k+1)!} , \forall x \in \mathbb{R} \mbox{ and } f \in {\mathscr R[0,b]}.$$  $$\int^{b}_{0} \frac{\sin nx}{nx}dx = \int^{b}_{0}f(x)dx=\sum^{\infty}_{k=0}\int^{b}_{0}(-1)^k\frac{(nx)^{2k}}{(2k+1)!}dx= \sum^{\infty}_{n=0}(-1)^k\frac{(n)^{2k}b^{2k+1}}{(2k+1)(2k+1)!}$$  $$\lim_{n \to \infty} \sum^{\infty}_{k=0}(-1)^k\frac{(n)^{2k}b^{2k+1}}{(2k+1)(2k+1)!}=  ?$$
89197
How can I prove that the function  $$   f(x) = \left\{\begin{array}{l l}     x &amp;\text{if }x \in \mathbb{Q} \\     -x &amp; \text{if } x \notin \mathbb{Q}    \end{array} \right. $$ is discontinuous for $x \neq 0$, using $\epsilon$'s and $\delta$'s?  I see that is truth. But I cannot prove using only $\epsilon$'s and $\delta$'s.
72478
Let $f$ be an entire function such that $ |f(z)| \to \infty$ as $|z| \to \infty$. Prove that $f$ is a polynomial.
4428
Let $f$ be a $C^2$ function in $[0,1]$ such that $f(0)=f(1)=0$ and $f(x)\neq 0\,\forall x\in(0,1).$ Prove that  $$\int_0^1 \left|\frac{f{''}(x)}{f(x)}\right|dx\ge4$$
97927
Prove that the function : $f(x) = x$ when $x$ is rational and $f(x) = 1 - x$ when $x$ is irrational is continuous at $x = \frac{1}{2}$  To prove the continuity it is necessary to prove that for every $\epsilon$ there exists a $\delta$ such that $|f(x) - f(\frac{1}{2})| &lt; \epsilon$ whenever $|x -c| &lt; \delta$ $- (1)$ Now if we choose $a  € (\frac{1}{2} - \delta, \frac{1}{2} + \delta)$ and also $a $ is rational and choose $\epsilon &lt; |a - \frac{1}{2}|$ then condition in $(1)$ is not satisfied. How to move from here?
46557
I just wanted to ask, if my proof is correct. I haven't seen the equation before, but I think it's quite useful.  Let $f$ be an bijective differentiable function. Then the inverse function $f^{-1}$ exists and the following equation holds:     $$\int\limits_{a}^{b} f(x) dx = b \cdot f(b) - a \cdot f(a) - \int\limits_{f(a)}^{f(b)} f^{-1}(x) dx$$   Proof.  $f$ is an bijective differentiable function and $F$ is an antiderivative of $f$.  First we need to find an antiderivative of $f^{-1}$.  $\int f^{-1}(x) dx$ with substitution $x = f(y)$ yields:  $$\int y \cdot f'(y) dy = y \cdot f(y) - \int f(y) dy = y \cdot f(y) - F(y)$$  resubstitution yields:  $$\int f^{-1}(x) dx = x \cdot f^{-1}(x) - F(f^{-1}(x))$$ hence $\int\limits_{f(a)}^{f(b)} f^{-1}(x) dx = \left[x \cdot f^{-1}(x) - F(f^{-1}(x)) \right ]_{f(a)}^{f(b)}$  $$=b \cdot f(b) - F(b) - (a \cdot f(a) - F(a)) = F(a) - F(b) + b \cdot f(b) - a \cdot f(a)$$ $$= \int\limits_{b}^{a} f(x)  dx + b \cdot f(b) - a \cdot f(a) = -\int\limits_{a}^{b} f(x)  dx + b \cdot f(b) - a \cdot f(a)$$  All in all:  $$\int\limits_{f(a)}^{f(b)} f^{-1}(x) dx = -\int\limits_{a}^{b} f(x)  dx + b \cdot f(b) - a \cdot f(a)$$   which is equal to     $$\int\limits_{a}^{b} f(x) dx = b \cdot f(b) - a \cdot f(a) - \int\limits_{f(a)}^{f(b)} f^{-1}(x) dx$$   q.e.d.
45862
Prove that the function $f: R\to R$ given $f(x)=x\cdot |x|$ is a bijection  Proof 1: f is injective  Suppose $$f(a) = f(b) \implies a\cdot |a| = b\cdot |b|$$ $$\implies (a\cdot |a|)^2 = (b\cdot |b|)^2$$ $$\implies a^4 =b^4$$ $$a=b$$ Thus, f is injective  Proof 2: f is surjective  Pick $n\in R$, we want to show $n=f(x)$ for some $x\in R$  $$\implies n= a\cdot |a|$$ for some $a\in R$  I dont know how to complete the surjective proof
112713
Just wanted to verify this proof was correct. Trying to make sure I have a handle on these subjects before the exam. Any style comments also more than welcome.    Statement: Let $f : A \mapsto B$ be a function. $f$ is injective if and only if for all $X \subseteq A$, $f^{-1}(f(X)) \subseteq X$.  Proof:  $\Rightarrow$ Suppose that $f$ is injective. Let $X \subseteq A$, and let $x \in X$. Since $f$ is injective, $f^{-1}(f(X)) = \{x\} \in X$, so $f^{-1}(f(X)) \subseteq X$.   $\Leftarrow$ Suppose that for all $X \subseteq A$, $f^{-1}(f(X)) \subseteq X$. Let $a_1, a_2 \in A$ and $X = \{a_1\}$. Suppose $f(a_1) = f(a_2)$. If $a_1 \neq a_2$, then $f^{-1}(f(X)) = \{a_1, a_2\} \not \subseteq X$, a contradiction. Thus $a_1 = a_2$, so $f$ is injective.
29377
Let $f$ be differentiable and let $\lim_{x \to \infty}f'(x)=+\infty$ prove that: 1) $\lim_{x \to \infty}(f(x)-f(x-1))=+\infty$ and 2) $\lim_{x \to \infty}f(x)=+\infty$.  1) I'll prove by contradiction. Let $\lim_{x \to \infty}(f(x)-f(x-1))=a$, where $a$ is from and $\Bbb R$. $\lim_{x \to \infty}(f(x)-f(x-1))=\lim_{x \to \infty}\frac{f(x)-f(x-1)}{x-(x-1)}=\lim_{x \to \infty}f'(c)=a$, where c is from $(x-1,x)$ (mean value theorem). As $x$ goes to infinity so does $c$ i.e. we have that $\lim_{x \to \infty}f'(c)=\lim_{c \to \infty}f'(c)=a$. Which is a contradictition since we have that  $\lim_{x \to \infty}f'(x)=+\infty$ .   2) I will also prove this by contradiction. We have that $\lim_{x \to \infty}f(x)=a$ This is equivalent to $\lim_{x \to \infty}f(x)-f(x-1)=0$. I use mean value theorem and get $\lim_{x \to \infty}(f(x)-f(x-1))=\lim_{x \to \infty}\frac{f(x)-f(x-1)}{x-(x-1)}=\lim_{x \to \infty}f'(c)=0$, where c is from $(x-1,x)$. As $x$ goes to infinity so does $c$ i.e. we have that $\lim_{x \to \infty}f'(c)=\lim_{c \to \infty}f'(c)=0$. Which is a contractition since we have that  $\lim_{x \to \infty}f'(x)=+\infty$ .