Suppose that $$ f (x_0) = g(x_0) $$ and that $$ f'(x)\le g'(x), \qquad x \ge x_0. $$ Then show that $$ f(x)\le g(x), \qquad x \ge x_0. $$   This is an &quot;obvious&quot; fact if one assumes that the second equation means that $g$ grows faster than $f$  on $[x_0, \infty)$ (or at least as fast). But how would you go about to prove this from the definitions? Thanks in advance!
38171
I have to do the following exercise:     Let $f$ and $g$ two differentiable functions such that $f(0)=g(0)$ and $f'(x)\leq g'(x)$ for all $x$ in $\mathbb{R}$. Prove that $f(x)\leq g(x)$ for any $x\geq0$.   Now, I know this is true because the first derivative of a function is the angular coefficient of the function in a point $x$. So, $f'(x)\leq g'(x)$ means, in other words, that the function $g(x)$ grows faster than $f(x)$. I think this is the base for a more formal proof, could someone help me to figure out a more formal proof?
80454
More specifically, I would like to know if there is a way I can express  $$\int \frac{x g'(x)}{f'(x)} dx $$  In terms of $f(x)$ and $g(x)$. Both $f(x)$ and $g(x)$ are non-negative and known to be increasing in $x$. If that helps.
90475
Let $F(x)=(1+\frac{1}{x})^x$.  How do we prove $F(x)$ is increasing when $x&gt;0$?
59446
Assume that $f(x),g(x)$ are positive and are in $L^1$. Moreover, they are differentiable and their derivative is integrable. Let $h(x)=f(x)*g(x)$, the convolution of $f$ and $g$. Does the derivative of $h(x)$ exist? If yes, how can we prove that $$ \frac{d}{dx}(f(x)*g(x)) = \left(\frac{d}{dx}f(x)\right)*g(x)$$  Thanks
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.
110796
Let $f$ be defined on the interval $I$. all we know about $f$ is that there is a constant $K\gt 0$ so that $|f(a)-f(b)|\le K|a-b|^2$ for all $a$,$b$ on $I$.  Q: show that $f$ is constant on $I$  hint: start with calculating the derivative using the definition?  how do I go about differentiating this inequality using the definition, and can I use $a=x$ and $b=x+h$ ? if yes then how
112764
Let $f(x)\in C^2:[0,1]\rightarrow\mathbb{R}$ satisfy $f(0)=0,f(1)=1,f'(0)=f'(1)=0$, prove that: $$\max_{x\in[0,1]}|f''(x)|\ge4$$  By using Taylor series I can prove that $\max_{x\in[0,1]}|f''(x)|\ge2$, but I cannot go further.
103888
I am working on proving the below inequality, but I am stuck.  Let $g$ be a differentiable function such that $g(0)=0$ and $0&lt;g'(x)\leq 1$ for all $x$. For all $x\geq 0$, prove that $$\int_{0}^{x}(g(t))^{3}dt\leq \left (\int_{0}^{x}g(t)dt  \right )^{2}$$
31277
Suppose $f$ and $g$ are entire functions, and $|f(z)| \leq |g(z)|$ for all $z \in \mathbb{C}$, Prove that $f(z)=cg(z)$.   My try :  I consider $h(z)=\frac{f(z)}{g(z)}$. If I prove that $h(z)$ is entire, using the fact that $|h(z)|\leq 1$, the result follows immediately from Liouville's theorem.  To prove that $h(z)$ is entire, I have to prove that $h(z)$ has removable singularities at possible zeros of $g(z)$.  Suppose $g(z_0)=0 \Rightarrow |g(z_0)|=0 \Rightarrow |f(z_0)|=0$, BUT what if $z_0$ be a root of of order $k$ of $g(z)$ (i.e. $f^{(n)}(z_0) = 0 $ for all $ 0 \leq n \leq k-1$) and be a zero of order $m$ of $f(z)$ and $k&gt;m$ ?  And if that's not the case, how to rigorously show that $z_0$ is a removable singularity ?  Thank you all in advance.
121433
Let $g: \mathbf R \to \mathbf R$ be a function which is not identically zero and which satisfies the equation $$ g(x+y)=g(x)g(y) \quad\text{for all } x,y \in \mathbf{R}. $$ Show that $g(x)\gt0$ for all $x \in \mathbf{R}$.
67960
I proved as follows:  Let $f(x) = \tan x -  x - \frac{x^{3}}{3}.$ Then $f '(x) = \sec^{2}x - 1 - x^{2}$ and $f''(x) = 2\sec^{2}x\tan x-x &gt; \tan x - x,$ since $\\tan x &gt; x,$ $f''(x) &gt; 0$ so $f'(x)$ is increasing. Since $f '(0) = 0,$ $f '(x) &gt; 0$ for all $x \in (0, \frac{\pi}{2}).$ So $f(x)$ is increasing. since $f(0) = 0,$ $f(x) &gt; 0$ for all $x \in (0, \frac{\pi}{2}).$  But my proof seems complex. Is there any simpler way to prove this?
96766
Let $f\ge0$ be integrable on $[a,b]$. Let $f$ be continuous on $x_0\in (a,b)$ and let $f(x_0)&gt;0$.   Prove $$\int_{a}^{b}{f(x)dx}&gt;0$$  It is easy to see it, graphically, because $$f\ge 0 \Rightarrow\int_{a}^{b}{f(x)dx}\ge0$$  and if $f&gt;0$ somewhere between $a$ and $b$, then the zone of $F$ between those point is "real" (positive) and $\ne 0$.  My problem is, how do I show it formally?  I would appreciate your help.
19802
If $f(x)$ and $g(x)$ are two periodic functions, that is, $f(x+T_1)=f(x)$ and $g(x+T_2)=g(x)$ for every $x \in \Bbb R$. Now that $\displaystyle\lim_{x\to\infty}(f(x)-g(x))=0$.  Conjecture: $f(x) \equiv g(x)$.
114923
If $f'(x) = g(x)$ and $g'(x) = - f(x)$ for all real $x$ and $f(5) =2 =f'(5)$ then we  have to find $f^2$$(10) + g^2(10)$  I tried but got stuck
101755
Suppose $g$ is infinitely differentiable function such that for all natural $n$, there exists $c_n$ and $\delta_n$ such that if $|x|&lt;\delta_n$, then $|g(x)|\leq c_n|x|^n$. Show  $g^{(n)}(0)=0$ for all $n$.  Could someone provide a hint on where to start? I was thinking of a proof by induction, but even the base case eludes me.
69535
Let $f(x) = |x|$ for $x \in [-1,1]$ and extend $f$ to all of $\mathbb R$ by requiring $f(x+2)=f(x)$. Prove that $|f(x)-f(y)|\le|x-y|$ for all $x$ and $y$.   This is easy to observe on a graph but my proof is long and a bit tedious. Just curious if there is a quick/elegant way?
94331
Problem: Hi, I'm trying to solve the following proof:  If $a &gt; b \geq 0$, Prove that $a^2 &gt; b^2$.  But I'm a bit new with proofs and I don't manage to write a proper one. I would be very grateful if someone could explain me which are the steps required to demonstrate the proof. Thank you.