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Is e^x is strictly increasing or monotonically increasing or both

Since $f'(x) = e^{x}$ and $e^{x}>0$ for all real $x$, it follows that $f(x)$ is a strictly increasing function on the real line. It also follows that $f'(x)\geq 0$, so based on your definition of a ...
PH_1729's user avatar
2 votes

How to map identity to a sigmoid?

I don't know what you would consider as a better alternative, but maybe one example could be the following function for $k\geq 1$: $$f_1(x,k) = \begin{cases} 0, \quad x \leq 0; \\ 1,\quad x\geq 1; \\ \...
Joako's user avatar
  • 1,404
1 vote

Is there an analog for factorials in division?

As some of us have indicated in the comments to the OP, repeated division without grouping is ambiguous because division is not associative. That is, $$ (a \div b) \div c \not= a \div (b \div c) $$ ...
Brian Tung's user avatar
  • 34.5k
1 vote

Definition of correspondence

A correspondence is considered to be a generalization of a binary-relation. A function is a binary-relation (a correspondence); there is a Domain and Codomain (there are two sets being mapped between)....
tylerbakeman's user avatar
2 votes

Given a functional equality prove that the function is never $0$

If $u(t)=0$ then $0=t^2+t+2=(t+1/2)^2+7/4$. This never happens. Note that the hypothesis $u(0)=1$ was useless.
Anne Bauval's user avatar
  • 39.6k
1 vote

Why can't I find the range of the function $\frac{1}{\sqrt{x-5}}$?

I recommend thinking about this a little less computationally and more as this function being collection of simpler functions composed with each other that move the domain around for the next function ...
Santiago Beck's user avatar
1 vote
Accepted

Why can't I find the range of the function $\frac{1}{\sqrt{x-5}}$?

For the function: $$ x=\frac{1+5y^2}{y^2} $$ has a domain of $\mathbb R-\{0\}$ as you said. But it does not mean the function $$y=\frac{1}{\sqrt{x-5}}$$ has the same range. Because, if you see: $$ x=\...
M.Riyan's user avatar
  • 134
1 vote

Characterize a specific measurable space

$f+\frac 1 f \ge 2$ so integrability of $f$ and $\frac 1 f$ implies integrability of the constant $2$ which means $\mu (E) <\infty$. Conversely, if $\mu (E)<\infty$ then we can take $f\equiv 1$...
geetha290krm's user avatar
  • 39.2k
-1 votes
Accepted

Determine all possible values ‚Äčof the sum $S=a+b+c+d$

Continuing on your answer. Since, $f(f(a)) - f(b) = 2022 = c^2a - cb + cd$ $b-a=20$ $d-c=25$ (mistake in your answer) Expanding this equation further gives us: $c(ca-b+d) = 2022$ and on using the ...
HopelessBhai's user avatar
3 votes
Accepted

Do $\operatorname{arcsec}\left(\frac{2x}{5x+3}\right)$ and $\operatorname{arccos}\left(\frac{5x+3}{2x}\right)$ have the same domain?

$$f(x)=\operatorname{arcsec}\left(\frac{2x}{5x+3}\right) \quad\text{and}\quad g(x) = \operatorname{arccos}\left(\frac{5x+3}{2x}\right).$$ I found the domain of $f(x)$ to be $\left[-1,\frac{-3}{7}\...
ryang's user avatar
  • 39.9k
0 votes

Is $f(x)=|x|$ a convex function?

Let $f_n(x) = \sqrt{x^2+{1 \over n}}$ and note that $f(x) = \lim_n f_n(x)$. Since the $f_n$ are smooth and $f_n''(x) >0$, hence convex. Since $f_n(tx+(1-t)y) \le t f_n(x) + (1-t)f_n(x)$ for all $t \...
copper.hat's user avatar
  • 174k
0 votes

How to find the range of a quadratic function we can't use the quadratic formula?

Alternative approach: To analyze the behavior of $~f(x) = \dfrac{x}{1 + x^2},~$ start by examining its 1st derivative. Then, either also examine its 2nd derivative, or look for alternate approaches. ...
user2661923's user avatar
  • 37.3k
0 votes

How to find the range of a quadratic function we can't use the quadratic formula?

According to the AM-GM inequality, one concludes that \begin{align*} 1 + x^{2} \geq 2\sqrt{x^{2}} = 2|x| & \Rightarrow \frac{|x|}{1 + x^{2}} \leq \frac{1}{2}\\\\ & \Rightarrow \left|\frac{x}{1 ...
Átila Correia's user avatar
1 vote

How to find the range of a quadratic function we can't use the quadratic formula?

You can only use the quadratic formula for $ax^2+bx+c=0$ when $a\neq0$. So, you need to consider separately the possibility $y=0$, and in that case you get $x=0$, consistent with $y=x/(1+x^2)$. Hope ...
ultralegend5385's user avatar
0 votes

How to find the range of a quadratic function we can't use the quadratic formula?

What you assumed in your question is wrong. You can't claim $y \neq 0$ because it is zero if $x=0$. Also, notice that $f(x)$ is a continuous function in $\Bbb R$. But to find the range, you need to ...
Afntu's user avatar
  • 2,219
1 vote

Find minimum value of $f(x)= x^{1.5} + x^{-1.5} -4(x + x^{-1})$

Another way without using derivatives. I will use the sum of cubes formula and the completing the square method. $\begin{align}f(x)&=x^{1.5}+x^{-1.5}-4(x+x^{-1})=\\[3pt]&=\left(\sqrt x\right)^...
Angelo's user avatar
  • 12.4k
0 votes

Let $f \colon \Bbb{N} \to \Bbb{N}$ such that $f^{f(m)}(n) = n + f(m)$. Prove that $f(0) = 1$.

We can see that $f$ is injectif : $$S(0,m): f^{f(m)}(0)=f(m)\implies f^{f(m)-1}(0)=m$$ Thus $f$ is surjectif . So let $a\in\mathbb N: f(a)=1$. So plugging $a$ into $f^{f(a)-1}(0)=a\implies f^0(0)=a\...
Ahmed Mallek's user avatar
0 votes

Maximum value of $2$ variable function $f(u,v)=\frac{\left(1-\sqrt{uv}\right)^2}{\frac{1-u^2}{2u}+\frac{1-v^2}{2v}}$

After simplications $$f(u,v)=\frac{2\, u\, v \,\left(1-\sqrt{u\, v}\right)^2}{(u+v) \,(1-u\, v)}$$ If we use the symmetry $f(u,v)=f(v,u)$ the problem is simple. Since $u>0$ $$f(u,u)=\frac{(1-u)\, u}...
Claude Leibovici's user avatar
1 vote
Accepted

Why is interpolating $y=g(x)$ then applying $h(y)$ not equivalent to interpolating $h(g(x))$?

First, an objection: I don't think the title formulation matches the context provided. In the first case, you approximate $V(I)\approx V_L(I)$ via linear interpolation and then compute $V_L(I)/I$. In ...
Semiclassical's user avatar
3 votes

Why is interpolating $y=g(x)$ then applying $h(y)$ not equivalent to interpolating $h(g(x))$?

Interpolation treats your function as if it's a line (i.e. linear function) between your two interpolation endpoints. If your function $h( )$ preserved "being a line" then you would get the ...
JonathanZ's user avatar
  • 11.2k
1 vote

Maximum value of $2$ variable function $f(u,v)=\frac{\left(1-\sqrt{uv}\right)^2}{\frac{1-u^2}{2u}+\frac{1-v^2}{2v}}$

We start from $\displaystyle f(u,v)=\frac{(1-\sqrt{uv})^2}{\frac{1-u^2}{2u}+\frac{1-v^2}{2v}}, u,v>0$ by substituting $uv=a^2$ where $a>0$. We further assume $a\ne 1$ for obvious reasons. ...
Mostafa Ayaz's user avatar
  • 32.6k
1 vote
Accepted

Prove that a functional equation has at least one root

First of all we need to define $g\circ$$f:[0,1]\rightarrow[-1,4]$ where $g(x)$=$x^3$+$\beta$$x^2$+$\gamma$$x$ just as you did. Now we can apply the Intermediate Value Theorem since the defined ...
Elfryionnn's user avatar
3 votes
Accepted

Find minimum value of $f(x)= x^{1.5} + x^{-1.5} -4(x + x^{-1})$

set $y=x^{1/2}$ and get $$f(x)=y^3+y^{-3}-4(y^2+y^{-2})$$decompose,get $$f(x)=(y+1/y)(y^2+1/y^2-1)-4(y^2+1/y^2)$$noticed $$(y^2+1/y^2)=(y+1/y)^2-2$$now you can use your $\lambda = y+1/y$ and get$$f=\...
UnnamedUser's user avatar
0 votes

Find minimum value of $f(x)= x^{1.5} + x^{-1.5} -4(x + x^{-1})$

The derivative of the function after setting $u=\sqrt x$ turns out to be $$\frac{3 u^6-8 u^5+8 u-3}{2u^5}=\frac{(u-1) (u+1) \left(u^2-3 u+1\right) \left(3 u^2+u+3\right)}{2u^5}$$ The real positive ...
Math-fun's user avatar
  • 9,667
0 votes
Accepted

How to write this in a rigorous way?

I would proceed as follows: First, prove that the image of $f$ is bounded (I omit the details of that step). Then, since $M := \sup_{\mathbb R} f$ exists and is not zero (why?), use the definition of $...
user1149748's user avatar
2 votes

Derivation of Legendre Polynomials from only orthogonality

I think I figured it out. I was inspired by the technique in this answer. They used the inner product $$(f,g) = \int_{-1}^1fgdx$$ Then $(xf,g)=(f,xg)$, and using this, they applied this to the ...
Thomas Blok's user avatar
1 vote
Accepted

(How) can I determine the point of tangency of $k\cdot x$ and $\sin{x}$?

The slope of $kx$ is $k$ so one wants to find the $x$ where the slope of $\sin x$ is also $k$ and where both curves meet, i.e., $kx = \sin x$ and $(d/dx)\sin x=k = \cos x$. This leads to $x=\tan x$ ...
R. J. Mathar's user avatar
  • 3,263
2 votes

Maximum value of $2$ variable function $f(u,v)=\frac{\left(1-\sqrt{uv}\right)^2}{\frac{1-u^2}{2u}+\frac{1-v^2}{2v}}$

1. Question Copy to Refer to its Equations in the Solution Finding maximum value of $$\displaystyle f(a,b)=\frac{\bigg(1-\sqrt{\tan\frac{a}{2}\tan\frac{b}{2}}\bigg)^2}{\cot a+\cot b},\,\text{ Where }...
Stephen Elliott's user avatar
0 votes

Does there exist a widely-used operator $\boxdot$ such that $(\theta \boxdot \phi)(x) := \theta(x) \circ \phi(x)$?

"Widely used", not that I'm aware of. But this can be expressed in combinatory logic as $\mathsf{SB}$. This works because $\mathsf{S}$ is the combinator $(a,b,c,d \mapsto a (b d)(c d))$, and ...
Deusovi's user avatar
  • 2,895
0 votes

What is the maximum of $ \frac{\sin(n(x+a))}{\sin(x+a)} + \frac{\sin(n(x-a))}{\sin(x-a)}$?

1. Question Restatement to Reference its Equations and Link We know the global maxima of the function $$f\left(n,x\right) =\frac{\sin(nx)}{\sin(x)} \text{ is } n $$ $$\tag{Eq. 1.1}$$ (thanks to this ...
Stephen Elliott's user avatar
1 vote

What is the maximum of $ \frac{\sin(n(x+a))}{\sin(x+a)} + \frac{\sin(n(x-a))}{\sin(x-a)}$?

Based on the plots I have shown in the question, it seems the below equation gives an approximation of the maximum$$N + \frac{sin(2Na)}{sin(2a)}$$ . The cyan colored plots are plots of the above ...
RajaKrishnappa's user avatar
1 vote

Show that $f(x):=\cos(x^2)$ is not periodic.

If a continuously differentiable function $f(x)$ is periodic, so is $f'(x),$ hence $f'(x)$ is bounded. However $(\cos(x^2))'=-2x\sin(x^2)$ is unbounded, hence it is not periodic.
Ryszard Szwarc's user avatar
1 vote

Solving for the range of a function algebraically

I am trying to find the range of: $$ \frac{x}{x^2-16} $$ Alternative approach: The first thing to realize is that both $~f(x) = x,~$ and $~f(x) = x^2 - 16~$ are continuous functions. Therefore, so ...
user2661923's user avatar
  • 37.3k
0 votes

If $f(x)f''(x)\le0$, and $f(a)=0$, can we comment about the sign of $f(a+h)f''(a-h)$?

You can take cases Case-1 F(x)>0 therefore F''(x)<0 Which means concavity of graph is downward Case-2 F(x)<0. Therefore F''(x)>0 Which means concavity of graph is upward If you do rough ...
NN EDITZ's user avatar
3 votes
Accepted

Solving for the range of a function algebraically

Your range is the set of values of $y$ such that the equation $$ yx^2-x-16y=0 $$ has a solution $x\neq \pm 4$. When $y=0$ your equation is linear in $x$ and is satisfied by $x=0$. Otherwise, it is ...
GReyes's user avatar
  • 17k
1 vote
Accepted

Solutions to $y'=y$

This approach gives you all the solutions: First, note that $e^x$ is a solution. Suppose that $f$ is any other solution and consider $g(x)=f(x)e^{-x}$. Then $g'(x)=f'(x)e^{-x}-f(x)e^{-x} = 0$, so $g(x)...
jjagmath's user avatar
  • 18.9k
1 vote
Accepted

Maximum value of $2$ variable function $f(u,v)=\frac{\left(1-\sqrt{uv}\right)^2}{\frac{1-u^2}{2u}+\frac{1-v^2}{2v}}$

$$\displaystyle f(a,b)=\frac{\bigg(1-\sqrt{\tan\frac{a}{2}\tan\frac{b}{2}}\bigg)^2}{\cot a+\cot b}$$ $$\displaystyle f(a,b)=\frac{\bigg(1-\sqrt{\tan\frac{a}{2}\tan\frac{b}{2}}\bigg)^2}{\frac{1-\tan^2\...
whatamidoing's user avatar
  • 2,879
1 vote

Maximum value of $2$ variable function $f(u,v)=\frac{\left(1-\sqrt{uv}\right)^2}{\frac{1-u^2}{2u}+\frac{1-v^2}{2v}}$

Andrea Marino has already provided a good answer. One can avoid using derivative. We have $0\lt u\lt 1$ and $0\lt v\lt 1$. By AM-GM inequality, we have $u+v\ge 2\sqrt{uv}$ which implies $$\frac{1}{u+v}...
mathlove's user avatar
  • 144k
0 votes

Back to basics: Why do we care about symmetry of functions?

I think we can get some information from the symmetry of a function. e.g. a simple sin function, it has period and we can use it to solve equations. and we can only draw a period instead of a whole ...
TaylorZzz's user avatar
1 vote
Accepted

How to say these two distinct functions have the same structure?

It is possible to phrase this in very abstract terms along the lines I indicted in my comment but I think that would obscure the underlying idea. Here's a simple definition that I think captures the ...
Qiaochu Yuan's user avatar
-2 votes

Prove that assuming $f:S\rightarrow T$, $f$ is a bijection iff there is $g:T\rightarrow S$ such that $f\circ g$ and $g\circ f$ are identity maps

Reverse implication seems right, but we have a problem with the forward implication, let's discuss it. Since you have done the reverse implication right (or so we have assumed) let's just focus on the ...
Unknown 21103907's user avatar
1 vote

What is the maximum of $ \frac{\sin(n(x+a))}{\sin(x+a)} + \frac{\sin(n(x-a))}{\sin(x-a)}$?

Discussion on the special case in @user64494's answer. The maximum of $$f(x) := \frac{\sin 4(x + 1)}{\sin(x + 1)} + \frac{\sin 4(x - 1)}{\sin(x - 1)}$$ is given by $$f_{\max} = \frac{8\sqrt{6}}{9} (2 -...
River Li's user avatar
  • 40.3k
0 votes

Difference between $y = |x|$ and $|y| = x$

In $y = |x|$ you can see that if we replace x by -x then value of y not change which implies that their graph will be like even function so if you make the graph for x > 0 then reflect the graph in ...
JAYENDRA JHA's user avatar
0 votes

Find all the functions f: $\mathbb R^{2}\to \mathbb R$, where $\frac{\partial^2 f}{\partial y\partial x}=0$

Your hypothesis $\frac{\partial^2 f}{\partial y\partial x}=0$ means that $$\frac{\partial f}{\partial x}(x,y)=\frac{\partial f}{\partial x}(x,0),$$ or equivalently: $$f(x,y)-f(x,0)=f(0,y)-f(0,0),$$ i....
Anne Bauval's user avatar
  • 39.6k
1 vote

Finding a pair of functions that meets a few requirements

To find a pair of functions $ f(x) $ and $ g(x) $ that meet the given requirements, let's summarize the requirements and solve them step by step. Here are the requirements: $ f(x) = x \cdot g(x) $ $ ...
warashi nguyen's user avatar
0 votes

Functions that gradually slow down as X gets bigger

I'll elaborate on @PierreCarre's answer. What we need is a function $f$ that takes a number of steps $s$ and tells us what milestone we have reached ($f(s)$). I will first derive its inverse, because ...
Aldoggen's user avatar
  • 564
3 votes

Maximum value of $2$ variable function $f(u,v)=\frac{\left(1-\sqrt{uv}\right)^2}{\frac{1-u^2}{2u}+\frac{1-v^2}{2v}}$

You can substitute $s = (u+v)/2$ and $p =\sqrt{uv}$ and you get, using the sum-difference rule on the $(1-p^2)$ in the denominator: $$f(u,v) = g(s,p) = \frac{p^2(1-p)}{s(1+p)}$$ The two variables are ...
Andrea Marino's user avatar
1 vote
Accepted

Determining $t(x)$ from $\frac{dx}{dt}$?

Numerically, the calculation of $x(t)$ is a table with two rows, $t$, and $x$, and it simultaneously calculates $t(x)$: calculating $t(x)$ from is simply a matter of reading the table inversely. ...
TheAlertGerbil's user avatar
1 vote

Can a function $f$ exist such that $f(x)f(y) = xy + 1$ for positive real $x$ and $y$

There seems to be no useful "next best thing" that is true! To see this, consider an integral domain (containing a multiplicative neutral element $ 1 $) $ R $, and assume that $ x , y , u , ...
Mohsen Shahriari's user avatar
1 vote
Accepted

Proving $f + c = O(f)$ doesn't always hold- where is my mistake?

Your proof is correct. To simplify it you can just give the counterexample you stated: Taking $f(x)=1/x$, and $c=1$, there is no $k$ such that $f(x)+c < kf(x)$ for all $x$. The statement as given ...
Slugger's user avatar
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