The curve of the function $f (x)$ is given.  Find the real roots of the equation $f (4-3x^2)=0.$  I know the possible solution:  From  the equations $4-3x^2=-2$  and  $4-3x^2=4$ we have $ x=\left\{0,\sqrt 2,- \sqrt 2\right\} $.  Now, I want to create a little proof for the alternative solution before moving on to the calculation. It is obvious that, we don't know directly what the explicit form of $f(x)$ is. Goal:  I want to find a new $ g (4-3x ^ 2) = 0 $ equation that will definitely give the same real solutions as the equation $f (4-3x ^ 2) = 0.$  Let's define the function $g:\mathbb{R}\longrightarrow \mathbb {R}$. For any $x\in\mathbb {R}$ if $f(x)=0$, then $g(x)=0$ and if $f(x)≠0$, then $g(x)≠0$ and there exist at least one $x=r, r\in \mathbb{R}$, which gives $f(r)≠g(r)$. According to our definition of $g(x)$, we deduce that for any $x\in\mathbb{R}$, if  $f(4-3x^2)=0,$ then $g(4-3x^2)=0$  and if  $f(4-3x^2)≠0,$ then $g(4-3x^2)≠0.$ Finally, we can deduce that solving the equation $ f (4-3x ^ 2) = 0 $ is equivalent to solving the equation $ g (4-3x ^ 2) = 0. $ In other words, the real roots of the equation $ f (4-3x^2) = 0 $ are the same as the equation $ g (4-3x^2) = 0$. So, it is enough to solve the equation  $g (4-3x^2) = 0$.   Let, $g(x)=(x+2)(x-4)$. Then we have, $$g(4-3x^2)=0 \Longrightarrow  (4-3x^2+2)(4-3x^2-4)=0 \Longrightarrow -3x^2(6-3x^2)=0 \Longrightarrow x=\left\{0,\sqrt 2,- \sqrt 2\right\} $$  Question: Is my proof-solution correct?
The graph of the function $f (x)$ is given.  How many different real roots does the $f (4-3x^2)=0$ equation have?  Here is possible solution:  $$\Bigg [{4-3x^2=-2 \\ 4-3x^2=4} \Longrightarrow x=\left\{0,\sqrt 2,- \sqrt 2\right\} $$  Then, here is different solution:  $$f(x)=x^2-2x-8$$ $$f(4-3x^2)=0\Longrightarrow (4-3x^2)^2-2(4-3x^2)-8=0 \Longrightarrow 9x^4-18x^2=0 \Longrightarrow x=\left\{0,\sqrt 2,- \sqrt 2\right\} $$   Problematic point. Here, I assumed the function is quadratic. But, obviously, this function may not be quadratic.  My questions:  Question $-1:$ Is the first solution completely correct?  Question $-2:$ Can we say that, the second solution is completely and definitely wrong? If so, is it possible to add something to this method and turn it into the right solution?
I'm currently using the below script to timestamp when an edit is made to column 7 on a sheet named "Sheet 1".  function onEdit(e) {    if ([7].indexOf(e.range.columnStart) != -1) {    e.range.offset(0, 7).setValue(new Date()).setNumberFormat("dd.MM.yyyy at HH:mm");    }   }     I have a second sheet "Sheet 2" which I need to run the same script on but against column 5.  function onEdit(e) {    if ([5].indexOf(e.range.columnStart) != -1) {    e.range.offset(0, 7).setValue(new Date()).setNumberFormat("dd.MM.yyyy at HH:mm");    }   }     I've tried .getSheetByName("Sheet1") but just cant seem to get it working correctly. I'm struggling to understand the concept of quantum entanglement. I've distilled my understanding into an analogy, and I need your help to validate it. Here it is:  Let's say I receive two envelopes. Both envelopes have this written on them:     Hello! Open this letter...   If the paper inside is red, then other envelope has blue paper.   If the paper inside is blue, then other envelope has red paper.   So, if I open one letter, and the paper inside is red, then I know the other letter contains the blue paper.  Is this equivalent to quantum entanglement? The fact that I will know the color of the other paper when I open (i.e. observe) one envelope?  In theory, if one of the letters "appeared" in a different galaxy, I would still know the color of the paper inside that letter, just by opening my letter, instantly, correct?  Before I open the letter, is the paper inside each envelope in superposition (??), i.e. 50/50, red or blue?  In this analogy, I am assuming that what's written on the letters is true. Is this a correct assumption in quantum physics? I think so yes?  In real quantum physics, would I be able to change the color of one paper, so that I change the color of the other one? No right? Definitions of "linearity" across branches of mathematics or levels of math education Linearity is a ubiquitous concept in mathematics; however, each branch of mathematics appears to have its own definition of what a linear map (function, functional, functor, transformation, form, operator, or whatsoever) is. These different definitions can lead to confusion, especially when people apply conflicting definitions of linearity to the same mathematical object (just  to figure out what I mean).  One of my favorite examples with potential for conflict is the natural logarithm $\ln : \mathbb{R}_{&gt;0} \to \mathbb{R}$, which the majority of people, as I could learn during hot disputations not only  but also with colleagues in my field, regard as clearly non-linear, even though it fully complies with the definition of a linear map from the vector space $\mathbb{R}_{&gt;0}$ to the vector space $\mathbb{R}$ (cf. Definition 4 below). A similar conflict arises analogously for the exponential function $\exp :\mathbb{R} \to \mathbb{R}_{&gt;0}$.  I do not want to initiate yet another discussion here, about what the right and what the wrong definition of linearity is! Rather, ...  ... I would like to compile a list of existing definitions, to get a better overview of the different uses of the term "linearity" across the different branches of mathematics and levels of education.    Let be begin with the following four, which are certainly far from constituting a complete list. What about bilinear or multilinear forms, linear morphisms and linear functors in category theory, linear metrics, or other concepts of linearity? I am looking forward to see your answers.  At school I was taught that     Definition 1:   A function $f$ is linear if it is of the form $f(x) = ax + b$.   Later I learned that $f(x) = ax + b$ is affine, rather than linear, because     Definition 2:   A function $f$ is linear if it is of the form $f(x) = ax$.   From real analysis and systems theory, I know the following     Definition 3:   A map $f$ is linear if         $f(x + y) = f(x) + f(y)$   $f(a \cdot x) = a \cdot f(x)$.      This definition also applies to the expected value in statistics, as        $\mathrm{E}(X + Y) = \mathrm{E}(X) + \mathrm{E}(Y)$   $\mathrm{E}(a \cdot X) = a \cdot \mathrm{E}(X)$.      In a linear algebra course, I learned for the first time that we must not forget to take into account that the vector-space operations of the domain and the codomain of a map $f$ may be dissimilar. In fact, if $(V, \oplus, \odot)$ and $(W, \boxplus, \boxdot)$ are two $\mathbb{K}$-vector spaces, then     Definition 4:   A map $f : V \to W$ is linear if         $f(x \oplus y) = f(x) \boxplus f(y)$   $f(a \odot x) = a \boxdot f(x)$.         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