Consider $$y= \frac{(x^2-9)}{x-3}$$ For y to be meaningful x must not equal 3 , So, 3 is not included within the domain of the function y (Call it f(x) if you will)  It follows that,  $$yx-3y = x^2 -9 $$ Plug in 3 to x out of curiosity, I find, $$ 0=0 $$ A perfectly valid relation! How is this possible?  This problem is even more predominant in partial fraction decomposition. Have a look  at this example I found on the web,  x cannot equal -1 or 2. How is it giving us the right answers then? What is the math behind this? When I do these kind of problems it feels like something's is happening mathematically, hidden from my view  All help is appreciated
120638
Write $\dfrac{4x+1}{x^2 - x - 2}$ using partial fractions. $$ \frac{4x+1}{x^2 - x - 2} = \frac{4x+1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x-2} = \frac{A(x-2)+B(x+1)}{(x+1)(x-2)}$$ $$4x+1 = A(x-2)+B(x+1)$$ $$x=2 \Rightarrow 4 \cdot2 + 1 = A(0) + B(3) \Rightarrow B = 3$$ $$x = -1 \Rightarrow 4(-1) +1 = A(-3)+ B(0) \Rightarrow A = 1$$ Thus, $$\frac{4x+1}{x^2-x-2} = \frac{1}{x+1} + \frac{3}{x-2}\textrm{.}$$  The substitution of $x$ ($x = 2, -1$) is a common method to find out the coefficient of the partial fractions. However, the equation on the third line is obtained by multiplying $(x+1)(x-2)$, which is assumed to be nonzero. Here we have a contradiction. Furthermore, the original function is not defined at $x=-1,2$. How can we substitute these value for $x$? So is this method valid and rigorous? How to modify it so that it is rigorous?
6750
I have the following question:    Show that if $z=x+ct$ and $y=x-ct$ then:  $$\frac{\partial^2 u}{\partial t^2}-c^2\frac{\partial^2u}{\partial x^2}=0\text{ implies } \frac{\partial^2 u}{\partial z \partial y}=0$$    If I had $u$ it would follow trivially I imagine, but I don't have $u$. It would seem that $u$ is linear in $t,x$ and since $t,x$ are linear in $y,z$ it seems logically consistent, but I am not sure what to do. Perhaps this is related to harmonic functions.  What should I first do to get this started?    Perhaps I notice that $c=\frac{z-x}{t} \text{ and } c=\frac{x-y}{t}$  and thus $\frac{z-x}{t} =\frac{x-y}{t}\implies x-y=z-x\implies 2x=y+z$
39864
Let $$f(x)= \frac{x}{x-9} \text{ and } g(x) = \frac{-6}{x+5} $$   Find the domain of $f \circ g.$    Please show the steps and substitution/algebra to the problem. Thank you.
5814
If $0\leq x,y\leq\frac{\pi}{2}$ and $\cos x +\cos y -\cos(x+y)=\frac{3}{2}$, then how can I prove that $x=y=\frac{\pi}{3}$? Your help is appreciated.I tried various formulas but nothing is working.
70913
If I have, for example, the function $$f(x)=\frac{x^2+x-6}{x-2}$$ there will be a removable discontinuity at $x=2$, yes?  Why does this discontinuity exist at all if the function can be simplified to $f(x)=x+3$?  I suppose the answer is that you can't simplify it because you can't divide by something that could potentially equal $0$.  But what if you start with $f(x)=x+3$? What's stopping you from multiplying it by $1$ in the form $\frac{x-2}{x-2}$? Can multiplying by $1$ really introduce a discontinuity to the function?
98449
$M(x,y)dx + N(x,y)dy=0$ is said to be a perfect differential when   $$\frac{\partial (M(x,y))}{\partial y}=\frac{\partial (N(x,y))}{\partial x}$$  Let $M_y=\frac{\partial (M(x,y))}{\partial y}$ and $N_x=\frac{\partial (N(x,y))}{\partial x}$.  In case if:  $\frac{M_y-N_x}{N(x,y)}$ is only a function of x only (say $f(x)$) then it has a integrating factor $e^{\int f(x)dx}$.  But if $\frac{M_y-N_x}{N(x,y)}$ is a constant then what will be the integrating factor?     Say for this problem:   $(axy^2+by)dx+(bx^2y+ax)dy=0$   Is there any other method to solve this differential equation?
114686
How to deal with $m+\frac{4}{m^2}\geq3$ for every $m &gt; 0$ ? I multiplied both sides by $m^2$ and got $m^3+4-3m^2\geq0$ and have no idea how to continue with this
33120
The task is as follows:     Given: $$F(x,y,z) = 0$$    Goal:  Show    $\frac{\partial  z}{\partial y}|_x \frac{\partial  y}{\partial x}|_z \frac{\partial x}{\partial z} |_y = -1$   Here is my work so far:  (1) Differentiate with respect to y, I get:  $0 + F_2 + F_3  \frac{\partial z}{\partial y} = 0$  So $ F_3  \frac{\partial z}{\partial y} = - F_2$  (2) Differentiate with respect to x, I get:  $F_1 + F_2 \frac{\partial y}{\partial x}  + 0 = 0$  So $ F_2  \frac{\partial y}{\partial x} = - F_1$  (3) Differentiate with respect to z, I get:  $F_1  \frac{\partial x}{\partial z}    + 0  + F_3 = 0$  So $ F_1  \frac{\partial x}{\partial z} = - F_3$  (4) After some manipulations with the $F_i$, I get to the conclusion that $\frac{\partial z}{\partial y}* \frac{\partial y}{\partial x} * \frac{\partial x}{\partial z}  = -1$, so when evaluated with x, z, y respectively, conclusion is still true  My questions are:  (1) Is my proof correct?  (2) For example, when I differentiate with respect to y, I "let" $F_1$ be 0 and find partials for other coordinates.    I had a hard time trying to explain to my friend on the reason(s) why I can do such "let be 0" thing.  Although I think if I can't do that, then there is no way that I can reach the conclusion, but I somehow feel confused about the fact too.  Since my book is doing it that way, my understanding is that I can do such "let be 0" thing based on the independece of x with respect to y, when I differentiate with respect to y.  But is my thought ok?  Would someone please help me on this question? Thank you very much ^_^
13705
Suppose f is a function such that $f'(x) = \frac{1}{x}$ for all $x &gt; 0$. Prove that if $f(1) = 0, then f(xy) = f(x) + f(y)$ for all $x,y &gt; 0$  As stated above, how do i go about doing this question?   Hi all, i would like to apologize if the question is missing context or other details.   This question was part of my calculus tutorial. I was on the topic of application of differentiation when this question was given to me.  Thus far i have learnt about intermediate value theorem, mean value theorem, rolle's theroem, Fermat's thereom.  As of this point in my learning, i have not touched upon integration. The only reason why i accepted the answer with integration in it was because i understood it.  Thus i believe i am suppose to use one the few theorems that i've stated to solve this question. But i have no idea where to start.  At first i differentiated the equation $f(xy)=f(x)+f(y)$ with respect to x  i believed the answer to be $f'(xy)\frac{dy}{dx}=f'(x)+f'(y)\frac{dy}{dx}$  which would be $\frac{1}{xy}\frac{1}{x}=\frac{1}{x}+\frac{1}{y}\frac{1}{x}$  At this point a few questions came to my mind  1) am i suppose to differentiate the eqn in the first place 2) what has $f(1)=0$ got to do with anything 3) what theorem am i suppose to use to solve this question  This was my thought process throughout the question and i hope someone would point out my mistakes. Thank you
25534
I'm reading Spivak's Calculus:  2 What's wrong with the following &quot;proof&quot;? Let $x=y$. Then $$x^2=xy\tag{1}$$ $$x^2-y^2=xy-y^2\tag{2}$$ $$(x+y)(x-y)=y(x-y)\tag{3}$$ $$x+y=y\tag{4}$$ $$2y=y\tag{5}$$ $$2=1\tag{6}$$  I guess the problem  is in $(3)$, it seems he tried to divide both sides by $(x-y)$. The operation would be acceptable in an example such as:  $$12x=12\tag{1}$$ $$\frac{12x}{12}=\frac{12}{12}\tag{2}$$ $$x=1\tag{3}$$  I'm lost at what should be causing this, my naive exploration in the nature of both examples came to the following: In the case of $12x=12$, we have an imbalance: We have $x$ in only one side then operations and dividing both sides by $12$ make sense. Also, In  $\color{red}{12}\color{green}{x}=12$ we have a $\color{red}{coefficient}$ and a $\color{green}{variable}$, the nature of those seems to differ from the nature of $$\color{green}{(x+y)}\color{red}{(x-y)}=y(x-y)$$ It's like: It's okay to do the thing in $12x=12$, but for doing it on $(x+y)(x-y)=y(x-y)$ we need first to simplify $(x+y)(x-y)$ to $x^2-y^2$.
76355
Given $x,y \in (-2,2)$ and $xy=-1$  Minimum value of $$f(x,y)=\frac{4}{4-x^2}+\frac{9}{9-y^2}$$  My try:  Converting the function into single variable we get:  $$g(x)=\frac{4}{4-x^2}+\frac{9x^2}{9x^2-1}$$  $$g(x)=\frac{4}{4-x^2}+1+\frac{1}{9x^2-1}$$  Using Differentiation we get:  $$g'(x)=\frac{8x}{(4-x^2)^2}-\frac{18x}{(9x^2-1)^2}$$  $$g'(x)=2x\left(\frac{4(9x^2-1)^2-9(4-x^2)^2}{(4-x^2)^2(9x^2-1)^2}\right)$$  $$g'(x)=70x\frac{9x^4-4}{(4-x^2)^2(9x^2-1)^2}=0$$  So the critical points are:  $x=0, x=\pm \sqrt{\frac{2}{3}}$  But $x \ne 0$ since $xy=-1$  $$g'(x)=70x \frac{(3x^2-2)(3x^2+2)}{(4-x^2)^2(9x^2-1)^2}$$  By using derivative test we get Minimum occurs when $x=\pm \sqrt{\frac{2}{3}}$  Hence $$x^2=\frac{2}{3}, y^2=\frac{3}{2}$$  Min value is $$\frac{4}{4-\frac{2}{3}}+\frac{9}{9-\frac{3}{2}}=\frac{12}{5}$$  Is there any other approach?
50786
I was reading a solution when I came across this statement.     So $$\frac{\partial f_i}{x_j}=\frac{\partial f_j}{x_i}.$$ Then there exists a differentiable function $g$ on $\mathbb{R}^n$ such that $\frac{\partial g}{\partial x_i}=f_i$.   Why is this true?
46125
I want to know whether $\lim_{(x,y)\to (0,0)}\dfrac{x^2y^2}{x^3+y^3}$ exists or not. I tried to approximate to (0,0) from different "paths" and the result was always 0. For example,  $f(x,mx^2) = \dfrac{m^2x^3}{1+m^3x^3}$  But that doesn't show that the limit is 0.
113259
I know this is impossible, but why is the following not possible:     $y + x = 3$ is the same as $y^2 + x^2 = 9$   They're meant to be equivalent.
30520
I haven't found any useful method to solve the following problem: Prove that if $x,y,z\in\mathbb{Z}$ and $x^3+y^3=3z^3$ then $xyz=0$.  Source:
30226
Do we rely on certain intuition or is there an unofficial general crude checklist I should follow?  I had a friend telling me that if the sum of the powers on the numerator is smaller then the denominator, there is a higher chance that it may not exists. And if the sum of powers on the numerator is higher then the denominator, most likely, it exists. Also if there is a sin or cos or e-constant, it most likely exists.  How can I know what to do at the first glance given so little time exists for me in exam to ponder? If I spend all my time on figuring out a two-path test when the limit exists, that would be a huge disaster. Is this one of those cases where practice makes perfect? Example: $$\lim_{(x,y)\to(0,0)}\frac{(\sin^2x)(e^y-1)}{x^2+3y^2}$$ Please give me a hint and where do you get the hint. Example: $$\lim_{(x,y)\to(0,0)}\exp\left(-\frac{x^2+y^2}{4x^4+y^6}\right)$$ I need a hint for this too.  Common methods I have learnt for reference: Two-Path test, Polar Coordinates, Spherical Coordinates, Mean Value Theorem using inequalities.
13058
Suppose $f=f(x,y(x))$.  Then applying the chain rule we get $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial x}$.  From this it seems that it always holds that $\frac{\partial f}{\partial y}\frac{\partial y}{\partial x}=0$.  Where's the mistake?