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This is possible if and only if $\tilde x$ is an isolated point of the domain. Example 1: For example, the domain could be $$(-\infty,\tilde x - \epsilon)\cup \{\tilde x\}\cup (\tilde x +\epsilon,\infty)$$ if it is a subset of $\mathbb R$ (which, incidentally, is not specified) and the function rule could be $$f(x) = \begin{cases} 0 & \textrm{ if }x ... 12 This function cannot exist, if it must be continuous. Suppose such a function f:R \to R exists, and for some x \in R, f(x) = c \ne 0. Let y \ne x. By the Intermediate Value Theorem, then \exists z \in [x, y] (or [y, x]) such that 0 < f(z) < c. 7 Good question. The functions \frac{x^3-8}{x-2} and x^2+2x+4 are equal for all x except at x=2, where the first function is undefined, and the second function is 12. However, if you are taking the limit of the first function as x \to 2, all you care about are the values of the function at x near 2; the value of x at 2 is irrelevant when ... 5 The difference is that \mapsto denotes the function itself. Thus you need not name the function. a\mapsto b fully describes the action of the function. \to, on the other hand, describes only the domain and codomain. Thus one might say x\mapsto x+1 is equivalent to f(x)=x+1, in which case we would say f:\mathbb{R}\to\mathbb{R}. 5 The \to arrow points from the domain of a function to its codomain or target set. The \mapsto arrow (fittingly called \mapsto in TeX) shows what an individual element of the domain will be mapped to, i.e. it shows what the function does while \to shows where it operates (so to say). So, the elaborate way to write a function definition is ... 5 The answer is no. Take a look at this paper. The corollary on p. 353 ensures that for any two disjoint countable dense subsets A, B \subset \mathbb{R}, there exists an everywhere differentiable function H such that H'>0 on A and H'<0 on B. Note that this function is monotone on no subinterval of \mathbb{R}. I should give credit to user ... 5 We say f:A\to B is a function if, for any a\in A there exists exactly one b\in B such that f(a)=b. Since \infty is not an element of \mathbb Z^+, if you want f:\mathbb Z^+\to \mathbb Z^+, you can't have f(n)=\infty for any positive integer n. 4 The following is maybe too fancy. Let a=-1 and b=1. Let$$f(x)= \begin{cases} x^2\sin(1/x) & \text{if } x \ne 0, \\ 0 & \text{if } x=0. \end{cases}$$The derivative of this function exists at 0, but is not continuous there. If you want greater smoothness at 0, replace x^2 by x^{77} If you want the same sort of thing on a general ... 4 Letting t=2^{1/3}\Rightarrow t^3=2, we have$$(x-2)^3=(t^2+t)^3\Rightarrow x^3-6x^2+12x-8=t^6+3t^5+3t^4+t^3=4+6t^2+6t+2.$$Then, we have$$\begin{align}x^3-6x^2+6x&=(6t^2+6t+6)+8-6x\\&=6t^2+6t+14-6(t^2+t+2)\\&=14-12\\&=2.\end{align}$$4 You are right, f and g are different functions since they don't have the same domain. And, yes, f is not continuous at 2 because it is not defined at 2. If we talk about limits then you can use the fact that if f(x) = g(x) except at x=a, then$$ \lim_{x\to a} f(x) = \lim_{x\to a} g(x). $$So now you note that for any x\neq 2, you indeed ... 4 There is no continuous function f:\mathbb{R}\rightarrow\mathbb{R} such that f(x)=0 for all x\not=x_0 and f(x_0)\not=0, since by definition of continuity$$f(x_0)=\lim_{x\rightarrow x_0,x\not=x_0} f(x)=0$$3 As @Josh Keneda points out, a characteristic function won't work in general if L is infinite. But we can use the following slight modification:$$ f(x) := \sum_{n=1}^\infty \frac{1}{n} \cdot \chi_{x_n} (x). $$It is clear that f is discontinuous at every x_n, because the set where f(x) = 0 is dense. Below is a proof that f does what you want ... 3 I'm guessing that \;x=(x_1,x_2)\; , and then$$f(x)\ge 0\;\;\forall\;x\in\Bbb R^2\;,\;\;\;and\;\;\;\; f(2,1)=0$$so... 3 Yes, since differentiability is sufficient for continuity. Not necessarily. Consider the function $$f(x,y)=\left\{ \begin{array}{lr} 0 & x=0\textrm{ or }y=0\\ 1 & otherwise \end{array} \right.$$ Since \frac{\partial f}{\partial x}(0,0)= \lim_{h \to 0} \frac{f(h,0)-f(0,0)}{h}=\lim_{h \to 0} ... 3 As stated in the comments, we have x=y=0 implies 2f(0)=4f(0) so f(0)=0. Also we must have an even function since with x=0 we see that f(y)+f(-y)=2f(y) implying f(y)=f(-y). Now for k\in\mathbb N we can prove by induction that f(ky)=k^2f(y). This is clearly true for k=1. So if we assume this holds for k and proceed to k+1 we see that ... 3 You are close. I think some people use B^A to denote the set of all functions from A to B. So you could write$$\left\{f \in B^A: \left|f^{-1}(\{k\})\right|=c_k\right\}$$to denote the set of all functions such that c_k elements of A are sent to k \in B. 3 Is this what you're looking for?$$\min(a,\,b) \; = \; \frac{a\,+\,b\; - \; |a-b|}{2} \max(a,\,b) \; = \; \frac{a\,+\,b\; + \; |a-b|}{2} 3 I'll just count the number of times the sum=A[i]+A[j]+A[k] statement gets executed: \begin{align*} \sum_{i=1}^n \sum_{j=1}^i \sum_{k=1}^j 1 &= \sum_{i=1}^n \sum_{j=1}^i j \\ &= \sum_{i=1}^n \frac{i(i+1)}{2} \\ &= \frac{1}{2}\left[\sum_{i=1}^n i^2 + \sum_{i=1}^n i\right] \\ &= \frac{1}{2}\left[\frac{n(n + 1)(2n + 1)}{6} + ... 3 If you say x^2+y=\text{constant}, you're defining a curve in the xy-plane, which is a parabola y=-x^2+\text{constant}. If you move along that curve, both x and y are changing. But if you write z=y^2+x, defining z as a function of those two variables, both of which can vary freely, then the expression \partial z/\partial x means the rate of ... 3 In this context, \mathbb{R}-\{x\} means the set of all real numbers that are different from x. Injective: if f(s)=f(t) then \frac{5s+1}{s-2}=\frac{5t+1}{t-2} so that (5s+1)(t-2)=(5t+1)(s-2)\implies 5st-10s+t-2=5st-10t+s-2 $$which simplifies to imply that s=t. Surjective: suppose that y is any number in \mathbb{R}-\{5\}, let us demonstrate ... 2 Using L'Hospital rule for \infty/\infty format, if \lim_{x\to\infty}f(x)=\infty:$$\lim_{x\to\infty}\frac{f(x)}{x}=0\implies\lim_{x\to\infty}\frac{f'(x)}{1}=0$$See if you can use the same for second part similiary. 2$$ (2k + 1)^2 + (2l + 1)^2 = 4\left(k^2 + l^2 + k + l\right) + 2 $$This means that \left(a^2 + b^2\right) = 4 \lambda + 2, meaning that 4 could not possibly divide \left(a^2 + b^2\right), since there's always a remainder of 2. ...by the way, this means that for 4 to divide a^2 + b^2, both a and b must be even, since certainly if one is odd ... 2 Continuity of f can be seen as "if you move just a bit, the image through f will move a bit too". In your case f^{-1} being not continuous would imply that you can find two points on the codomain "close enough" such that their f-counter image are actually far away. Think of a rectangle and glue two opposite sides together, and think of the obvious ... 2 It reads as: "Show that if f of x equals f of y for arbitrary x and y in A with x different than y, then x equals y". The comma as nothing to do with pairs, it is just a comma from common language. Moreover f \colon A \to B should be read as: f is a function from the set A into the set B. 2 The first part should be read "Suppose that f is a function from the set A into the set B." The second part doesn't make sense, as written. It should say "Show that if f(x)=f(y) for arbitrary x,y\in A, then x=y." That is, if f(x)=f(y) implies that x=y for any elements x and y of A (not necessarily different elements, just with ... 2 Assume you can do it for n=2^k, specifically n=4, then f(\frac{(x_1+x_2+x_3+(x_1+x_2+x_3)/3)}{4})\leq \frac{f(x_1)+f(x_2)+f(x_3)+f((x_1+x_2+x_3)/3)}{4} rearrange to get n=3 case. Same approach can be generalized to any non 2^k number. 2 Let me answer this by way of a different problem since there is no way to give a hint without giving the answer directly. Suppose we want to find out how far the ball falls in .5 seconds. Our distance function d(t) tells us this. In order to find out what the distance is, we need only to plug in the time (in seconds). Thus the distance the object fell ... 2 As this is a question about notation allow me a few remarks: You won't wind up with a "four-element set". As B has only two elements, f[A] - i.e. the image of A - will have at most two elements. This also means that your \{5,5,6,5\} and \{5,6,6,5\} are the same set, namely just \{5,6\}. The functions will differ, though. If you write your ... 2 Your operative definition of \longrightarrow and \longmapsto are correct. In your example you are dealing with vectors in \mathbb{R}^n, which are elements of a set, so you should use \longmapsto. see also this wikipedia article: http://en.wikipedia.org/wiki/List_of_mathematical_symbols 2 You have$$ A\cap B\subset A\implies f(A\cap B)\subset f(A),\\ A\cap B\subset B\implies f(A\cap B)\subset f(B)  so it follows that $f(A\cap B)\subset f(A)\cap f(B)$.