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First, we want to make sure that the term inside the logarithm is always positive, as $\log{x}$ is defined only for $x > 0$. $$x-x^2 = x(1-x) > 0 \implies 0 < x < 1$$ Next, we need the term inside the square root to be positive or equal to 0, so we get: \begin{align*} \log_{0.4}(x-x^2) &\geq 0\iff\\\frac{\ln(x-x^2)}{\ln{0.4}} &\geq ... 0 You have here two conditions on the domain: x-x^2 >0 \tag{1}$$because \log_{0.4}(t) is defined only for t>0, and$$\log_{0.4}(x-x^2)\geq 0 \tag{2}$$because \sqrt{t} is defined only for t\geq 0. From (1) we want that x>x^2 and so x\in (0,1) because x^2 \geq 0 and x^2\geq x for x\geq 1. For condition (2), we know that ... -1 First, I have to rewrite the question. If X and Y are two non-empty sets where f:X\to Y is a function. A function F:P(X)\to P(Y) is defined such that$$F(C) = \{f(x):x\in C\}\text{ for }C \subseteq X$$and$$F^{-1}(D) = \{x:f(x)\in D\}\text{ for }D \subseteq Y$$for any A \subseteq X and B \subseteq Y then: F^{-1}(F(A))=A ... 1 It seems you are having more conceptual difficulties. The problem is fairly straightforward in this case, the solution (as mentioned by others) is quite easy to come up with. I just wanted to outline how one might approach this problem, since you did not know how to do just that. First we are asked to find a bijection between the sets \{1,2,3,...\} and ... 0 Try the function n \mapsto -n 1 The question seems garbled, but I'll try to clear it up. We're given a function f\colon x\mapsto x^2 - 3, \colon A\to \Bbb R. defined on A=\{x\mid x\ge 0\}, the nonnegative reals. So A = \Bbb R_{\ge 0}. On \Bbb R_{\ge 0}, f sends 0 to -3, its strictly increasing, unbounded and continuous, and more. It's an injection, with range B=\{x\mid ... 0 I have proved it as follows. Taking the expression for f'(x), multiplying through by (1+x^{2} \sin^{4} \frac{1}{x}), we need to prove the following inequality, call it (a):$$\left(1+x^{2} \sin^{4} \frac{1}{x}\right) \arctan(x \sin^{2} \frac{1}{x}) + x\sin^{2}\frac{1}{x} > 2\sin\frac{1}{x} \cos\frac{1}{x} \qquad \mbox{(a)}$$Observe that since ... 3 The domain is correct; the solution to f(x)=1 is correct too: indeed$$ \frac{2x-1}{x^2}=1 $$becomes$$ (x-1)^2=0 $$Now for the inequality it is essentially the same. You want to see for what value of x you have f(x)\le1, that is,$$ \frac{2x-1}{x^2}\le1 $$This translates into$$ (x-1)^2\ge0 $$so every x\in D_f satisfies the inequality. Or you ... 0 We can first calculate the 1st derivative.$$\frac{df}{dx} = \frac{d(2x-1)}{dx} \cdot x^{-2} + \frac{d(x^{-2})}{dx} \cdot (2x-1) = \frac{2}{x^2} - \frac{2(2x-1)}{x^3}$$Set \frac{df}{dx} = 0, we have x =1. So the turning point is x =1. By calculating the second derivative, we can find out that this point is the maximum of the function. f(x=1) = 1. ... 0 Simple, use derivative, not approximations. You get: (x\arctan(x\sin^2(\frac{1}{x})))'=\arctan(x\sin^2(\frac{1}{x}) + \frac{x}{1+x^2\sin^4(\frac{1}{x})}(sin^2(\frac{1}{x}) + \frac{x}{1+x^2\sin^4(\frac{1}{x})}(-2x\sin\frac{1}{x}\cos\frac{1}{x}(-\frac{1}{x^2})) so all are positive for x \geq 1. 3 f(x)= \frac{x+a}{x^2-1} a \in {1,2,3,.....,100} When you take a=1, then f(x) will not be able to achieve 0 as denominator also becomes 0. In other words x= \pm 1 are not in domain of function. For all other values of a, f(x)=0 for x=-a Now set \frac{x+a}{x^2-1}=1 and obtain quadratic in x Try to use the fact that Ax^2+Bx+C=0 ... 1 Here's an intuitive way to think of it. Consider a more general case:$$y=\sin(\frac{1}{x})$$When x gets closer and closer to 0, there will be an infinite amount of times it crosses the x-axis. This is because \frac{1}{x} approaches -\infty from the left and \infty from the right. When \frac{1}{x} gets very close to the y-axis it will grow ... 1 Since \sin x is zero at \pi,2\pi,\ldots, (\sin x)(\sin x^{-2})  is zero at \dfrac{1}{\sqrt \pi},\dfrac{1}{\sqrt {2\pi}},\ldots , so the answer to your question is "yes". \sin\left(\dfrac1x\right) is a simpler example with the same property. 0 First, this function f can be defined at 0 by continuity, with value 0. Now take x_k = \frac{1}{\sqrt{2\pi k}}, k >0. All x_k are in [0,1], and f(x_k)=0, an infinity of times in the interval, and increasingly denser near 0. A lot of them are quite easy to build, and motivate of lot of exercices for students to study continuity, ... 1 Yes. Def'n : A linear order <_S on a set S is order-dense iff \forall x,y\in S\;(x<_S y\implies  \exists z\;(x<_Sz<_Sy)). Theorem.(Cantor). If S is countably infinite and <_S is an order-dense linear order on S with no end-points (no <_S max or min) then there is an order-isomorphism from S to Q . So let S be the ... 0 Have you learnt what the Fourier Transform does? it takes a function g(t) in the time domain, and gives a new function \mathcal{F}\{g\} = G(f), which for different values of f (which are frequencies), provides the amplitude of the given frequency in the original g(t). You can take the signal you have, apply the fourier transform on it, take as many ... 1$$p(2,4)= p((1,2)+(1,2))=p(1,2)+p(1,2)=1+1=2\neq3,$$so p is non linear. 3 Two of the most frequently used step functions are the "floor" and "ceiling" functions. The floor function takes a real number and returns the largest integer(whole) number less then or equal the real number. For a positive real number this is equivalent to cutting off all the numbers after the decimal place. The ceiling function returns the smallest ... 1 Your question is the same as asking: "are there scalars a and b so that a \cdot 1+ b \cdot 2 = 1 and a \cdot 2 + b \cdot 4 = 3." You should be able to show easily that this is impossible, so the function is not linear. 1 You want x+2y=1 and 2x+4y=3\implies x+2y=\frac{3}{2} This is impossible. -5 Step functions are regularly used to describe cumulative densities for discrete random variables. For example, consider a random variable, X, to be the throw of a single die. The possible values that X can take are \{1, 2, \dots, 6\}. The cumulative density function P(X\leq x)={F_X}(x) is easily drawn. At each score of the die the function jumps, ... 2 Presumably you're allowed to assume that f is differentiable. The Idea: If you draw a picture you convince yourself that if, say, y<z<x then$$\frac{f(y)-f(z)}{y-z}\le\frac{f(y)-f(x)}{y-x}.$$Now let z\to y, the left side of the inequality tends to f'(y) and you're done. Now to make that idea into an actual proof from what we're given we write ... 0 Consider any plane simple differentiable loop. We say that this is convex if one can draw a straight segment connecting any two points, without leaving the "inside" of the loop. Convexity is just the name of this property, a way -if you like- to spend less time conveying the meaning. Now, by Dini's theorem the support (or graph, the actual line in the ... 0 This is what I tell my students. We know what a convex set is, and we need a name for functions satisfying the condition above. By (verbal) analogy we call them convex, too. But in that case the curve is the bottom (down) part of a convex region, so we can say that convex means convex down. But then concave up should equal convex down, i.e., the curve is the ... 0 Here is an example of modulo function using recursion - var modulo = function(x, y) { var result = 0; if (x === 0 && y === 0) { return NaN; } else if (x < y) { return x; } else { result = modulo(x - y, y); } return result; }; // example: modulo(23, 4); //3 2$$f(x)=\begin{cases} 2x-4,&x\gt0\\ -3x+1,&\text{otherwise} \end{cases},$$means that f(x)=2x-4 when x satisfies the condition “x\gt0”, and f(x)=-3x+1 when x doesn't satisfy the condition “x\gt0”, i.e. when x\leqslant0. So to determine the value of f for a certain x we would first have to know whether this number satisfies the first ... 2 First, you can define \phi however you want on F\setminus Im(f), it does not matter for your problem. Then for any y\in Im(f), write y=f(x) and define \phi(y) = g(x). To see that this is well-defined, you have to check that g(x) does not depend on the choice of x but only on y. But if y=f(x') then by hypothesis g(x')=g(x). So \phi is ... 0 Well, erf(x)\le1; that's probably not the bound you want. But if you're talking about large x that's the best bound on erf(x) that you're going to get. Seems to me what matters is how small 1-erf(x) is for large x. For that you want to bound the "complementary error function": If x>0 then$$\int_x^\infty ...

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Corrected Question The question was incorrect to begin with. It should ask: $\def\rr{\mathbb{R}}$ $\def\less{\smallsetminus}$ Given a function $f : \rr\less\{\frac12\} \to \rr$ such that $f(\frac{x-2}{2x}) = 2x+5$ for every $x \in \rr\less\{0\}$, find $f^{-1}(3)$. Correct Solution The solution you gave is also logically incorrect. Given $y \in ... 0 This can me asked as a graph theory problem: how many graphs with vertices$\{a,b,c,d\}$exist such that each vertex lies in exactly one cycle of length at most$2$? Case 1: All cycles are of length 1. This is equivilant to the identity function, which we know is unique. However from a graph theoretic point of view this is equal to$\binom{4}{4}$because ... 0 The graph is made up of two parts: When$x$is odd, it is the graph of$f(x) = 3x + 1$. When$x$is odd, it is the graph of$f(x) = x/2$. That's why you get two lines: the first with slope$3$on top and the second with slope$1/2$on the bottom. Note that the lines between the dots should not be there. The domain is only on the integers. 7 The number of such functions is the number of ways of dividing our set, in this case$\{1,2,3,4\}$into$1$or$2$element subsets. For given such a subdivision, we can define$f(x)$to be$x$if$x$is a singleton in the subdivision, and by$f(x)=y$,$f(y)=x$if in the subdivision$x$and$y$are a "couple." Conversely, a function$f$such that$f(f(x))=x$... 2 Note that a function does not necessarily have a closed expression which defines it. that is you to not necessarily need to express it$f(x)=5-x$or$f(x)=x$, but rather we may just state where each element go. So one function is if$f(1)=4, f(4)=1, f(2)=3, f(3)=2$. Each function satisfying the above condition must be of the form$f(x)=y$if and only if ... 2 I just wish to contribute a "quicker" development of Ragib Zaman's derivation. Just as he had shown, $$\cos^k \theta = \left( \frac{ e^{i\theta} + e^{-i\theta} }{2} \right)^k = \frac{1}{2^k} \sum_{n=0}^k \binom{k}{n} (e^{i\theta} )^n (e^{-i\theta})^{k-n} = \frac{1}{2^k } \sum_{n=0}^k \binom{k}{n} e^{i(2n-k)\theta}$$ Now, assuming that$\theta$is real, we ... 0 Assume by contradiction that one of the roots$z_0$is not simple. This means that $$f(z)=(z-z_0)^2g(z)$$ for some$g$which is analytic at$z_0$. Then $$f(z_0)=0 \\ f'(z_0)=2(z_0-z_0)g(z_0)+(z_0-z_0)^2g'(z_0)=0$$ This gives$15z_0^4-3z_0^5=0$. Therefore$z_0=0$, which is not possible, or$z_0=5$which doesn't lie in$D(0,1)$. 0 I am not familiar with the notation$D(0;1)$. I am going to interpret it as$z_0$is one of the root with multiplicity 2, which should exists if the solutions are not distinct. Hence$z_0$should satisfy both$f(z_0)=0$and$f'(z_0)=0$. $$f(z_0)=e^{z_0}-3z_0^5=0$$ $$f'(z_0)=e^{z_0}-15z_0^4=0$$ subtracting the equations gives us $$15z_0^4-3z_0^5=0$$ ... 1 A quadratic approximation will usually get worse the further out that you go, and so the$R$term will only get bigger. This is because$R = f(x,y) - \hat{f}(x,y)$, where$\hat{f}$is the quadratic approximation (this is not standard notation). Therefore if$|x-x_0|,|y-y_0| \leq \epsilon$then$|R| \leq |f(\epsilon + x_0,\epsilon + y_0)-\hat{f}(\epsilon + ...

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The first equality is true: \begin{align*} f^{-1}(Y\setminus F)&=\{x\in X:f(x)\in Y\setminus F\}\\ &=\{x\in X:f(x)\in Y\text{ and }f(x)\not\in F\}\\ &=\{x\in X:f(x)\in Y\}\cap\{x\in X:f(x)\not\in F\}\\ &=\{x\in X:f(x)\in Y\}\cap\left(X\setminus\{x\in X:f(x)\in F\}\right)\\ &=f^{-1}(Y)\cap\left(X\setminus f^{-1}(F)\right)\\ ...

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Yes to both. To see why the first is true, observe that $x\in f^{-1}(Y\setminus F)$ if and only if $f(x)\in Y\setminus F$ if and only if $x\notin f^{-1}(F)$.

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Solve the inequality $$2x>\frac{x^3}{3}$$ that can be rewritten as $$x(x-\sqrt{6})(x+\sqrt{6})<0$$ The expression on the left changes sign only at $-\sqrt{6}$, $0$ and $\sqrt{6}$ (where it vanishes). Since its value at $1$ is $-5$ you have the diagram $$\begin{array}{ccccccc} \_\!\_\!\_\!\_\!\_\!\_\!\_ & -\sqrt{6} & ... 2 Noting that \int_0^a x(a-x) \; dx = a^3/6, let$$ f_n(x) = \cases{ 9 n^3 x (1/n - x) & for $0 \le x \le 1/n$\cr 0 & otherwise\cr} $$1$$ f_n(x) = \cases{ 3nx & for $0<x<\frac{1}{2n}$\cr 3-3nx & for $\frac{1}{2n} \leq x < \frac{1}{n}$ \cr 0 & otherwise\cr} $$0 You forgot the solution x=0. By the Intermediate Value theorem the relative positions of the curves can change only at intersection points. Now, for x\gg 1, f(x)<g(x), hence we conclude if x>\sqrt 6, f(x)<g(x), if 0<x<\sqrt 6, f(x)>g(x), if -\sqrt 6<x< , f(x)<g(x), if x<-\sqrt 6, f(x)>g(x). 0 The roots, x = 0, -\sqrt{6}, +\sqrt{6} show merely the crossover points. You must test the four regions defined by those three points: -\infty < x \leq -\sqrt{6} -\sqrt{6} \leq x \leq 0 0 \leq x \leq +\sqrt{6} \sqrt{6} \leq x < \infty 1 Well, you're looking for period p\in\mathbb R such that \forall x\in \mathbb R we have$$ f(x+p)=f(x) $$In the case of$$ f(x)=\sin(x)+\cos(x) $$we have$$ f(x+p)=\sin(x+p)+\cos(x+p) $$but since both \sin(x) and \cos(x) are periodic to p=2\pi you are finished here and f is periodic to p=2\pi as well. Remark: As mentioned in the comments by ... 0 I think "composed" has a consistent usage ("f composed with g" = f \circ g) but "precomposed" doesn't. I'm happy with people using "precompose", but the "pre" here can seemingly refer to either: "before it in the order we write it": so "f precomposed with g" is the same as f \circ g. "before it in the order we apply functions": so "f ... 3 Step functions can be used in the process of defining the Riemann integral, and in general for approximating continuous functions. 1 I'll assume f is defined in a neighborhood of a. Note that you cannot use a particular f for proving the result. For the case \gamma>0 (in particular \gamma=1), the squeeze theorem suffices: from \lim_{x\to a}|x-a|^\gamma=0 it follows that$$ \lim_{x\to0}|f(x)-f(a)|=0 $$so also \lim_{x\to0}(f(x)-f(a))=0 and, finally,$$ ...

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$$\lim_{x \to a}\frac{f(x) - f(a)}{x - a} = 0$$ means that $f$ is differentiable at $a$ and $f'(a) = 0$ (by definition). If $f'(a) = 0$, then $f$ doesn't necessarily have a (global or local) extremum at $x = a$. For instance, take $f(x) = x^3$ and $a = 0$.

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