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0

You just need to find the absolute maxima or minima of $f(x)$ in its domain $[-3,3]$. Write $f(x)=√g(x)$, where $g(x)=9-x^2$. Now, $g(x)$ has an extrema if $g'(x)=0$ which gives $x=0$ as the point of maxima as $g''(0)<0$. Hence $f(0)=√g(0)=3$, $f(-3)=f(3)=0$. Clearly, $Range(f)=[0,3]$.

1

$$9-x^2\ge 0\implies D_f=[-3,3]$$ set $x=3\sin\theta$ $$y=\sqrt{9-9\sin^2\theta}=3\,|\,\cos\theta\,|\implies R_f=[0,3]$$

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The usual convention is that the principal square root function $f(x) = \sqrt x$ returns only the non-negative root. So in this case, the range corresponds to the semicircular upper half of the circle defined by $x^2 + y^2 = 9$. Which gives the range as $[0,3]$.

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You correctly found the domain, although you meant that the expression that's under the square root, which is called the radicand, must be non-negative. Observe that $y = \sqrt{9 - x^2} \implies y \geq 0$. Moreover, if we square both sides of the equation, we obtain $y^2 = 9 - x^2$, which is equivalent to $$x^2 + y^2 = 9$$ This is the equation of a circle ...

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Hint: Since a square is non-negative, $0\le 9-x^2\le9\;$ on $\;[-3,3]$, and $\;\sqrt x\;$ is a continuous increasing function.

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Let: $$f(x)={x+\tan{x}\over A +B(x+\tan{x})^{2n}}$$ Set $x:=-u$ \begin{align} f(-u)&={-u+\tan{-u}\over A +B(-u+\tan{-u})^{2n}} \\ &={-u-\tan{u}\over A +B(-1)^{2n}(u+\tan{u})^{2n}} \\ &=(-1){u+\tan{u}\over A +B(u+\tan{u})^{2n}} \\ &=-f(u) \end{align} Since x and u are dummy variables, $f(-x)=-f(x)$. Thus $f(x)$ is an odd function. Your ...

1

Let us call the integrand as $$f(x)={x+\tan{x}\over A +B(x+\tan{x})^{2n}}$$ Then it is quite evident that it is an odd function of $x$ $$f(-x)=-f(x)$$ and hence you can easily conclude $$\int_{-n}^{n}f(x)dx=0$$

0

HINT: Use $$I=\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$$ $$\implies I+I=\int_a^b[f(x)+f(a+b-x)]\ dx$$ Here $a=?,b=?$ and $\tan(-x)=-\tan x$

1

I see what you're getting at based on our comment discussion now. The conditions also change. For example, say we have $f(x) = x^2$ if $x > 10$. Then $f(2x) = (2x)^2$ if $2x> 10$. (Other pieces are irrelevant for this discussion and the same thing happens to them, so it's sufficient to consider one piece.) Similar reasoning for horizontal ...

1

The composition of $g$ and $f$ is defined. Let's examine the definition of a composite function. Let $u$ and $v$ be any two functions. We define a new function $u \circ v$, the composition of $u$ and $v$, by $$(u \circ v)(x) = u(v(x))$$ The domain of $u \circ v$ is the set of $x$ in the domain of $v$ such that $v(x)$ is in the domain of $u$. Since $f: (... 2 Sorry, I got it! For anyone liking future help I figured it out with my textbooks: $$f(g(x))\\ g(x) = x + 3\\ f(x + 3) = f(g(x)) = (x + 3)^2 -3$$ $$g(f(x)) = g(x^2 - 3) = x^2 - 3 + 3 = x^2$$ 1 The minimum of$|f(z)|$on$D$is not$1$but$1/e$, which is achieved at$z=-1$on the boundary. 4$\sin x=0$is rejected because when$\sin x=0$, the value$\csc x$does not exist, so$(\sin^2 x)(1+\csc x)$does not exist, so it can't be equal to$0$or to anything else either. 2 Let's say$(\sin x)^2=0$. Then, take square root of both sides to get$\sin x=0$. Now, we have either$x=0$to$x=\pi$. Then, substitute back in: $$x=0 \implies (\sin 0)^2(\csc 0+1)=\text{undefined because} \csc 0 \text{ is undefined}$$ $$x=\pi \implies (\sin \pi)^2(\csc \pi+1)=\text{undefined because} \csc \pi \text{ is undefined}$$ Always remember to ... 0 It can be simply defined as an infinite product $$f(x)=\prod_{n=1}^{\infty}\cos^{2n}(\frac{\tau x}{2})$$ Where$\tau=2\pi$1 There exists an infinite family of analytic solutions. But most are not clear until you have the machinery of Q-calculus! We start by "standardizing" your invariance equation into a form that is ripe for analysis using "q" series. So $$f(x) = a f(bx) \rightarrow \frac{f(bx) - f(x)}{(b-1)x} = \frac{(1-a)}{(b-1)x} f(x)$$ You can observe that ... 1 Your question doesn't make sense as stated. The notation$f:\mathbb{R}^2/\lbrace(0,0)\rbrace \to \mathbb{R}$means that$f$is a function with domain$\mathbb{R}^2/\lbrace(0,0)\rbrace$and codomain$\mathbb R.$You surely didn't mean to ask "is this function a function?" Now the question "Is the expression$\arctan (x/y)$well defined for all$(x,y) \ne (0,...

2

$$x^{ log_{ 2 }x }+\frac { 16 }{ x^{ log_{ 2 }x } } =17\\ { x }^{ 2\log _{ 2 }{ x } }-17x^{ log_{ 2 }x }+16=0\\ \left( x^{ log_{ 2 }x }-16 \right) \left( x^{ log_{ 2 }x }-1 \right) =0\\ x^{ log_{ 2 }x }=16\Rightarrow \log _{ 2 }{ x^{ log_{ 2 }x } } =\log _{ 2 }{ 16 } \Rightarrow { \left( \log _{ 2 }{ x } \right) }^{ 2 }=4\Rightarrow \log _{ 2 }{ x } =\pm ... 2 let$$y=x^{log_2x}$$your equation becomes,$$y+\frac{16}y=17$$solve it, you get two solutions: 1 and 16. Now it becomes less horrible,$$x^{log_2x}=1 ~or~ 16 $$This leads to solution x =1, 2^2 and 2^{-2} 0 Adding to @Adriano's answer, if you want to reflect y=f(x) across y=mx+b you have to reflect the function by the x-axis and then rotate it by \theta=\pi+2\tan^{-1}(m) radians. So if we have y=f(-x), then by using the rotation matrix in linear algebra, we can set x=x'\cos(\theta)+y'\sin(\theta) and y=y'\cos(\theta)-x'\sin(\theta). So the ... 1 The expression \arctan(\frac{x}{y}) is not defined for y=0, so with that as the definition your f will only be defined on \mathbb R\times(\mathbb R\setminus\{0\}). But on that set it assign exactly one value to each pair (x,y) and that makes it a function. If you want a function that has the values you mention in a comment, you need to define it ... 2 Well, first notice that \frac{x}{y} is not defined for y=0, so you should not consider in the domain the set A = \{(x, y) : y = 0\}. Now, you should verify when the function f(z) = arctan(z) is well-defined. 1 Not sure what to say for minimum length. http://www.wolframalpha.com/input/?i=arctan%28x%2Fy%29 -1 for graph Y = x^2 + Kx -2 to cut x - axis y coordinate must be zero thus, x^2 + Kx -2 = 0 Now for this equation for solution to be finitly 2 , b^2 - 4ac > 0. here b = K , a = 1 and c = -2. for every value of K , b^2 - 4ac will be greater than zero. for example k = 0 => (0)^2 - 4(1)(-2) = 8 for k = -1 => (-1)^2 - 4(1)(-2) = 9 hence for all value of K ... 0 We are basically attempting to show that$$ y = x^2 +kx - 2$$Equals 0, for exactly two, distinct x, for any choice of real number k. A trick here is to basically factor the expression, and show that the factors give rise to distinct roots. So we want to find an a,b such that$$ (x-a)(x-b) = x^2 + kx - 2$$And show that regardless of the k, a \... 2 Use the quadratic function, with a= 1, b= k, and c= -2. There's three different outcomes for a quadratic: {b^2-4ac}\gt 0 \rightarrow we get two solutions, which means we have two x-intercepts. {b^2-4ac}=0 \rightarrow we get one solution, which means we have one x-intercept {b^2-4ac}\lt 0 \rightarrow no real solutions, which means no x-intercepts. ... 2 We are looking for n such that 3\pi is a period, but not necessarily the smallest period. We have 2\cos nx\sin\frac{5x}{n}=\sin(\frac{5x}{n}+nx)+\sin(\frac{5x}{n}-nx). So we require 3(n\pm\frac{5}{n}) to be an even integer. Clearly that is impossible for |n|>15 or for n not a factor of 15. But equally clearly any factor of 15 (positive or ... 2 You cannot ignore the exponent or the multiplication. In this case, set temporarily t=\log_3x so the equation becomes$$ t^2-3t+2=0 $$which has roots t=1 and t=2. Thus you get the two equations$$ \log_3x=1 $$and$$ \log_3x=2 $$Can you finish them? Your method would be sound, too, provided you did the decomposition right. 5 should be$$\left( \log _{ 3 }{ x } -1 \right) \left( \log _{ 3 }{ x } -2 \right) =0\log _{ 3 }{ x } =1\Rightarrow \quad x=3\\ \log _{ 3 }{ x } =2\Rightarrow x=9$$2 Graph it using Desmos.$$\tan\left(\frac{\pi}{2}\text{floor}(x)\right)$$Now, it's easy to see that this is 0 when \text{floor}(x) is even and undefined when \text{floor}(x) is undefined, so we have a period of 2. 1 You must be talking about bounded linear operators, not limited. In normed vector spaces it holds that a linear operator is bounded if and only if it is continuous. This even is given as a definition sometimes. Now, if your definition of bounded linear operator is that it maps bounded sets to bounded sets then: From your expression for ||A|| you get that ... 0 Let f(n) and g(n) be two function,then if \lim \limits_{x \to \infty} \frac{f(n)}{g(n)} =0 then f(n)=O(g(n)) and g(n)=\Omega(f(n)) if \lim \limits_{x \to \infty} \frac{f(n)}{g(n)} =\infty then f(n)=\Omega(g(n)) and g(n)=O(f(n)) if \lim \limits_{x \to \infty} \frac{f(n)}{g(n)} =k(constant),then f(n)=\theta(g(n)) and g(n)=\theta(f(n)) ... 1 Your approach is lacking. You say Now,above equation could hold only if c \lt 0 ,which contradicts our assumption Which is the correct idea, but it is not entirely true. The above equation, for example, holds if c=1 and n=10. Now, I assume you know that my counterexample is not good, but it works because you forgot to say that the equation ... 1 The idea being discussed doesn't need a rigorous definition of wavelet, or even to use wavelets at all; the notion of a wavelet is, I think, mainly a clever way of generating a convenient basis in a systematic fashion starting from a basic shape. (the particular basis you listed could probably stand to be scaled so as to be orthonormal basis rather than just ... 0 a homogeneous function is a polynomial function which all the terms have the same degree. then in your example (in this case of one dimension) v \rightarrow av + z is not a homogeneous polynomial since z is a vector constant, not a variable. 0 as you go solving for f^{-1} if y = f^{-1} then: X = \frac {y}{1-y^2}\\ xy^2 + x - y = 0 Now here you divided through by x in order to apply the quadratic formula. But in so doing, you have obliterated x = 0 and must make a note to yourself to return to this before you are complete. y = \dfrac{-1+\sqrt{1+4x^2}}{2x} \text { if } x\ne 0, or x ... 0 f^{-1} is actually defined at 0, and f^{-1}(0) = f^{-1}(f(0)) = 0. The problem is that you arrived at the expression of f^{-1} under the (implicit) assumption that y \neq 0. In fact,$$y = f(x) = \frac{x}{1 - x^2} \iff yx^2 - y+ x = 0$$This is a quadratic equation in x iff y \neq 0, so the obtained expression is valid only when y \neq 0. If ... 0 let us assume that the following equation is discontinuous at point c... f(x) + f(2x) + f(4x) where f is a continuous function. If that is the case, then the limit as x goes to c should not exist. However, the limit operation is linear. This means we have the following three limits that should exist: limit x -> c : f(x) = f(limit x -> c : x) = f(x) It ... 0 Well you shouldn't care actually. As n gets bigger, all terms n^i with i<4 are negligible when compared to n^4 (This is because \lim_{n \rightarrow \infty} \frac{n^i}{n^4} = 0 when i<4). Hence, you know that starting from a certain n_0 (you don't have to know its value yet), n⁴ + 100n² + 50 < c n^4 (for any c>1 you want). And ... 1 The biggest term in this polynomial is n^4, so you have to choose a term bigger than n^4. Since 2n^4 > n^4, you can choose 2n^4 to find some constant n_0 where n > n_0 \implies f(n) \leq 2n^4. In this case, n_0=11. However, you didn't have to pick 2n^4. For example, you could've picked 3n^4 since 3n^4 > n^4. In this case, we ... 0 Well, you can basically take a guess for c. Notice if you were to take a larger n, you could make c smaller. But 2 seems right, so let's try that. You could then graph 2n^4 and n^4 + 100n^2 +50 and see that our desired inequality holds for all n\geq 11. 1 If the indicated homeomorphisms are h_A and h_D, just take$$g = h_B\circ f\circ h_A^{-1}$$It's continuous and onto because each of the elements of the composition are. 4 This one is pretty straightforward. Intuitively, all we have to do is pick a function that maps set A to the image of your initial fuction, and a function that "extends" the image of f to set B, namely the identity function. Here, the functions are named "map" and "ext", respectively and are defined as follows: map:A\rightarrow Imf\subseteq B, ... 0 Rather a comment than an answer. The only thing missing in Georgy's answer is an integral:$$ \int_0^\infty n^2xe^{-nx} dx = -\frac{1}{n} \int_0^\infty n^2x\, d e^{-nx} = \left[ - \frac{n^2x e^{-nx}}{n} \right]_0^\infty + \int_0^\infty n e^{-nx} dx = 0-\left[e^{-nx}\right]_0^\infty = 1 $$And a picture, where the grey area is independent of \,n , i.e. the ... 1 The first equality is true, because f^2(x) is usually defined to be (f \circ f)(x) = f\big(f(x)\big), and (f \circ g)(x) is defined as f\big(g(x)\big). Because of these definitions, the second equality does not hold. However, if the notation f^2 or \circ is differently defined, any one or possibly both equalities may be true. 1 It is nowhere near as neat.$$\begin{align}\mathsf M_{X/\, Y}(t) =&~ \mathsf E(e^{tX/\, Y}) \\[1ex]=&~\mathsf E(\mathsf E(e^{tX /\, Y}\mid Y)) \\[1ex]=&~ \mathsf E(\mathsf M_{X\mid Y}(t/Y))\end{align}$$Now, if X and Y are independent (as required for the summation formula you gave) then this becomes:$$\begin{align}\mathsf M_{X/\, Y}(t) =&...

1

Just to better asses the solution given by Kitter Catter, premised that \eqalign{ & x = \left\lfloor x \right\rfloor + \left\{ x \right\} \cr & \left\lfloor { - x} \right\rfloor = - \left\lceil x \right\rceil \cr & \left\lceil x \right\rceil = \left\lfloor x \right\rfloor + \left\lceil {\left\{ x \right\}} \right\rceil \cr}...

2

Starting from $n=0$, even-$n =2k$ terms are $(4^k - 1)/3$ and the subsequent odd-$n = 2k+1$ terms are $2\times (4^k- 1)/3$. So overall: \begin{align} \sum_{k=0}^{49}\left[(1+2)\frac{4^{k} - 1}{3}\right] + \frac{4^{50}-1}{3} &= \sum_{k=0}^{49}(4^{k} - 1) + \frac{4^{50}-1}{3} \\ & = \sum_{k=0}^{49}(4^{k}) + \frac{4^{50}-1}{3}-50 \\ &= \frac{4^{50}-...

5

The sequence of $[2^n/3]$ for $n\in \mathbb{N}$ and $n\geq 2$ would be like this: $$1,\ 2,\ 5,\ 10,\ 21,\ 42,\ 85,\ \cdots$$ So, comparing the numbers in the sequence successively, you may guess that it is a recursive sequence like $a_1=1$, and for $n>1$, we have $a_n=\begin{cases} 2a_{n-1} & \text{if } n\ \text{ is even, }\\ 2a_{n-1}+1 & \... 9 I think I have an answer. Look at$\frac{1}{3}$in binary format and you will notice that it is$0.\overline{01}$. This tells us something, namely that$[2^{2i}/3]+[2^{2i+1}/3] = 4^i-1$. This immediately tells me that your sum reduces to:$\$\sum \limits_{i=0}^{50} [2^{2i}/3]+\sum \limits_{i=0}^{49} [2^{2i+1}/3]=\sum \limits_{i=0}^{49} [2^{2i}/3]+\sum \...

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