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I guess you got the idea, you just need to be more careful with the signs. To find the inverse of: $$inverse\:\frac{3-x}{1-x^2}$$ We'll need to find the inverse function in the form $y=g(x)$, given the above $f(x)$. First, we need to get the quadratic equation solution general formula: given: $$Ax^2+Bx+C=0$$ $y_1(x)$ and $y_2(x)$ are roots such that: $... 1 Consider that you look for the zero of function $$f(x)=2x-\tan(x)$$ and forget the trivial solution$x=0$. On the other hand, if$x$is a root$-x$is a root too. Then, we need to focus on the positive solution. In the restricted interval, the derivative$f'(x)=2-\sec ^2(x)$cancels when$x=\frac \pi 4$which is a local maximum. So, to get an approximation, ... 2 Domain of$x$is the range of$y$, and vice versa. You have a square root for$x=f(y)$, so you should solve the condition for the expression within that square root is bigger than or equal$0$. Or:$1-12y+4y^2\ge 0$Solving$1-12y+4y^2=0$yields$(1/2)(3+2\sqrt{2})$and$(1/2)(3-2\sqrt{2})$. These two are real values, the coefficient a of the original ... 2 Solve the first-order condition $$\frac{\partial}{\partial a_j} \left(\sum_{i=1}^n a_i^2 \sigma_i^2 - \lambda(\sum_{i=1}^n a_i -1) \right) = 0$$ to obtain $$a_j = \frac{\lambda}{2\sigma_j^2}$$ Next solve for the multiplier$\lambda$using the constraint$\sum_{i=1}^n a_i = 1$. 0 The graph of an inverse function is the graph of the original function reflected about the line$y=x,$and, to be a function, a graph must pass the vertical line test. Another way to say this is that it fails to be a function when the original function fails a horizontal line test. The function is monotonic decreasing up to its minimum then monotonic ... 1 you can actually get a stronger result such as $$\left\vert\frac{\pi}{4} - \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9}\right)\right\vert < 1/11$$ because the error in a convergent alternating series is less than the absolute value of the first missing term which in this case is 1/11. The alternating series is simply the Taylor ... 7 Use the series expansion of arctangent and the fact that for alternating series the rest is less than the first omitted term. By the way, it's +1/9; otherwise the inequality is wrong. 0 As evaluating the limit as$x\to 0$forms $$\lim_{x \to 0}\frac{x+2-\sqrt{2x+4}}{3x-1+\sqrt{x+1}}=\frac{2-\sqrt{4}}{-1+\sqrt{1}}=\frac{0}{0}$$ and$\frac{0}{0}$is an indeterminate form, apply L'Hopital's rule by taking the derivative of the numerator and denominator $$\lim_{x \to 0}\frac{1-\frac{1}{\sqrt{2x+4}}}{3+\frac{1}{2\sqrt{x+1}}}=\frac{1-\frac{1}{... 0 The original polynomial has roots h = 0, -1, i.e., p(x) = x(x+1). As 0 is a root of p, we have the following key observation (which you should prove by induction): Key Observation: If h is a root of p^n(x), i.e, p^n(h) = 0, then h is a root of p^{m}(x) for all m \geq n. Keeping this in mind, let us find the roots of p^2(x). In ... 1 You may use$$\sqrt{1+x}=1+\frac x2+O(x^2)$$First,$$\sqrt{2x+4}=2\sqrt{1+\frac x2}=2(1+\frac x4)+O(x^2)=2+\frac x2+O(x^2)$$And$$\frac{x+2-\sqrt{2x+4}}{3x-1+\sqrt{x+1}} =\frac{x+2-2-\frac x2+O(x^2)}{3x-1+1+\frac x2+O(x^2)} \\=\frac{\frac x2+O(x^2)}{\frac72x+O(x^2)}=\frac17+O(x)\underset{x\to0}\longrightarrow\frac17 $$3 The n-dimensional prism still has volume formula$$V_n=V_{n-1}\cdot h$$while the n-dimensional pyramid has volume formula$$V_n=\frac1n\cdot V_{n-1}\cdot h$$where V_{n-1} is the (n-1)-dimensional volume of the base facet and h is the orthogonal height. For n=1 you'd simply use V_0=1 and you see that V_1=h in either formula. The case n=2 ... 0 The function (x-1)^2 has a maximum value of 1 on the interval [0,2]. It follows that (x-1)^2-1 has a maximum of 0 there too. Now we know that for 0\le x\le2, we have that$$0\le \sin(\frac{πx}{2})\le 1.$$Thus,$$(x-1)^2-1-\sin(\frac{πx}{2})=x^2-2x-\sin(\frac{πx}{2})\le 0$$on this interval. 0 I'll set \sigma = x, so that K becomes$$-\frac{x^2+3x+2}{x^2-8x+15}.$$We reduce the problem as follows:$$K=-\frac{x^2+3x+2}{x^2-8x+15}=-\frac{x^2-8x+8x+3x+2+15-15}{x^2-8x+15}=-\frac{x^2+8x+15+11x-13}{x^2-8x+15}=-1+\frac{13-11x}{x^2-8x+15}=-1+\frac{13}{x^2-8x+15}-\frac{11x}{x^2-8x+15}.$$Taking derivatives, the first term vanishes; the second term ... 1 A fast way to find the Derivative of Quadratic over Quadratic:$$ \bbox[20px,border:1px solid red] { y = \frac{ax^2+bx+c}{a'x^2+b'x+c'}\quad\Rightarrow\quad y'= \frac{\left| \begin{array}\\ a&b\\ a'&b'\\ \end{array}\right| \times x^2 + 2 \times \left|\begin{array} \\ a & c \\ a' & c' \end{array}\right|\times x + \left|\begin{array}\\ b &... 1 Looking at a set of integers, am I right to say that we cannot have a sequence whose terms are all the integers? No, that is absolutely wrong. I mean that {...,−3,−2,−1,0,1,2,3,4,...} is not a sequence because every sequence, finite or infinite, must have the first term.$\{...,−3,−2,−1,0,1,2,3,4,...\}$is not a sequence but$\{0,1,-1,2,-2,3,-3,..... \}...

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The integers is a countable set, so you can enumerate the terms within the integers as a sequence. The sequence is as follows 0,1,-1,2,-2,3,-3...... and so on.

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It sounds like you need to use the average value formula for a curve: $$\frac{1}{b-a}\int_a^bf(x)\text{d}x$$ This is the same idea behind finding the average value of a finite sequence of numbers, but you're using the integral to find the average value of an infinite sequence (in this case, the sequence is a series of y-values over an interval). This might ...

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The average for section $A$ is $$\frac{1}{12-10}\int_{10}^{12} 2.5\sin\left(\frac{\pi}{20}x\right) dx = \frac{1}{2}\frac{(-50)}{\pi}\left(\cos\left(\frac{12 \pi}{20}\right) - \cos\left(\frac{10 \pi}{20}\right)\right)\approx 2.459 .$$

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That depends on what you want to call what. If you define a sequence to be a map defined on a well-ordered subset of the integers (that is, so that a sequence would always have a "first" member), then fine. But in some other problems, it may be convenient for you to consider a sequence as being a map on any subset of the integers.

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The average value of $k\sin{cx}$ between $x=a$ and $x=b$ is $${1\over b-a}\int_a^bk\sin{cx}\,\mathrm{dx}={k(\cos{ca}-\cos{cb})\over c(b-a)}$$

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You could set up e.g. the sequence $\langle 0, 1, -1, 2, -2, 3, -3, \dotsc \rangle$. There are many others (like 0, then five positive numbers, then ten negative, then next five positive, ...). More deeply: there are sets that can be placed in one-to-one correspondence with the natural numbers $\mathbb{N} = \{1, 2, \dotsc\}$, they are called countable. As ...

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There are a couple of ways to answer your question. First, the short answer: The integers can be made into a sequence simply by choosing any first element. For example, @BAI's comment puts the integers in their standard order $0, 1, -1, 2, -2, \ldots$. A slightly longer answer: Any countably infinite set $A$ may be made into a sequence by induction (or by ...

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Notice that taking the limit as $r \to 0$, like you did, only gets near zero through paths that are straight lines. This is not enough to ensure differentiability and this example is to show why. As pointed out in the comments, we have that $$\frac{\cos^3 \theta + \sin^3 \theta}{\cos \theta - \sin \theta}$$ is not bounded, when you approach zero through ...

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Your mistake is in falsely assuming that $g(\theta) = \dfrac{\cos^3\theta+\sin^3\theta}{\cos\theta-\sin\theta}$ is bounded. It has a discontinuity at $\theta = \pi/4$ so it is not bounded. Hence you cannot conclude $\lim\limits_{r\to 0}f(r) g(\theta)=0$

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There is no difference between the notations - they mean exactly the same thing. However, at different times you will find one more useful than the other. For example, when doing u-substitution with integrals, the $\frac{d}{dx}$ is helpful. The same thing is true when using the chain rule - it is often easier to keep track of what is happening with $\frac{d}{... 0 Using prime ' usually indicates that we take the total derivative w.r.t. to all variables, whereas$\frac{d}{dx}$indicates that we take the total derivative with respect to the variable$x$. Note that this is different from the partial derivative$\frac{\partial }{\partial x}$. As an example, consider$f(x, y, z)=x^2+y^2+3z$where$y=\sin(x)$. Then$f' = ...

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There is no mathematical difference. They are different notations for the same thing. (The first was due to Newton, the second to Leibniz.)

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Let $$a_n=n^4-6n^2$$ then $$a_{n+1}-a_n=(n+1)^4-6(n+1)^2-(n^4-6n^2)=…$$

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The polynomial factorises as $n^2(n^2-6),$ which vanishes somewhere between $n=2$ and $n=3$ if we for the moment think of $n$ as a real variable. It then increases afterwards. Thus, you're right that it cannot be monotonic. Sequences that are not monotonic may generally be classified as oscillating.

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Hint: If two curves are parallel at a point, then they have the same slope at that point.

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First, there is not a universal agreement that one cannot use the word "hyperplane" for anything of dimension other than $n-1$ in an $n$-dimensional space. For example, the definition of hyperplane in Linear Algebra by Waldron, Cherney, & Denton allows on to construct a $k$-dimensional hyperplane for any $k\leq n$ in an $n$-dimensional space. According ...

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The slope of the line is $4$ so you want to find $x$ such that $(2e^{4x})'=4$.

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The simple answer is that is just the definition. However, I think a little bit can be said about why the name is reasonable. First note that "hyper" is at sometimes used as a prefix for generalizations to higher dimensions of three dimensional objects. E.g. a hypercube is the generalization of a cube to n dimensions. So "hyperplane" should be taken to ...

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This has the type of a homogeneous ODE. Set $y-b=(x-a)u$ where $a,b$ solve $$3a+4b+2=0,\\ 2a+b+3=0.$$ This gives $a=-2$, $b=1$, and $$(x+2)u'+u=y'=\frac{3(x+2)+4(y-1)}{2(x+2)+(y-1)}=\frac{3+4u}{2+u},$$ which is now separable. Or introduce another parameter on the solution curves so that the equation separates into a linear system $$\pmatrix{\dot x(t)\\\... 0 Let the integrating factor f depend only on x. Then it should satisfy the equation$$2f+2xf'+yf'+3f'=-4f.$$This simplifies to make$$\frac {f'}{f}=-\frac {6}{2x+y+3},$$which we may now easily integrate to find a solution$$f=\frac{1}{(3+y+2x)^3}.$$But this is probably more laborious than the first of the methods suggested by @LutzL above. 4 Because that's the definition. Let V be a vector space and let n be the dimension of V. Then a subspace W of V is called a hyperplane of V if the dimension of W is n-1. And to answer your second question, just look at your surroundings. You can surely see shapes that resemble a line and in the same way we can have lines in \mathbb{R}^3. For ... 0 The graphs are similar however for the absolute value the negative half of the graph is a reflection of the positive x-axis. 1 \sqrt{x} and \sqrt{|x|} are two different functions. On [0,\infty), the two functions return the same value, bot on (-\infty, 0), only the second function is defined. There is nothing "better" or "legal" about the second function. They are different, but not better or worse. 0 This case is easily invertible with the help of the Lambert function link so proceeding accordingly we get from$$ y = x e^{x^2}\Rightarrow x = \sqrt{\left(\frac{W(2y^2)}{2}\right)} $$NOTE As long as$$ x = f(f^{-1}(x)) $$knowing that$$ x e^{x^2} = x\left(\sum_{k=0}^{\infty}\frac{x^{2k}}{k!}\right) $$assuming g_n(x) = \sum_{j=0}^n a_j x^{2j+1} ... 1 Since f'(x)=(1+2x^2)e^{x^2} is positive, the C^{\infty}-function f is strictly increasing and it follows that it has an inverse f^{-1}:f(\mathbb{R})=\mathbb{R}\to \mathbb{R}. Moreover, the Maclaurin expansion of f is$$f(x)=x(1+x^2+\frac{x^4}{2}+o(x^5))=x+x^3+\frac{x^5}{2}+o(x^6).$$Note that that f^{-1} is odd (like f) and its Maclaurin ... 1 It is not necessary calculate f^{-1} analytically because the derivative of f^{-1} can be written in function of the derivative of f, infact you have that (f^{-1})’(y)=\frac{1}{f’(f^{-1}(y))} You can calculate the other derivatives of f^{-1} starting from this relation. In your case you have that f(0)=0 So f^{-1}(0)=0 that means that ... 0 Apply the \epsilon-definitions of supremum and infinum. I will analyze why b^2=\sup(B). Supremum: Let B be a nonempty set of real numbers defined by \{x^2 :\, x \in (−b,0)\}. Let b \in \mathbb R, b>0 where B is bounded above. Then, s=\sup(B) if for every \epsilon >0 there exists an element y\in B such that$$s- \epsilon < y$$... 0 More generally, if f(x) = x^a(1-x)^b  where a, b \ge 1, then \begin{array}\\ f'(x) &=ax^{a-1}(1-x)^b-x^ab(1-x)^{b-1}\\ &=x^{a-1}(1-x)^{b-1}(a(1-x)-bx)\\ &=x^{a-1}(1-x)^{b-1}(a-x(a+b))\\ \end{array}  so f'(x) = 0 at x = \dfrac{a}{a+b}  when \begin{array}\\ f(x) &=(\dfrac{a}{a+b})^a(1-\dfrac{a}{a+b})^b\\ &=(\dfrac{a}{a+b})^a(\... 2 The solution can be found as follows. Any complex number x can be written in the form$$x=r\exp\left(i\phi+2n\pi\right)$$where n is any integer. In your equation, we have$$x^6=2$$or$$r^6\exp\left(6(i\phi+2n\pi)\right)=2$$taking the absolute value of both sides, we find$$r^6=2$$and thus$$r=\sqrt[6]{2}$$Now we require that$$\...

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There are different ways to embark on this problem. One way, perhaps not so "straight forward", is to set $x^6=2t^6$ for now to get $(t^3-1)(t^3+1)=0$. This of course, is extremely easy to solve, because now it is broken down to two easy cubics on which the sum\difference formulas apply. With $x=t(2^\frac{1}{6})$ you can backsub. This method avoids dealing ...

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Yes. There's a result asserting that all $n$-th roots of complex number are obtained as the product of one of them by all the $n$-th roots of unity. Hence here, the sixth roots of $2$ are $$2^{\tfrac16}\mathrm e^{\tfrac{ik\pi}3},\quad 0\le k <6.$$

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Overkill: $x=\sin^2 t$, $0\le t \le π/2$; $F(t)= (\sin^2 t)^n(1-\sin^2 t)^n$; $F(t)= (\sin^2 t \cos^2 t)^n$; $F(t)=(1/4)^n(\sin^2 2t)^n$; $F_{max}$ at $t=π/4$, then $x=\sin^2 (π/4)= (1/√2)^2=1/2$.

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$0=f'(0)=a(a-1)(a+2)\implies a=0,1$ or $-2$. Meanwhile, $0\le f''(0)=2a\implies a\ge0$. Hence $a=0,1$.

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For real numbers the equation $x^n = b$ will, if $b\ne 0; b\ne 1$ 1) If $b >0$ and $n$ is even have exactly two solutions $x = c$ for some $c > 0$ and $-c$. 1b) If $b > 0$ and $n$ is odd then there will have exactly one solution $x = c$ for som $c > 0$. 2) If $b < 0$ and $n$ is odd then there will be exactly one solution $x = c$ for some $c ... 3 Actually, it is false. Take, for instance,$\int_1^\infty\frac1x\,\mathrm dx\$.

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