New answers tagged

3

$$\frac {e^{2x}}{e^x+1} = \frac {e^{2x}+e^x-e^x}{e^x+1}=$$ $$\frac {e^{x}(e^x+1)-e^x}{e^x+1}=e^x-\frac {e^x}{e^x+1}$$


0

$$I=\int \frac{e^{2x}}{1+e^x}dx=\int \frac{e^{2x}-1+1}{1+e^x}dx\\ \\ =\int \frac{e^{2x}-1}{1+e^x}dx+\int \frac{1}{e^x+1}dx\\ \\ =\int (e^x-1)dx+\int \frac{e^{-x}}{1+e^{-x}}dx\\ \\ =e^x-x-\ln(1+e^{-x})+c$$


2

To get the result in your book, note that $$e^{2x}=e^x e^x=e^x(e^x+1-1)=e^x(e^x+1)-e^x$$


3

I think it is simpler to first use the substitution $$u=e^x$$ to obtain the integral of a rational function: $$\int\dfrac{u}{1+u}\,du$$ and then use division of polynomials.


0

Just think of $e^{2x}$ as $(e^x)^2$. Now your integrand takes the form $\frac{u^2}{1+u}$, so you can long divide those polynomials in $u$ as usual. That's how they got that decomposition. This does not handle the differential terms as in substitution. That would have to happen after the algebraic moves.


0

We have $$\begin{cases} \dfrac{2}{x-3} \gt \dfrac{5}{x+6} \\ \dfrac{1}{3} \lt \dfrac{1}{x-2} \end{cases}$$ We start by solving each inequality by itself. I will assume you can already do this, so we now have $$\begin{cases}x<-6,\ 3<x<9\\ 2<x<5\end{cases}$$ If we imagine drawing these on a number line then we get the following: -6 -5 -...


3

Let us consider the inequalties $(1) \quad \dfrac{2}{x-3} \gt \dfrac{5}{x+6} $ and $(2) \quad \dfrac{1}{3} \lt \dfrac{1}{x-2} $. Then compute the set $L_1$ of the solutions of $(1)$ and the set $L_2$ of the solutions of $(2)$. The set of solutions of $\left\{ \begin{aligned} \dfrac{2}{x-3} \gt \dfrac{5}{x+6} \\ \dfrac{1}{3} \lt \dfrac{1}{x-2} \end{...


3

One solution to this can be found when you input the equation in wolfram alpha: It appears that wolfram alpha considers negative values outside of the domain of the principal root. If you click on "Use the real-valued root" you will get the expected answer.


2

Let us consider the second inequality. There are two cases here: $x>2$: then$$\frac13<\frac1{x-2}\iff\frac{x-2}3<1\iff x-2<\frac13;$$ $x<2$: then$$\frac13<\frac1{x-2}\iff\frac{x-2}3>1\iff x-2>\frac13.$$ Now, you do the same thing with the other inequality. Finally, you take the intersection of your answers.


2

The first it's $$\frac{2}{x-3}-\frac{5}{x+6}>0$$ or $$\frac{x-9}{(x-3)(x+6)}<0,$$ which gives $$(-\infty,-6)\cup(3,9).$$ Solve the second inequality by the same way. Can you end it now? I got the following answer: $$(3,5)$$


1

WA always thinks that $x^{\frac{1}{3}}=\sqrt[3]{x},$ but it's wrong. By definition, the domain of $x^r$, where $r\in\mathbb Q$, is $(0,+\infty)$. The domain of $\sqrt[n]{x}$, where $n\in\mathbb N$, $n$ is even and $n\geq2$, is $[0,+\infty)$; The domain of $\sqrt[n]{x}$, where $n\in\mathbb N$, $n$ is odd and $n\geq3$, is $\mathbb R$.


0

Thr answer of Wolfram is not correct: in fact $\sqrt[3]{x}$ is defined $\forall x \in R$ (also negative or $0$). So in your case $\sqrt[3]{(x-1)^2(x+2)}$ is defined $\forall x \in R$


1

If you know an approximate of the solution, you can use the fixed-point method to find a less approximate (!) solution. Define $$\alpha_{n+1}=\sqrt[k+1]{{4.68-4.50\cos \alpha_{n}\over 1.23}\cdot \alpha_n^{k}}$$and find the limit. I tried $k=3$ and obtained $\approx 4.5166$. All you have to do is that using a scientific calculator press an approximate of ...


0

You can observe that $$ \lvert\sin x\rvert=\sqrt{1-\cos^2x}=\sqrt{1-\frac{24^2}{25^2}}=\frac{7}{25} $$ Since $\pi/2<x<\pi$, we conclude that $\sin x=7/25$. Also $$ \lvert\cos y\rvert=\sqrt{\frac{1}{1+\tan^2y}}=\frac{40}{41} $$ and from $\pi<y<3\pi/2$ we conclude that $\cos y=-40/41$. Thus $$ \sin y=\tan y\cos y=-9/41 $$


1

Your method is correct. Indeed, refer to the graphs: $\hspace{1cm}$ $$\cos x=-\frac{24}{25}=\frac{\overbrace{-24}^{adjacent}}{25}, x\in \left(\frac{\pi}{2},\pi\right); \quad \tan y=\frac{9}{40}=\frac{\overbrace{-9}^{front}}{\underbrace{-40}_{adjacent}},y\in \left(\pi,\frac{3\pi}{2}\right).$$ Also note: $$\cos x=-\cos(\pi -x)=-\frac{24}{25} \Rightarrow \cos ...


0

Christiane $$C_x=1200\cos{\frac{\pi}{6}}=600\sqrt{3}N$$ $$C_y=1200\sin{\frac{\pi}{6}}=600N$$ Hayley $$H_x=200\cos{\frac{7\pi}{6}}=-100\sqrt{3}N$$ $$H_x=200\sin{\frac{7\pi}{6}}=-100N$$ Since we want the ring not moving at all, then the sum of vectors in $x$-axis and $y$-axis should be equal to 0 $$B_x=-C_x-H_x=-500\sqrt{3}N$$ $$B_y=-C_y-H_y=-500N$$ Because ...


2

Let $\sqrt{x}=a$ $$a^2-8a+7=0$$ Find the factor $$(a-7)(a-1)=0$$ Then you have $$a=7\text{ and }a=1$$ Remember that $a=\sqrt{x}$, then $$x=1 \text{ or } x=49$$


0

Hint: This follows from Gauß's lemma .


0

Assume $r=\frac{q}{p}$ is rational and f is monic and $f(x)=x^n+a_{n-1}x^{n-1}+...+a_1 x+a_0$ so we have $0=p^{n-1} f(r) = \frac{q^n}{p}+a_{n-1} q^{n-1}+a_{n-2}q^{n-2}p+...+a_1 qp^{n-2}+a_0 p^{n-1}=0$ So fraction $\frac{q^n}{p}$ should be equal to integer $-(a_{n-1} q^{n-1}+a_{n-2}q^{n-2}p+...+a_1 qp^{n-2}+a_0 p^{n-1})$ which means $p=1$


1

hint It can be written as $$8x-7x-8\sqrt{x}+7=$$ $$8\sqrt{x}(\sqrt{x}-1)-7(\sqrt{x}-1)(\sqrt{x}+1)=$$ $$(\sqrt{x}-1)\Bigl(8\sqrt{x}-7(\sqrt{x}+1)\Bigr)=$$ $$(\sqrt{x}-1)(\sqrt{x}-7)=0$$ thus...


2

I feel like it's possible for $r$ to be a rational, because a polynomial with integer coefficients can have rational roots. Yes, you're right. The statement is false. For clearly, for example, $$(x-1)(2x-1)$$ belongs in your ring and has the factor $(x-1/2).$


1

$\displaystyle x-8\sqrt{x}+7=0$ or, $\displaystyle (\sqrt{x} -1)(\sqrt{x} -7)=0$ Hence, $x = 1$ and $49$


1

Why did things get compilicated? If you replace $x$ with $y^2$ and $\sqrt x$ with $y$ you should get $y^2 - 8y +7 =0$ and that should not be complicated. ======== Factor or use the quadratic formula so $(y - 7)(y-1) = 0$ so $y = 7$ or $y=1$. Or $y = \frac {-(-8) \pm \sqrt {64-4*7*1}}2 = \frac {8\pm {36}}2 = \frac {8\pm 6}2=4\pm 3=1,7$; so $y=7$ or $y =...


4

If $x - 8\sqrt x + 7 = 0, \tag 1$ then $x + 7 = 8\sqrt x; \tag 2$ then $x^2 + 14 x + 49 = (x + 7)^2 = (8\sqrt x)^2 = 64x; \tag 3$ thus $x^2 - 50x + 49 = 0, \tag 4$ which factors as $(x - 1)(x - 49) = x^2 - 50x + 49 = 0; \tag 5$ thus $x = 1 \; \text{or} \; x = 49; \tag 6$ it is now a simple matter to check that $1$, $49$ obey (1).


4

Isolate the square root, then square both sides: \begin{align*} x-8\sqrt x+7&=0\\ \implies x+7&=8\sqrt x\\ \implies (x+7)^2 &= 64 x\\ \implies x^2 -50x + 49&=0\\ \implies (x-49)(x-1) &=0\\ \implies x = 49 \text{ or } x&=1, \end{align*} which are the two solutions.


7

Let $y=\sqrt{x}$, therefore $y^2=x$ $y^2-8y+7=0$ therefore $(y-7)(y-1)=0$ hence $y=1,7$. Therefore $x=1,49$. These are indeed the only solutions.


1

You could have nonreal and complex coefficient for A and B. They're just numbers! $$ \frac{x-3}{x^2-7}=\frac{A}{x-\sqrt{7}}+\frac{B}{x+\sqrt{7}}$$ So by equality we have: $$ \frac{x-3}{x^2-7}=\frac{A(x+\sqrt{7})+B(x-\sqrt{7})}{(x+\sqrt{7})(x-\sqrt{7})}$$ S0: $$A+B=1 \; and \; (A-B)\sqrt{(7)}=-3$$ Finally : $$A=\frac{\sqrt{7}-3}{2\sqrt{7}}$$ $$B=\frac{\...


1

Yes, you can do it, provided you don't mind using irrational real numbers. That is, if $a$ is positive, then you have that $$x^2-a=(x-\sqrt a)(x+\sqrt a).$$


3

There are several ways to define $x^y$ for real $y$ and $x>0$. Here are three popular variations: $x^y=\exp(y\ln x)$, where $\exp$ is defined through, say, a power series $x^y=\lim_{n\to\infty}x^{y_n}$ where $y_n$ is a sequence of rationals converging to $y$, and $x^{y_n}$ is defined through roots and repeated multiplication The function $y\mapsto x^y$ ...


3

One can also define $2^{\sqrt{3}}$ as follows. First, $2^k$ for integral $k$ is clear. If $k>0$ is an integer, it is clear how to define $2^{1/k}$: as the unique positive real $k^{\mathrm{th}}$ root of $2$. Then, for $r$ any positive real, choose a sequence $a_n = \frac{p_n}{q_n}$, where $p_n, q_n$ are integers, converging to $r$. Then $2^r:=\lim_{n\to\...


0

The meaning is $e^{\ln(2)\sqrt 3}=\exp(\ln(2)\sqrt 3)$. The exponential is a well defined function.


0

$|x+y-1| + |2x + y + 1|=1$ As usual, you can do this in cases depending on whether $x+y-1$ and $2x+y+1$ are great or equal to zero, or less than zero. The trick is to realize that $x + y -1 = 0$ and $2x+y+1$ equal zero are two intersecting lines that cut the plane into four regions and each region will represent one of the four cases. For instance. The ...


0

If you know already a bit of linear algebra you may proceed as follows: You surely can draw $|u|+|v|=1$. Now, consider the mapping $$\begin{pmatrix}u \\ v\end{pmatrix} = \begin{pmatrix}1 & 1 \\ 2 & 1\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}+\begin{pmatrix}-1 \\ 1\end{pmatrix}$$ The inverse mapping (just solve for $\begin{pmatrix}x \\ y\end{...


1

From the denominator of the fraction $\frac{\frac{2-b}{2b}}{b-2}$ pick up $-1$ and obtain: $ \frac{\frac{2-b}{2b}}{-(2-b)}$. Now rewrite the fraction as: $\frac{2-b}{2b}\cdot \frac{-1}{2-b}$. $2-b$ simplifies and you have: $-\frac{1}{2b}$.


2

$\displaystyle \frac{\frac{2-b}{2b}}{b-2}$ $\displaystyle = \left(\frac{2-b}{2b} \right)\left(\frac{1}{b-2}\right)$ $\displaystyle = \left(\frac{(-1)(b-2)}{2b} \right)\frac{1}{(b-2)}$ Now cancel $(b-2)$ from the numerator and the denominator to get $\displaystyle -\frac{1}{2b} $ Is it clear now? Please let me know.


1

When you have a fraction, you often simplify it by multiplying it by $1$ in the form of $\frac aa$ for some convenient $a$. Here you have a fraction with $ \frac {2-b}{2b}$ in the numerator and $b-2$ in the denominator. A convenient $a$ is $\frac 1{b-2}$. Then we have $$\frac {\frac {2-b}{2b}}{b-2}=\frac {\frac {2-b}{2b}}{b-2}\cdot \dfrac{\left(\frac 1{b-...


0

The equation: $$ A |L_1| +B |L_2| =C $$, if $L_1,L_2$ are non-paralal lines, and $A,B,C >0$, represents a parallelogram with $L_1$ and $L_2$ as diagonals and their point of intersection is the center of the 'gram. If $L_1$ and $L_2$ are perpendicular it is square or rhombus. If $L_1,L_2$ are non-perpendicular it is rectangle or a parallelogram. Where ...


1

You have to distinguish several cases: Assuming $$x+y-1\geq 0$$ and $$2x+y+1\geq 0$$then we have$$2y+3x=1$$ or $$y=-\frac{3}{2}x+\frac{1}{2}$$ In the case of $$x+y-1<0$$ and $$2x+y+1\geq 0$$ you will get $$x=-1$$ Can you finish?


1

Assuming you mean: Fix positive natural numbers $s,t$. Let $(x_0,y_0)$ be the unique pair such that $0<x<1$ and $$\begin{align} y_0 &= x_0^t \\ y_0&= (1-x_0)^s \text{.} \end{align}$$ Let $(x,y)$ be the unique pair such that $0<x<1$ and $$\begin{align} y&=\exp\left(\tfrac{s}{\ln x}\right)\\ y&=\exp\left(\tfrac{t}{\ln (1-x)}\right)\...


1

The axioms used for this derivation don't assume that $$1 + (1+1) + (1+1+1) + (1+1+1+1) + \dots = 1 + (1 + (1 + (1 + \dots$$ In fact the freedom to arbitrarily regroup terms is not assumed for series in general, let alone tricky things like this.


0

Benjamin's application of force must be equal and opposite to the combined force of Christine and Hayley. $C = (1200 \cos 30^\circ, 1200\sin 30^\circ) = (600\sqrt 3, 600)\\ H = (200\cos 210^\circ, 200\sin 210^\circ) = (-100\sqrt 3, -100)$ Hayley and Christine are pulling in opposite directions! Ben is pulling in the same direction as Hayley, and pulling ...


1

You have done the problem the right way, but made two crucial arithmetic errors: $$1200\cos 30^{\circ} - 200 \cos 30^{\circ} = 1000\cos 30^{\circ} = 500\sqrt{3},$$ and similarly for the other set of equations. (You multiplied $\cos 30^{\circ}$ by itself for some reason.) If you make those changes and rework the problem, you should get an answer of $1000$ N ...


-1

So you are indeed trying to find integer solutions to the equation $$mn^3 - m^3n = D$$ It looks like you've flipped the sign of $n_1$, presumably because the associated root is always negative (if $m$ and $D$ are both positive, this equation will always have exactly one negative root, with the other two roots either positive or complex conjugates). Since you ...


1

Update: suppose you want to create a curve from $(0,a)$ to $(b,0)$. Then the following will do: $$ \varphi_{a,b}(x) := \frac{a}{b} \frac{b - x}{1 + ax} $$ for all $a,b > 0$. To expand on @J.W. Tanner's answer: For any $a > 0$ $$ f_a(x) := \frac{a(1 - x)}{a + x} $$ fulfills the condition: We have $f_{a}(1) = \frac{a(1 - 1)}{a + 1} = \frac{0}{a + 1} = ...


1

$$y=\dfrac2{x+1}-1=\dfrac{1-x}{1+x}$$ defines a curve that has $y$-intercept $1$ and $x$-intercept $1$ . (I started with $y=\dfrac1x$ and added $1$ in the denominator to make the $y$-intercept $1$. But that has no $x$-intercept, so I multiplied by $2$ and subtracted $1,$ to get $x$-intercept $1$ while maintaining $y$-intercept $1.)$


2

If you graph the functions (should be easy to do by hand. If you don't know how to graph the sine function, learn here), a solution is where the graphs intersect. Where c = 0: There is one solution between 0 and $\pi$. This proves option 3 and option 2 correct. Where c = 1: There are 3 solutions, two between $-\pi$ and 0 and one between 0 and $\pi$. This ...


2

Think of (cos(t), sin(t)) as (x,y) coordinates on the unit circle. I would visualize the point on Quadrant 1, and see what happened next. $$\begin{matrix} t → t+\pi & (x,y) → (-x, -y) & \text{Quadrant 3}\cr t → -t & (x,y) → (x, -y) & \text{Quadrant 4}\cr t → t+\frac{\pi}{2} & (x,y) → (-y, x) & \text{Quadrant 2}\cr t → -t+\frac{\pi}{2}...


1

If one doesn't count multiplicities of roots $$ f(x)=36x^4-76x^3+42x^2+1 $$ has one point at which $f(x)=1$, two points at which $f(x)=2$, three points at which $f(x)=3$, and four points at which $f(x)=4$.


4

Hints: For (a), use the fact that $\sin(t+π)=-\sin t$ and $\cos(t+π)=-\cos t$ as well. For (b), you need to know that $\sin$ is odd and $\cos$ is even. Thus, $\sin(-t)=-\sin t$ and $\cos(-t)=\cos t.$ For (c), use the facts that $\cos(t+π/2)=-\sin t$ and $\sin(t+π/2)=\cos t.$ You should now be able to do (d) by applying the facts I suggested in (c) and (b)...


10

Note that $(2c)!\sigma(c)=\sum_{n=0}^{c-1}\binom{2c}{2n+1}$ while $(2c)!\gamma(c)=\sum_{m=0}^c\binom{2c}{2m}$, so it reduces to the famous result that even-sized subsets of a given nonempty set of finite size are exactly as numerous as the odd-sized subsets.


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