# Tag Info

1

If this is a quartic then you're given the three roots of its derivative. You can integrate this cubic to recover the quartic and use the known point to find the leading coefficient and the constant of integration. Note that $x=0$ is a zero of both the quartic and the cubic. Solution: ! The derivative has roots $-3,0,3$ and so is $ax(x^2-9)$. The ...

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The equation of a circle (algebraically) is given as $$(x-a)^2+(y-b)^2=r^2$$ where $a$ is the x coordinate of its centre, $b$ is the y coordinate of its centre and r is its radius. Once you have input these values you can then convert this to a parametric form by simply substituting $x=r{sin}(t)$, $y=r{cos}(t)$.

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HINTS Verify radius (5) with your work. Use standard form to start, then use $\sin, \cos$ for the two parts.

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R is very good statistical software available free from www.r-project.org It does a lot of complicated stuff, but just connecting dots is easy to show in a demo: x1 = c(1,2,3,4,5,6,7) y1 = c(5.4, 5.7, 6.2, 6.9, 6.5, 5.8, 4.1) # 'plot' sets up axes, labels, etc. plot(x1, y1, type="l", col="blue", lty = "dotted", # 'ell' not 'one' ylim=c(0, 8), ...

1

Already you have obtained that $$Z=(x^2+y^2+x-2y+2)+(2x+y-2)i=0$$ Note that $$x^2+y^2+x-2y+2=(x+\frac12)^2+(y-1)^2+\frac34\gt0.$$ Hence $\Re( Z)\gt 0$ and $Z\not=0.$ Therefore, there are no such complex numbers.

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You have $(x^2 +y^2 +x−2y+2)+(2x+y−2)i=0$ so this complex number is identically zero, meaning that both real and imaginary parts must be zero. However, looking at the real part, you have$$x^2 +y^2 +x−2y+2=0$$ $$\Rightarrow(x+\frac 12)^2+(y-1)^2=-2+1+\frac 14<0,$$ so there are no real $x$ and $y$ satisfying this equation and hence no locus.

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For plotting points, especially points in 3D space, try Graphing Calculator 3D: http://www.runiter.com/graphing-calculator/

1

One option is gnuplot, which handles a lot of graph formats. More flexible in format (and easier to integrate with LaTeX) is asymptote, but then you have to use a library, and defining the graph is more work.

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Most office application suites have some tool to list data and draw plots accordingly. They also allow you to import your data easily. A free example is Libre Office, with its "Calc" application. The commercial equivalent would be Microsoft Office with Excel. Another type of program that can do this are Matlab, Mathematica, MathCAD, and the like, they ...

0

It is fairly hard to visualize the symmetry, because as was written in the other two answers the objects in the Complex case are in general four-dimensional. The symmetry can be displayed however, by dropping one dimension. To augment the two answers given and help you visualize it, here is the case for the two maps: $\exp$, the principal branch of $\log$ ...

1

As I mentioned in my comment above, you can figure out $\dfrac{d^2x}{dt^2}$ from your equation in $|\mathbf{v}|$, given that you know that the change of speed is 3cm/s/s. \begin{eqnarray} |\mathbf{v}| &=& (\sqrt{1+8x^2})\dfrac{dx}{dt}\\ \dfrac{d|\mathbf{v}|}{dt} &=& (\sqrt{1+8^2})\dfrac{d^2x}{dt^2} + ...

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Notice that, at the given instant: $x=1$ cm and \begin{align*} \frac{d^2x}{dt^2}&=\frac{d}{dt}\left(\frac{dx}{dt}\right)\\ &=\frac{d}{dt}\left(\frac{v}{\sqrt{1+8x^2}}\right)\qquad\text{where }v=||\mathbf{v}||\\ &=\frac{\sqrt{1+8x^2}\dfrac{dv}{dt}-\frac{8x}{\sqrt{1+8x^2}}\frac{dx}{dt}v}{1+8x^2} \\ ...

1

In the first place: Don't flip the plot, because it mixes up the names of the variables. When studying $f$ and $f^{-1}$ at the same time you see the graph of $f^{-1}$ in the original $f$-plot when you tilt your head $90^\circ$ sideways. For functions $f:\>{\mathbb C}\to{\mathbb C}$ and their inverses $f^{-1}$ there is an analogous mischievous symmetry; ...

0

$\newcommand{\Cpx}{\mathbf{C}}$If $g:\Cpx \to \Cpx$ is a function, you can reflect the graph $w = g(z)$ across the (complex) line $w = z$ to get $z = g(w)$, or $w = g^{-1}(z)$. (Of course, $g^{-1}$ is generally multiple-valued, particularly if $g$ is entire.) Oh right, there's that small obstacle of living in a three-dimensional universe, where the graph of ...

2

Hint: In a neighborhood of 6 $f(x)$ is continuous and we have $f(x)>2$ , so : $$\lim_{x \rightarrow 6}f(f(x))= \lim _{f(x) \rightarrow 2^+} f(f(x))=\lim _{y \rightarrow 2^+} f(y)$$ and you can see this limit from the graph.

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This will be a 2-dimensional graph, since the position vector r has 2 components. Draw an x,y plane. At time t, the x position is $\cos(\omega t)$ and the y position is $\sin(3\omega t)$.

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The V(x) you gave touches the x-axis but your graph is above x-axis. I have not "designed " roots like this before, but believe it is plausible. I outline a numerical procedure to start with. The way you have sketched, all the roots must be complex. They are type 1 and type 2 complex roots ordered alternately in a waveform well above the x-axis. If ...

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You can set that the first derivative should be $0$ at the points you are interested in. This gives one linear equation in the coefficients of the polynomial for each point. For example third order polynomial $p(x)$ should have extremum at point $x_0$: $$p(x) = ax^3+bx^2+cx+d$$ $$p'(x_0) = 3a(x_0)^2+2bx_0 + c = 0$$ This equation is linear in a,b, and c. ...

2

If we define the ramp function $r$ as $$r(t)= \begin{cases} t, & t\ge 0\\\\ 0, & t<0 \end{cases}$$ then the function $f$ plotted in the post can be represented as $$f(t)=r(t)-r(t-2)$$ Note that if one introduces (i.e., adds) a step function, the resulting plot would exhibit a jump discontinuity. Inasmuch as the plot exhibits no jump, then ...

0

Represent the graph of f parametrically by $x=t, y=f(t)$. If we reflect the point $(t, f(t))$ in the line $y=ax+b$ to get the point $(x,y)$, then $\color{red}{y-f(t)=-\frac{1}{a}(x-t)}$ since the line and the line segment between $(t, f(t))$ and $(x,y)$ are perpendicular to each other. We also have that $\frac{y+f(t)}{2}=a\left(\frac{x+t}{2}\right)+b$ ...

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Let $s_k = \log r_k$. Your problem is equivalent to $$\text{minimize} - \sum_k w_k s_k$$ subject to $$\log a \leq s_k \leq \log b_k$$ The optimal solution is easy to obtain, if $w_k$ is positive, you should take $s_k = \log b_k$. Otherwise you should take $s_k = \log a$. Your plot really depends on what you want to show or illustrate. Here we see that the ...

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As we increase the value of $p$ (starting from $1$), at specific value of $p$, the curve changes its shape from concave to convex No, it only seems that way, because of the restricted window in which you see the graph. The function $(\log_b x)^p$ is not convex for any value $p>0$. It suffices to consider $\ln x$, because the base only contributes a ...

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I have a clever (I believe) way of handling such a comparison. I compare the ratio of the hilbert transforms of the estimate and the true answer. Of course there are some practical concerns as to how one would perform hilbert for purely numerical estimates, however this can work well for closed form asymptotic estimates Example Comparing Bessel $J_0(x)$ ...

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We can implement this as a translation of the line to the origin, a reflection about the line through the origin, and then a translation back in the same direction. In particular: suppose we want to reflect a point $(x_0,y_0)$ across this line. First translate it to the point $(x_1,y_1) = (x_0,y_0 - b)$. Then, reflect $(x_1,y_1)$ across the line $y = ax$ ...

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$$f(x) = \begin{cases}1 & \text{ if } x\leq 0 \\ 0&\text{ if }x>0\end{cases}$$ is one such function. Also, for any $\alpha > 0$, the function $$\frac 1{(x + 1)^\alpha}$$ satisfies your conditions. Furthermore, take any function $f(x)$ which is not identically equal to $0$ and is defined on $(a, \infty)$ for some $a\in\mathbb R$. Also, ...

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There are many trillions of answers. The function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 0$ except for $f(0) = 1$ will do.

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Thanks! After more investigation, I made this. http://tube.geogebra.org/m/1500615

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$x^y-y^x=0$ has a general solution in terms of the Lambert W function. $x^y-y^x=a\neq0,$ however, has no such general solution. If you wish, you may write ${\color{red}y}=\sqrt[\Large x]{x^{\color{red}y}-a}$ , and then repeatedly iterate the expression with regard to y.

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If the graph is more rounded than a parabola at a point where it touches the $x$-axis, then its square-root would be rounded there too, having zero gradient. If it is less rounded, then its square-root would have a cusp there with infinite gradient on both sides. If it is equally rounded, then its square-root would have a straight corner there. The problem ...

1

we now that the domain of the function is: $$ax+b\gt 0\Rightarrow x\gt\frac{-b}{a}\text{so the domain is:}(\frac{-b}{a},+\infty)$$ $$f'(x)=\frac{a}{ax+b}$$ in the domain of the function since we have $x\gt\frac{-b}{a}\Rightarrow ax+b>0$ the sign of $f'(x)=\frac{a}{ax+b}$ will be dependent to the sign of $a$ so: if $a\gt 0\Rightarrow f'(x)\gt 0$ and ...

0

$$h^{ \prime }\left( x \right) =\frac { a }{ ax+b } >0$$ $1$.if $a>0$ $ax+b>0$ $\Rightarrow$ $ax>-b$ $\Rightarrow$ $x>-\frac { b }{ a }$ function is increasing $2$.if $a<0$ $x<-\frac { b }{ a }$ function is decreasing And about concavity you are right,find second derivative and check intervals

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No, sometimes we use a different 'count,' (in the sense of the guy in the video). If you memorize the value of trigonometric functions evaluated at, say, something like $\frac{\pi}{6}$, then we could use a count of $\frac{1}{6}$ instead. It just depends upon how close together we want the points we are graphing to be. A count of $\frac{1}{4}$ just means that ...

1

I suspect that what you're really trying to do is find $a$ and $b$ that best fit the equation $y = b (a-x)^{8/25}$, and further that "best fit" is in the least squares sense, i.e. you want to minimize $$\sum_i (y_i - b(a-x_i)^{8/25})^2$$ where $(x_i, y_i)$ are your data points. This is a non-linear least squares problem, so it is not easy, but methods are ...

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A very simple formul everyone should know avoids to always redo these computation: is a striaght line passes though the points $(a,0)$ on the $x$-axis and $(0,b)$ on the $y$-axis, an equation of the straight line is: $$\frac xa+\frac yb=1.$$ Furthermore , this formula generalises in $3$-space to an equation of a plane, given its intersections with the axes: ...

1

It is very simple to plot graph of a straight line, by finding out the points of intersection of the given line with the coordinate axes as follows $\color{red}{\text{Intersection point with the x-axis}}$: Setting $y=0$ in the equation of the line: $6x-3y-18=0$ we get $$6x-0-18=0$$$$x=\frac{18}{6}=3$$ Hence the point of intersection with the x-axis is ...

1

"Horizontal asymptotes" is a geometric statement of the limits of the following two limits: $$\lim_{x \to +\infty} \frac{1 - 2^x}{1 + 2^x} \text{ and } \lim_{x \to -\infty}\frac{1 - 2^x}{1 + 2^x}.$$ in which \begin{align*} & \lim_{x \to +\infty} \frac{1 - 2^x}{1 + 2^x} = \lim_{x \to +\infty}\frac{\frac{1}{2^x} - 1}{\frac{1}{2^x} + 1} = \frac{0 - 1}{0 + ...

0

Suppose we have the system of equations $$\left\{\begin{eqnarray}ay&=bx+c\\ ay&=bx+c'. \end{eqnarray}\right.$$ Dividing by $a$, we obtain $$\left\{\begin{eqnarray}y&=\frac{b}{a}x+c\\ y&=\frac{b}{a}x+c', \end{eqnarray}\right.$$ which are parallel lines with slope $\frac{b}{a}$. In the second example you've given, you do not have the same ...

0

To me, it is self-evident that two equations with the same slope will always be parallel. For example $y=2x+7$ is parallel to $y=2x+0.1$, which is parallel to $y=2x+b$, where $b$ is any positive or negative number. So perhaps you could expand a bit on what bothers you with this claim.

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Hint : Two lines are parallel if and only if they have the same slope. If the line is given in the form $y=mx+b$, the slope is $m$.

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To scale a function $f\left(x\right)$ down horizontally (in the $x$ direction), what you do is replace $x$ with $nx$, i.e. $f\left(nx\right)$, where it will be scaled down by a factor of $n$. Vertically, simply scale it down by taking $\frac{1}{n}f\left(x\right)$, where $n$ is the factor by which it will be scaled down. Simply combine these two methods to ...

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$$\begin{array}{lrr} \text{Dr: Cash}&\90d\\ \quad\text{Cr: Salary income}&&\90d\\ \hline \text{Dr: Travel expense}& 150p\\ \quad\text{Cr: Cash}&&150p\\ \hline \text{Dr: Laptop}&700\\ \quad\text{Cr: Cash}&&700 \end{array}$$ Consider only Emma's cash account over the summer period. If the days Emma works is just enough ...

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You need to define your terms for people to have a good chance of understanding the equation you've come up with. I infer that $d$ must be the number of days worked, so $90d$ is her money earned. Similarly, $p$ is her days travelled so that $150p$ is her money spent on travel. Is $c$ supposed to be money spent on the laptop? In that case is $c=700$? Try ...

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Both are same, one is rotated by $45^0$ with respect to the other. $x = (x_1+ y_1)/2, y =(x_1 - y_1)/2$ gets from one form to the other.

0

You know how to sketch $y^2- x^2 =1$ already in 2-D. It is symmetrical with respect to x and y axes, and never cuts the x-axis. So now using $r^2 = x^2 + y^2$ as a rotation of $r$, the given equation can be recognized as a surface of revolution obtained by revolving either of the two hyperbolas $y^2 -z^2=1$ , $x^2 -z^2=1$ about z-axis. The ...

1

a) One possible interpretation is that the student makes an initial deposit of USD 450 into his account, on the same day he gets his paycheck. Out of each successive paycheck he deposits USD 100 into the account. The $r$ is the number of paychecks he has gotten after the initial deposit. b) The slope of the graph is 100 and represents the deposit amount for ...

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Hint: Graph what $\sin ( 2x / (1+x^2))$, you'll see the only way you can have a line passing through $(0,0)$ and 2 other points of the graph is if the line has slope smaller than 0 and bigger than $-2$ Since $$\frac{2x}{1+x^2} = \sin(-px) \implies 2x \approx -px \quad \text{around zero}$$

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Hint The equation defining the surface can be written in the form $$f(x^2 + y^2, z) = 0,$$ and so the surface is symmetric about the $z$-axis. In particular, it is the surface of revolution generated by the intersection of the surface with the half-plane $$\{(x, y, z) : x \geq 0, y = 0\},$$ and substitution of the conditions $x \geq 0, y = 0$ in the ...

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Firstly, let's take a look at the surface described by the equation $z^2 = x^2 + y^2$. Since $z^2$ is always non - negative for every $z \in \mathbb R$ and $x^2 + y^2$ is non - negative as well for every $(x,y) \in \mathbb R^2$, there is no restriction for $x,y$ (see below). If we look carefully, $x^2 + y^2= r^2$ reminds us the equation of a circle centered ...

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You can change the definition of your function $f$ to $\quad \quad f(x,y)=\tanh(\sqrt{(a\, b)^2}-\sqrt{(b\, x)^2 + (a\, y)^2}\,)$ To visualize the resulting elliptical contours of $f$ in Mathematica, you can use the following code: With[{a = 2, b = 3}, Module[{f, c}, c = Max[a, b] + 1; f[x_, y_] := Tanh[Sqrt[(a b)^2] - Sqrt[(b x)^2 + (a y)^2]]; ...

0

The "arrow endpoints" just show that the curve continues in the same way forever, to $\infty$ or $-\infty$ as the case may be. In both graphs the arrow endpoint shows that the graph continues down, and in the second graph it also continues to the left. In your first graph, the range is correct but the domain is wrong. You see from the graph that $x$ can be ...

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