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13

Algebraically, if we set $y=0$, then this becomes $$x^2=x+1$$ which is the quadratic polynomial of which the golden ratio is a root. Generally, wherever the golden ratio appears, it's because this polynomial showed up.


6

There are a few things going on. Jump to the end for discussion of complex numbers and what exactly WolframAlpha is plotting. When you plot the function, WolframAlpha and most plotting systems don't restrict themselves to special inputs like "only rational numbers with odd denominator." Even if you try using an odd denominator, most computational software ...


4

The following is a variation of the visualization of the function $x^x$ that I described in this answer. It's not clear to me how to explain it to middle school kids, though. Specifically, if you want to explain the WolframAlpha output to middle schoolers, then they've got to know that $$(-2)^x = e^{x\log(-2)} = e^{x(\log(2) + i\pi)} = 2^x e^{xi\pi} = ...


4

The derivative of the function $f(x)=x-\frac{1}{x^2}$ is $f'(x)=1+\frac{2}{x^3}$. The tangent line to the graph of $f(x)$ at the point $x=a$ has slope $f'(a)$, so it will be parallel with the line $y=3x$ if and only if $f'(a)=3$. So, you are looking for the real numbers $a$ for which $$1+\frac{2}{a^3}=3$$


4

The slope of the line is $3$ . So, $\frac{dy}{dx}=3$ $1+\frac{2}{x^3}=3$$\implies x=1$ and $y=0$ Equation of the line is $(y-0)=3(x-1)$ $\implies y=3x-3$


4

Each of the following steps destroys some solutions, but we get them all back via Evgeny's comment above $ \begin{align*} \sin(\cos(x)) &= \cos(\sin(y))\\ \cos(x) &= \sin^{-1}\cos(\sin(y))\\ \cos(x) &= \frac{\pi}{2}-\sin(y)\\ \cos(x) + \sin(y)&=\frac{\pi}{2} \end{align*} $ you can graph this and see that you get a row of "circles" like what ...


3

This is precisely because the golden ratio is a solution to $$t+1=t^2.\tag{$\star$}$$ In the case that $x=0$ or $y=0$ (but not both), we have that whichever of the two is non-zero may be the golden ratio, or may be the other solution to $(\star),$ but cannot be anything else.


2

By definition: $$\Bbb R^2 := \Bbb R \times \Bbb R = \{(x,y) \mid x,y \in \Bbb R\},$$ that is, the euclidean plane that you're used to, in doing analytic geometry, graphing functions, etc. For your exercise, call $x(t) = \ln t$ and $y(t) = 3 \ln (6t)$. Use properties of $\ln$ and eliminate $t$ from these relations, get $y$ as a function of $x$. You'll see ...


2

the matlab meshgrid outputs two rectangular grids. If you just want to display the surface in zone $1$ and not the others, [x,y] = meshgrid(0:0.01:3); z = your_function(x,y).*((y > 0.5*x + 1) & (y > 6*x - 2)); surf(z) should do it (I'm just using a logical condition to mask off the area of interest, and set the function equal to zero ...


2

There is a problem with negative numbers having rational exponents which should give real results. Please have a look at math.stackexchange.com/questions/956541 In short one has: $$-27=(-27)^{((2/3)(3/2))}=((-27)^{2/3})^{3/2}=9^{3/2}=27$$


2

Yes, you're correct. For $t \geq 0$, $\gamma(t) = (t,t)$ is indeed the line $y = x$. But for $t < 0$, $\gamma(t) = (-t,t)$. It is the graph of $f(x) = |x|$ rotated around the origin $90$ degrees clockwise.


2

HINT: $$\frac{dy}{dx}=1+\frac2{x^3}$$ which needs to be $=3$ This will give us the value of $x, x_1$(say) and consequently the value of $y, y_1$ Now the equation of the required tangent will be $$\frac{y-y_1}{x-x_1}=3$$


2

Yes, you are thinking correctly. $(f+g)(1) := f(1)+g(1)$ $(f-g)(5) := f(5)-g(5)$ $(f/g)(-3) := f(-3)/g(-3)$ $(fg)(5) := f(5)g(5)$ Other than this, is just looking at the drawing. After squinting my eyes a bit, I found no mistakes in your work. You're on the right track. Good studies.


2

take the graph that is an the right half-plane and reflect it about the $y$-axis to obtain the whole graph. Indeed if $x\ge 0$ then $f(|x|)=f(x)$ so for $x\ge0$ (the right half-plane) the graph does not change. If $x<0$ then $-x>0$ and $f(|x|)=f(-x)$, which makes the points $(x,f(-x))$ and $(-x,f(-x))$ to both belong to the graph, and these points ...


2

If we write $y = \pi/2 + s$, then near $(x =0, y=\pi/2)$ we have $$\cos(x) + \sin(y) = \cos(x) + \cos(s) = 2 - (x^2 + s^2)/2 + (x^4 + s^4)/4! - \ldots$$ When $t > 0$ is small, and $x$ and $s$ are small, the terms in fourth and higher powers of $x$ and $s$ are negligible, so the equation $\cos(x) + \sin(y) = 2 - t$ is well approximated by $x^2 + s^2 = 2 ...


2

I do like Desmos. I want to like it more, but it does have limitations. In this particular case, you can see how the loops arise quite clearly by plotting the 3D graphs of $z=\sin(\cos(x))$ and $z=\cos(\sin(y)$ on the same set of axes. The result looks like so:


2

Interesting curve. Try $$\Large x=r\left(1-2^{\frac y2}\right)$$ which is equivalent to $$\Large2^y=\left(1-\frac xr\right)^2$$ where $r$ is the radius or width of the asymptote. Not exactly a parabola per the classical definition, which has to fit a quadratic function. If you want to tweak the curvature, you might want to try a generalized version of ...


2

You can use this online tool. The demand curves are a-x (blue). For different values of a, you can draw the corresponding demand curves. If you want additionally different supply curves, you can draw up to thee different supply curves. My inputs: a-x (blue) x-2 (red) x-1 (green) a from 8 to 10; incrementing by 2. The graph can be very well improved by ...


2

Note, as fjardon mentioned, that we get $$ \begin {eqnarray*} x &=& 6 \sin \theta \cos \theta, \\ y &=& 6 \sin \theta \sin \theta. \end {eqnarray*} $$Now, rewrite $(x,y)$ as $ \left( 3 \sin 2\theta, 3 \cos 2\theta + 3 \right) $. Now, it is clearly that $$ \left( 3 \sin 2\theta \right)^2 + \left( 3 \cos 2\theta \right)^2 = 3^2, $$ from the ...


2

I like the program Asymptote. See a tutorial and a gallery of examples. Also, you can see the drawing in Calculating line integrals via Stokes theorem.


2

The point of drawings or other physical models in mathematics isn't to perfectly model the object, but just to produce something that looks close enough that it can act as a visual aid and help develop some reasonably accurate intuition. (Likewise, most mathematical writing is intended to convey meaning to other human beings, which is why theorems and proofs ...


2

You can regard $K$ as a parameter and do a linear fit to each set of data. When I put it in Excel I get $y=-0.1304x+13.858$ for the $K=1800$ data and $y=-0.3949x+38.892$ for the $K=4000$ data, where $x$ is in percent in both cases. Now you can view the constants as points on a line with $K$ the independent variable, so the global fit is $y=\left( -0.1304 ...


2

May be you could consider that the model is $$y=a+b x$$ and assume a linear dependency of $a$ and $b$ to $K$. This will give $$y=(a_0+a_1 K)x+(b_0+b_1K)=\alpha+\beta x + \gamma K+\delta x K$$ and so perform a multi linear regression in which the independent variables would be $x$,$K$ and $xK$. Then, a global fit could possibly be better since taking into ...


1

$\text{Dom}(f) = (-\infty,\infty)$, and $\text{Ran}(f) = (-1,\infty)$. the graph is the graph of $y = 4^x$ shifted $1$ unit down.


1

Here's an example of how you might do something like this. The key is setting up an $x$–$y$ grid on which to evaluate your function. Because, in your case, this is between two non-horizontal (or non-vertical) lines things can be trickier as you need to ensure matrix dimensions are consistent. This is just one way of doing this: x = 0:0.05:0.25; y1 = 2*x+1; ...


1

It is from the definition. However you may be intereseting in that what it come from or why we called it by this name. In fact: Pick a point $(x,y)$ from the space $\Bbb R^2$, you will find that $x$ is just the projection of the point $(x,y)$, and $y$ is also!


1

Hint: the first implication that I can think of is that the functional inverse would be equal to the function.


1

The asymptotes you found are correct, but I'd write down more formally how you found the horizontal asymptotes, something like: "$\lim_{x\to\pm\infty}\frac{\frac12x+4}{-\frac14x-1}=\ldots=-2$, and therefore the horizontal asymptote is $y=-2$".


1

It depends on the definition of polar coordinates you are using. For example this article claims that What's more, one often allows negative values of r under the assumption that $(-r,\theta)$ is plotted identically to $(r,\theta\pm\pi)$. However, that makes the question rather nonsensical since obviously you cannot plot the entire plane. It's better ...


1

The simple function $4\cos(x)$ will have a relative maximum at $(0,4)$. However, your function seems to be $4\cos(x-1)-2$. The phase shift I believe is $\frac{\omega}{\phi}$. In this case, $\omega = 1$ and $\phi=1$. So the phase shift is $\frac{1}{1}=1$.



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