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0

We want to find $a$ and $b$ such that, say, $y=a\ln(b\,x)$, passes through the given distinct points $(x_1,y_1)$ and $(x_2,y_2)$. This leads to the $2\times 2$ system \begin{align} y_1&=a\ln(b\,x_1),\\ y_2&=a\ln(b\,x_2). \end{align} Solving for $a$ and $b$, we find $$ a={y_1-y_2\over \ln(x_1/x_2)}, \qquad b=\exp\left({y_2\ln(x_1)-y_1\ln(x_2)\over ...


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} = ...


1

The problem is that Maple takes the complex root with less complex argument, so the cubic root of $-1$ is understood to be $e^{\frac{\pi}{3}}$. One possibility is to ask Maple to print a piecewise function, which in the $x\leqslant 4$ region should be defined as $-\frac{1}{(4-x)^{\frac{1}{3}}}$.


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 ...


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: $$-27=(-27)^{((2/3)(3/2))}=((-27)^{2/3})^{3/2}=9^{3/2}=27$$


1

The reachability relation $R$ for that digraph is the set of all ordered pairs of vertices $\langle x,y\rangle$ such that there is a directed path from $x$ to $y$ in the graph. For example, $\langle b,d\rangle\in R$ by virtue of the path $b\to f\to d$, and $\langle c,c\rangle\in R$ by virtue of the path $c\to d\to c$. Start with vertex $a$; what vertices ...


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 $y=-2$".


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.


0

It is correct. Because $(f+g)(x)=f(x)+g(x)$ $(fg)(x)=f(x)g(x)$ and so on. These properties can be proven by using limit and continuity. But this requires $f(x)$ and $g(x)$ to be continuous first.


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

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 ...


0

Why not just plot it as an implicit function? Maple, for example, has the implicitplot command. plots[implicitplot](y^2 - log(y)^2 = 4*log(x) + 4/x + 1, x = 0 .. 5, y = 0 .. 5); But as for your search for a second solution: $y^2 - \log(y)^2$ is increasing on $(0,\infty)$, so there is never more than one $y$ for a given $x$.


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 ...


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

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 ...


0

So here's our restatement (see discussion in comments above): $$ (ma)^2 + (nb)^2 = (pa)^2 $$ Please note that this means that $(ma, nb, pa)$ must form a Pythagorean triple, with hypotenuse $pa$. So you're trying to find all Pythagorean triples where the hypotenuse and one of the legs have a common divisor, $a$. ...


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; ...


0

Here is some basic info for exponential functions: $y=a(b)^x$ is the general form of an exponential function. $a$ is the value where the graph hits the $y$ axis. The graph never hits the $x$ axis because $a^x$ is always positive (for $a>0$) or always negative (for $a<0$) An exponential function grows faster than a power function, a function ...


0

Wolfram alpha is handy for these. You just type "plot 4^x -1" in the box and you get the plot:


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.


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.


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.


0

I'm not sure which axis time and the number I enter in the program should go on. While the choice is entirely up to you, a typical format would be to put the number you enter on the horizontal ('x') axis, and the result (speed of program) would go on the vertical ('y') axis. This is because you can think of the speed of the program as the result of ...


0

If a function's graph is symmetric with respect to the line $y = x$, then a point $(a, b)$ is on the graph if and only if the point $(b, a)$ is on the graph. Example. $f(x) = \dfrac{1}{x}$ To find the image of a function reflected in the line $y = x$, interchange the values of $x$ and $y$. Example. The image of the graph of $f(x) = \sqrt{x}$ is the ...


0

Interchanging $x$ and $y$ yields the same expression.


1

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


1

A cube ! With one vertex at $(0,0,0)$ and the "opposite" one in $(1,1,1)$.


-1

HINT: Define a co-ordinate plane, along with an origin. Use paint.


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 ...


0

Shasheff polytopes aka Associahedra are geometric representations to study associativity, after Jim Stasheff who first studied them in the 1960s. The associahedron Kn is a polytope of dimension n whose vertices are in one to one correspondence with the parenthesizings of the word x0x1 . . . xn+1. From the 2005 introduction by Loday "The multiple ...


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 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 ...


1

I'd use Share$\LaTeX$ with pgfplots. It takes some time to get used to, but the folks down at $\TeX$ Stack Exchange can help you out.


0

Since your curve does two very different things on the two intervals, you will need to define a piecewise function. When $x \leq 7$, your function would be $f(x)$ = $2^x$ When $x > 7$, your function would be $f(x)$ = ${0.5}^x + 10$.


0

It was parabolic but I should have put the exponent as a variable on y apply a scaling on the x axis i then used an evolutionary solver to find both scaling factor and exponent: $$\Large x = t a$$ With scaling factor a = 0.17369 $$\Large y = t^b$$ With exponent b = 0.2787 The domain of t is 0 to 1


1

If $x$ and $y$ are cartesian coordinates and $r$ and $\theta$ be polar coordinate of one point, the above equation becomes $(x^2+y^2)=6\,y ,$ or, $x^2+y^2-6\,y +9 =9 $ or, $ x^2 + (y-3)^2 = 3^2 $ It a circle on y-axis radius 3, displaced by 3 , so it is tangential to x-axis.


0

Let $x$ and $y$ be Cartesian coordinates of a point and $r$ and $\theta$ be polar coordinate of that point,hence $r=(x^2+y^2)^{\frac{1}{2}}$ and $\sin\theta=\frac{y}{r}$ Hence above equation is $r^2=6y$ where $r=(x^2+y^2)^{\frac{1}{2}}$, hence applying we get answer


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 ...


1

When converting to rectangular coordinates you get the following equations: $$ \left\{ \begin{align} x = r \cdot \cos(\theta) \\ y = r \cdot \sin(\theta) \\ \end{align}\right. $$ You can then replace $r$ by its value and obtain: $$ \left\{ \begin{align} x &= 3 \cdot 2 \cdot \sin(\theta) \cdot \cos(\theta) = 3\cdot \sin(2\cdot\theta) \\ y &= 3 ...


1

Let $x_{max}$ the values where the function reaches its maximum. Then, $$\begin{array}{rcl} \text{max value: }4&=&6\cos\bigg(\frac{2\pi}{14}x_{max}\bigg)-2\\ 6&=&6\cos\bigg(\frac{2\pi}{14}x_{max}\bigg)\\ 1&=&\cos\bigg(\frac{2\pi}{14}x_{max}\bigg)\\ \Rightarrow 2k\pi&=&\frac{2\pi}{14}x_{max}\qquad k\in\mathbb{N}\\ \Rightarrow ...


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 $\frac{1}{1}=1$.


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

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 ...


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 ...


1

we have $f(x)=\ln(x)$ by differentiating we get $$y''=-\frac{1}{x^2}<0$$ thus we can apply Jensen's inequality $$\frac{f(a)+f(b)+f(b)}{3}\geq f\left(\frac{a+b+b}{3}\right)$$


1

This is related with the convex of the function $\ln x$.


0

The short answer: the graph where $x \geq 0$ gets mirrored in the vertical axis. As for the difference between $f(|x|)$ and $|f(x)|$. Suppose $f(x) = x-4$. Then $f(|0|) = f(0) = -4$ whereas $|f(0)| = |-4| = 4$.


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

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 ...


0

Either you use a computer, or you sketch the curves based on recognizing the equation as something you already know. In the first case you should recognize $x^2+y^2=k$ as the equation for a circle with radius $\sqrt k$ rather than try to rewrite it to get $y$ as a function of $x$. Similarly for $e^{x^2-y^2}=k$ you would first take the logarithm on both ...



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