# Tag Info

3

Hint: For the first function use the identity $$\cos(x)=\sin\left(x+\dfrac\pi2\right),$$ for the second it is useful to note that $\arcsin(\sin(x))=x$, since $\arcsin x$ and $\sin x$ are inverse functions. Also be careful about the domains and ranges of those functions, and so $\arcsin(\sin(x))=x$ isn't true for all $x$.

3

The implicit equation is \begin{align} r^2 &=36\cos(2\phi)\\ &=36[\cos^2(\phi)-\sin^2(\phi)]\\ r^4 &=36[r^2\cos^2(\phi)-r^2\sin^2(\phi)]\\ x^4+2x^2y^2+y^4 &=36x^2-36y^2 \end{align} But parametrically, to plot the curve, \begin{align} x&=6\cos(\phi)\sqrt{\cos(2\phi)}\\ y&=6\sin(\phi)\sqrt{\cos(2\phi)} \end{align} Since we ...

3

The function $ce^{-x^2}$ works.

3

Another answer for the sake of completeness : if you don't know what $e$ is or if you want a single expression for the whole $\mathbb{R}$, you may use $f : x \rightarrow \dfrac{c}{1+x^2}$.

3

Let $g(x)$ be any continuous function with $g(x)\to+\infty$ as $x\to\pm\infty$ (for example $g(x)=x^4+42x^3+\sin x$ or $g(x)=e^{x^2}$). Then $g$ has a global minimum, say $g(x)\ge a$ for all $x\in\mathbb R$. Then for any $b>-a$, the function $f(x)=\frac{c(g(0)+b)}{g(x)+b}$ is a solution to your problem. (For example, with $g(x)=e^{x^2}$, we have $a=1$ and ...

2

What about this function? $f(x)=-x+c, 0\le x \le 1$; $f(x)=0, x>1$ $f(x)=x+c, -1\le x \le 0$; $f(x)=0, x<-1$

2

You could call them 'level sets'. Imagine you have a function $f(x,y) = |y| - |x| - \tan(y)\frac{-\sin(1/x)}{\tan(1/y)}$ then you're looking for the set of points where $f(x,y) = 0$. In python: import matplotlib.pyplot import numpy as np delta = 0.05 xrange = np.arange(-10.0, 10.0, delta) yrange = np.arange(-10.0, 10.0, delta) x, y = ...

2

In order you have a solution, the function $$f(x)=k\sqrt{x^2-1}-\log(x)$$ must go though a minimum and the value of the function at that point must be negative. So, $$f'(x)=\frac{k x}{\sqrt{x^2-1}}-\frac{1}{x}$$ The minimum occurs when $$\frac{k x}{\sqrt{x^2-1}}=\frac{1}{x}$$ (this reduces to a quadratic in $x^2$ and we need to only consider the positive ...

2

For the first : The graph is symmetry both about $x$-axis and $y$-axis. So, we only need to consider the case when $x\ge 0$ and $y\ge 0$. Then, note that $$|||a|-2|-1|=\begin{cases}-a+1&\text{if 0\le a\lt 1}\\a-1&\text{if 1\le a\lt 2}\\-a+3&\text{if 2\le a\lt 3}\\a-3&\text{if a\ge 3}\end{cases}$$ For the second : Since ...

2

The upper (or lower) half of any circle with center in the first quadrant and radius < distance center-axes, like $$y = 2 + \sqrt{1-(x-2)^2}.$$

2

How about $y=\frac{1}{\sqrt{x}}$?

1

I had to do my own version, using @Tharindu's knowledge for my assignment. Here it is if anyone wants a more detailed version. The coordinates are different, since I had a different question: We need to find an equation that will create a line that is perpendicular to the equation $x=3.58$, given that one point on the line B. As shown in steps beforehand, ...

1

You could do this in many ways. 1) It is obvious for some. 2) You can visualize this on your head, or even plot this graph and see clearly. 3) This is the only method that I could think at this time of which you could present if you were asked to prove it. $x= 10$ is a vertical line, A perpendicular line to this must be parallel to the x axis. Hence ...

1

Taking a look at your data, I guessed that maybe a function of the form $f(x)=a/x^p+b$ might work. If we then parameters $a$, $b$, and $p$ to minimize $$\sum_{(x,y)\in \text{data}} (f(x)-y)^2,$$ we find $f(x)=\frac{2106.91}{x^{2.11436}} + 0.463718$. The sum of the squared errors is $0.000480789$, which seems pretty good. The graph of $f$ with the data ...

1

The $x$ axis is in fact one of the tangent lines to the graph of $f$ at the origin. The problem is that the tangent is not uniquely defined and as such is analytically uninteresting. Tangents at points where a function is differentiable have very interesting properties (given in part by their uniqueness) which just do not apply to tangents at points where ...

1

A cusp is understood to be "pointy", such as the graph of the function $y=\sqrt[3]{x^2}$ at $x=0$. See also cusp catastrophe Or in other contexts, consider the upper half plane $\mathbb H$ modulo the action of $PSL(2,\mathbb Z)$: It is essentially a curved polygon with sided glued together, thereby producing two non-pointy corners and one cusp (because ...

1

a) Critical values are places where the derivative is zero or where the derivative fails to exist. Your graph shows three such points. b) A relative extreme is a point where the graph changes from increasing to decreasing, or vice versa. In terms of the derivative, it is a place where the derivative switches from positive to negative or vice versa (i.e., ...

1

Using $r^2=x^2+y^2,x=r\cos\phi,y=r\sin\phi$, and the trig identity $\cos(2\phi)=\cos^2\phi-\sin^2\phi$, we have $$r^2=36\cos(2\phi)$$ $$\implies r^4=36r^2(\cos^2\phi-\sin^2\phi)=(6r\cos\phi)^2-(6r\sin\phi)^2$$ $$\implies(x^2+y^2)^2=36x^2-36y^2$$ $$\implies {x}^{4}+2\,{x}^{2}{y}^{2}+{y}^{4}-36\,{x}^{2}+36\,{y}^{2}=0$$ Now let $x^2=a,y^2=b$. We now have ...

1

Keep in mind that the domain of the arcsin function is $[-1,1]$ and its range is $[-\pi/2,\pi/2]$. This is important because even though $\sin x$ and $\arcsin x$ are inverse functions, it's not correct to say that $\arcsin(\sin x)=x$ for all $x$. This might explain why your graphing calculator is giving you "pointed" curves.

1

Well... at $x=5$, it is clear that $$f(5)=(5-5)^2(5+6)=0^2*11=0$$ So from here, you need to observe that if you pick some number greater than $5$ for $x$ (we can write this as $x=5+x_0$, where $x_0 >0$) then we get $$f(5+x_0) = ((5+x_0)-5)^2((5+x_0)+6)=(x_0)^2(11+x_0)>0$$ The $>$ above is true because $(x_0)^2$ is always positive and $11+ x_0$ ...

1

you want to use Y = Asin(BX+C) such that the upper part (positive) will be your dayTime, OK: sin(x) varies in the positive Y-axis range when x varies between [0, $\pi$] in radians Assume the day is regulated to fit [0, 24] hours $$sunrize = sr$$ $$sunset = ss$$ length of the day (bright side) will be [ss-sr] (whatever the unit of time you use) this ...

1

It's a problem of definition of the function. Every function is not given by the relation $f$ but by three things: 1) a set $A$, the domain; 2) a set $B$, the codomain; 3) a relation $R$ which is a subset of $A\times B$ and has the property that it is a function, i.e. for each $x\in A$ there is exactly one couple $(x,y)\in R$. So a function is really a ...

1

A quotient is defined when both its numerator and denominator are defined and the denominator is nonzero. The quotient $\frac1{\frac1x}$ is defined when $1$ is defined (always true) and $\frac1x$ is defined, and $\frac1x$ is nonzero (always true). The quotient $\frac1x$ is defined when $1$ is defined (always true) and $x$ is defined (always true), and $x$ ...

1

Not sure if it's really a math question, anyway here's what I think. It's a quite complex situation because you have a (nearly) $5$-dimensional vector and you also want to represent for each vector the probability of the vector belonging to a certain class. I suppose that you have these probabilities and that you have graphing capabilities. One solution ...

1

It looks like the problem doesn't have a closed form answer, but I think I have a slightly more elegant way of formulating the problem, anyway. What you're trying to do is to maximize $k$ subject to the equations $y = k\sqrt{x^2-1}$ and $y = \ln x$ for $x > 0$. Solving for k gets you $k = \frac{y}{\sqrt{x^2 - 1}} = \frac{\ln x}{\sqrt{x^2 - 1}}$, so just ...

1

for $x=-1$ we get $-2-5+18/3=-7+6=-1$ and for $x=-7$ we obtain $-19-6=-25$

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Your $y$-coordinate calculations are not correct: $f(-1)=2(-1)-5+\frac{18}{-1+4}=-1\ne-5$ $f(-7)=2(-7)-5+\frac{18}{-7+4}=-25\ne-17$

1

You could use discriminant theory. Consider the equation as a quadratic in x, and find the set of values of y for which there are no real roots

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