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6

Using the identity $\sin^2 t + \cos^2 t = 1$, we can conclude that $(3 \sin^{3} t)^{2/3} + (3 \sin^{3} t )^{2/3} = 3^{2/3}$, that is, $x^{2/3} + y^{2/3} = 3^{2/3}$. From that point, you can differentiate implicitly to find the critical points, inflection points, regions of increase/decrease, etc. (as you normally would when attempting to graph a function ...


5

This one fulfills your requirements: $$f(x)=\frac{x^2+x^3}{1+x+x^2}.$$ We have: $$\forall x>0,\ 0<f(x)<x$$ $$f(0)=0,$$ $$f(x)-x^2=-\frac{x^4}{1+x+x^2}$$ so that $f(x)$ and $x^2$ are very close for small values of $x$, and $$f(x)-x=-\frac{x}{1+x+x^2}$$ so that $f(x)$ and $x$ get closer and closer as $x\to+\infty$. It's also cheap to compute with 2 ...


5

You could use a common hyperbola $y = \sqrt{a^2 + x^2} - a$. Example with $a = 10$:


5

Let's have some questions: 1- Is one of $x=0$ or $y=0$ allowed to be happen? No 2- So let $x\neq 0$and $y\neq0$. What we have if $x>0$ and $y<0$? Indeed, $1=-1$ which is forbidden. 3- what we will be there if $x>0$ and $y>0$ or $x<0$ and $y<0$. Indeed, we will face to the identity $1=1$. Conclusion: All $(x,y)$ wherein ...


4

That the graph of $\,\rm y=k(x-1)^2\,$ passes through the point with coordinates $(3,10)$ means that the pair $(3,10)$ satisfies that equation. So if you plug in those values you'll get: $$\rm\underset{\underset{\displaystyle y}{\displaystyle\uparrow}}{10}=k(\underset{\underset{\displaystyle x}{\displaystyle\uparrow}}3-1)^2\ \ \Rightarrow \ \ 10=4k \ \ ...


3

What if $x > 0$ and $y > 0$? Then we have $1=1$. This means that $(x,y)$ is a point in your set provided $x$ and $y$ are positive. If both are negative, then we find $-1=-1$. This does not work if $x>0$ and $y<0$ or the other way around since $1\neq -1$. Neither $x$ nor $y$ are allowed to be zero. Since the relation would be undefined. Thus ...


3

You can use a weighting function $w(x)$ that allows you to create a mixture between the functions $x^2$ and $x$, as $\frac{x^2+w(x)x}{1+w(x)}$. Ensure $w(0)=0$ so that the initial behavior is $x^2$ and $w$ growing sufficiently fast that the term $x$ supersedes it. $$w(x)=x^3\to y=\frac{x^2+x^4}{1+x^3}.$$ $$w(x)=e^x-1\to y=\frac{x^2+x(e^x-1)}{e^x}.$$ Note ...


3

Just because it works in Mathematica, doesn't mean it'll work in Wolfram|Alpha. From the FAQ: Can I use Mathematica code in Wolfram|Alpha? Yes. Most small pieces of Mathematica code will work in Wolfram|Alpha. In most cases, you can mix the Mathematica code with pseudocode and math. If you have access to Mathematica, you need to use == instead of ...


3

It might help to rewrite $g(x) = |x|$ as a piecewise function: $$g(x) = \left \{ \begin{array}{lr} -x & x < 0 \\ x & x \geq 0 \\ \end{array} \right.$$ Then the product is also a piecewise function. $$(fg)(x) = \left \{ \begin{array}{lr} -x^2 & x < 0 \\ x^2 & x \geq 0 \\ \end{array} \right.$$


3

One function that will produce the given graph is $$ f(x)= \sum_{k=1}^{\lfloor x \rfloor} k = \frac{\lfloor x \rfloor \cdot \lfloor x + 1 \rfloor}{2} $$ provided $x \geq 0$ and $f(x)=0$ if $x < 0$. Here $\lfloor x \rfloor$ denonotes the floor function, i.e., decimal truncation. If you restrict the domain of $f$ to set bounded from above then you can ...


2

The idea is to let go of the impression that graphics must be all nice and continuous. I'll give the ideas locally. For a) think of the graphic of $\frac{1}{x}$ near zero. The function is not defined on zero, and the limit does not exist there. At the point $x = -5$, make something similar, make the lateral limits different, that is, the graph should, say, ...


2

If you know (or can make good estimates for) other points on the curve, you can make decent approximations using triangles. E.g. if you divide it up into 1, 2, or 4 triangles (you can use however many you want; more triangles is more accurate), it might look like this: !! To find the length of the curve, just add the lengths of the hypotenuses of the ...


2

The graph of $f(x)=x$ is: So,we can reject the answer $B$. $2-x^2$ is a parabola..$2-x^2$ is always negative for $x>1$,for example,for $x=2$, $2-x^2=-2<0$. So,we can reject the answer $D$. Also, for $x>1$, $f'(x)=-2x<0 \forall x>1$,so $f$ is decreasing for $x>1$.So,the right answer is $C$,since at graph $A$,the function is increasing ...


2

How do you like $f_k(x) = \dfrac{1}{1+e^{-kx}}$, where $k\in(0,\infty)$ is a parameter?


2

The first thing you should do is factor the polynomial completely. Here, we have $(x^{2} - 1)$ can be factored into $(x - 1)(x + 1)$ because it is a difference of squares (since it is equal to $(x^{2} - 1^{2}))$. Similarly, $(x^{2} - 9)$ is also a difference of squares since it can be written as $(x^{2} - 3^{2})$, and so it factors into $(x - 3)(x + 3)$. ...


2

Divide into four cases; $(x,y)$ in first, second, third and fourth quadrant, you will get your graph. With this particular question, you will be getting some easy equations. Regarding the use of $techniques$ you were looking for, here also it can be applied, but becomes a bit tedious. Sometimes its better to use the old school method. Presence of both $|x|$ ...


2

If $x$ is in the interval $\left[\dfrac{\pi} 4, \dfrac{\pi} 3\right]$ then $4x$ is between $\pi$ and $\dfrac{4\pi} 3$, which puts $4x$ in the third quadrant. Hence $\cos 4x$ is negative and so is $x \cos 4x$ and its graph is below the $x$-axis.


2

Since the point $(x,y)=(3,10)$ satisfies the equation we can substitute into the equation: $$10=k(3-1)^2$$ $$10=4k\Rightarrow k=\frac{5}{2}$$


2

This is a table of the $y$-values you're looking for. As an example computation: when $x = -4$, we have $$ f(-4) = \frac{-(-4)^2 - 10(-4)}{2} = \frac{-16 + 40}{2} = \frac{24}{2} = 12 $$


2

We can get a better idea of what the graph might look like by using the fact that $|x| = x$ if $x\geq 0$, and $|x| =-x$ if $x\lt 0$. $$ |x-y| = m \implies \begin{cases} x-y = m & \text{when}\;x-y\geq 0 \iff x\geq y\\ x-y = -m & \text{when}\;x-y \lt 0\iff x < y\end{cases}$$


2

For any rational function of the form $ f(x) = \frac {ax+b}{cx+d} $, the vertical asymptote is $ x = \frac{-d}{c} $ because as $ x $ approaches $\frac{-d}{c} $, the function diverges to $ \infty $ or $ - \infty $. The horizontal asymptote is $ y = \frac {a}{c} $ because $ \lim_{x\to\infty}f(x) = \frac {a}{c} $ and $ \lim_{x\to-\infty}f(x) = \frac {a}{c} $. ...


2

Start with a graph of $\tan x$. For $\lceil\tan x\rceil$, mark all points of $\tan x$ where the $y$ coordinate is integer. The graph of $\lceil\tan x\rceil$ is a staircase with these points as right ends (because $\tan$ is monotonically increasing ). For $\tan\lceil x\rceil$, mark th epoints in $\tan x$ with integer $x$ coordinate instead. Again, thes ...


1

You can try to draw curvature circles at your graph and see where the radius is smallest or largest. $\hskip1.5in$ It is natural to define the curvature of a straight line to be identically zero. The curvature of a circle of radius R should be large if R is small and small if R is large. Thus the curvature of a circle is defined to be the reciprocal of ...


1

Case 1: Both $x$ and $y$ are positive(1st quadrant). Then, the equality changes to $x+y=m$ So draw a line in the first quadrant joining $(0,m)$ and $(m,0)$. Case 2: Both are negative (third quadrant) Proceed as in 1. $x+y=-m$. Case 3: 2nd quadrant $(x<0,y>0)$ Note: if $|x|<|y|$, then $x+y$ is positive. On the other hand, negative. Case 4: ...


1

Here is a drawing of the situation . The surface $S$ is shown in tan, the yellow arrow points in the direction of increasing x . The unit normal to the surface is the vector shown in red , the plane $R$ , $x=0$ is shown in white. I believe you need to develop a projection formula of the form : $$ \int_S F \cdot \hat n dS = \int_R \frac{F \cdot \hat ...


1

A couple of ideas: For each point, plot a dot where the point's pdf is maximized. Then plot a closed curve around that point so that the point's pdf has the same value on all the points on the curve and there is a fixed high probability (say, 90%) that the point is inside the curve. This is easy enough to do if $x_1$ and $y_1$ both have normal distributions ...


1

Let's find the derivative: we have $$ g(x) = \cos(x)\ln(\sin 2x) + \sin x \frac{2 \cos 2x}{\sin 2x}\\ = \cos(x) \ln(\sin 2x) + 2\sin x \cot(2x) $$ In order to find where the function has relative extrema, set this equal to zero $$ \cos(x) \ln(\sin 2x) + 2\sin x \cot(2x) = 0 \implies\\ \cos(x) \ln(\sin 2x) = -2\sin x \cot(2x) \implies\\ -\ln(\sin 2x) = ...


1

We can take a) $x - 2 = 2\cos t$, and $y - 1 = 2\sin t$, $t \in [0,\pi]$, and for b) $x + 3 = 3\cos t$, and $y-3 = 3\sin t$, $t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$


1

The easiest software you can used for that is Graphing Calculator 3D. It also generats nicer graphs than even Mathematica. You can modify and rotate your graphs live so it's great for inspecting the visual properties of various graphs. But before you can use it you need to convert your 3D equations from implicity to explicit. That is, all equations should ...


1

Consider the derivative of your function. It has to be positive,increasing and $f'(x→inf)=1$ $f'(x)=tanh(x)$ fits. $ \int \tanh(x) dx = \log (\cosh (x))+c$ Since we want $f(0)=0$ $\log (\cosh 0)+c=0\Leftrightarrow c=0$ Therefor $f(x)= \log (\cosh (x)) $ works.



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