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9

The function $f(x) = x (x^2 - 1) \left(1 + \dfrac{16}{27} (x^2-1)^2\right)$ has its $x$ intercepts $0, \pm 1$, its local extrema $(\pm 1/2, \mp 1/2)$, and its inflection point $(0,0)$.


5

$f(x)=x^3 - ax^2$ has intercepts at (0.0) and (a, 0) critical points at ( 0, 0 ) and $(\frac{2}{3}a, -\frac{4}{27}a^3)$ and an inflection point at $(\frac{1}{3}a, -\frac{11}{27}a^3)$ you get all integers if a is a multiple of 3.


3

$$\newcommand{\t}[1]{\text{#1}} |z-(-3/2)|\ge 2\\ \t{matches with}\\ |z-z_0|\ge r\\\iff |(x+iy)-(x_0+iy_0)|\ge r\iff|(x-x_0)+i(y-y_0)|\ge r\\ \iff \sqrt{(x-x_0)^2+(y-y_0)^2}\ge r\iff (x-x_0)^2+(y-y_0)^2\ge r^2 $$ Which is actually the exterior of circle with center $z_0$ and radius $r$, since distance from point is greater than r (think.think.)


3

Two surfaces are tangent at the point $P\in\mathbb{R}^3$ iff they are both containing $P$ and sharing the same tangent plane at $P$. equivalently: The graphs of $f,g\in\mathbb{R}^2\rightarrow \mathbb{R}$ are tangent to each other at $P\in\mathbb{R}^2$ iff they are sharing the same tangent plane at $P$, i.e. $f(P)+\nabla f(P)(X-P) = g(P)+\nabla g(P)(X-P)$, ...


3

A function is concave up (also called convex) on an interval $I\subset\mathbb R$ if $$f(ta+(1-t)b)\leq tf(a)+(1-t)f(b)$$ for all $a,b\in I$ and for all $t\in [0,1]$. Geometrically, this simply means that the line connecting two points $(a,f(a))$ and $(b,f(b))$ does not dip below the graph of the function over the interval with endpoints $a$ and $b$. ...


3

The ambiguity is in the shrink step. The origin of the shrink has to be specified - points to the right of this origin are moved to the left towards the origin and points to the left of the origin are moved to the right towards the origin. I believe that you are shrinking around 3, and the answer is shrinking around zero. Since the shrinking origin is not ...


3

When we do a horizontal shrink by a factor of $b$ we replace $x$ with $bx$, rather than multiplying the whole expression by b. So: $$g(x) = f(2x-3) + 1$$ not: $$g(x) = f(2(x-3)) + 1$$


3

There is actually a really easy way to test which function is correct. First, for $f(x)=\sqrt{x}$, you know you have the ordered pair $(4,2)$. Now consider the transformations to obtain $g(x)$: Shift right by $3$ units. Horizontal shrink by a factor of $2$. Shift up by $1$ unit. Now consider what happens to the ordered pair $(4,2)$ when you apply 1-3 ...


3

Firstly, apologies for the somewhat bad sketch. If you sketch $y=|x+2|$ and $y=|x-3|$ on the same set of axes, notice the three intervals where the function $y=|x+2|+|x-3|$ will be defined differently. The dotted arrows in the sketch above should help with that. Can you see where to go from here? Edit: Your function (and what I'm fairly sure the ...


3

But I am not entirely sure what this entails. It entails investigating the nature of $y=|x+2|+|x-3|$ by using your knowledge of $|x|$ which is defined as follows: $$ |x|= \begin{cases} x &\text{if $x \geq 0$},\\ -x &\text{if $x<0$}. \end{cases} $$ Similarly, for $|x+2|$, we have $$ |x+2|= \begin{cases} x+2 &\text{if $x \geq -2$},\\ -(x+2) ...


2

Any equation which can be written $$A+Bx+C\log(D+Ex)=0$$ has solutions expressed in terms of Lambert function. In the case of $x^y=y^x$, this writes $$y = -\frac{x}{\log x}\,W\left(-\frac{\log x}{x}\right)$$ and what Lambert and Euler showed is that, in the real domain, the function $W(z)$ exists if $z\geq -\frac 1e$. So, for the argument $-\frac{\log ...


2

For most points on $y=x$, you have $dy/dx=1$. There is one point where $dy/dx$ could be $1$ or $-1$. Call that point $(x,y)=(a,a)$. $$x\ln y=y\ln x\\ \ln y+\frac xy\frac{dy}{dx}=\frac{dy}{dx}\ln x+\frac yx\\ \ln a+\frac{dy}{dx}=\frac{dy}{dx}\ln a+1$$ This has a unique solution unless $\ln a=1$


2

The word for the measure you're looking at is "curvature". I'd probably call the thing you're looking for "the point of maximum curvature"; there may be a better name but I don't know it.


2

You can compute the RHS value to be $|1+i|=\sqrt{2}$. Your equation then has the form $$|z-m|=r,$$ with $z, m \in \mathbb C$, $r \in \mathbb R^{\ge 0}$. This corresponds to the circle of points centered at $m$ with a radius $r$.


2

Perhaps you really mean the image of the function $f:\mathbb R\to \mathbb C$ with $f(t)=5e^{it}$? It lies in $\mathbb C$, and is a circle of radius $5$ centered at $0$. That's distinct from the graph of the function, which lies in $\mathbb R\times \mathbb C$. Addendum 1: The graph is actually easy to visualize as well. As $t$ increases through $\mathbb R$, ...


2

The formula you propose is only valid for asymptotes of the form $$y = mx + q,$$ where $m \in \mathbb R \backslash \{0\}$ and $q \in \mathbb R$. Since $m \in \mathbb R$, these lines cannot be vertical. On the other hand, the asymptotes you seek are of the form $x = q$. You can find them by researching all the points $\alpha$ for which $$\left|\lim_{x \to ...


2

Through supremely literal, I guess just as we can think of the first derivative as 'how quickly the function changes' and the second as 'how quickly the function of how quickly the function changes changes', we can say that the third is just 'how quickly the function of how quickly the function of how quickly the function changes changes changes'. I would ...


1

Assume $z=x+iy$, where $x$ and $y$ are real numbers. Notice that the inequality $|z-1|<2$ can be written as $$\sqrt{(x-1)^2+y^2}<2$$ Squaring both sides yields $$(x-1)^2+y^2<4$$ which is the set of points inside (and not on) the circle centered at $(1,0)$ with radius $2$.


1

In addition to what others replied , consider its inverse function $$ y = f(x) = \tan^{-1} x $$ $$\left|\lim_{x \to +\infty} f(x)\right| = \pi/2 $$ which is a horizontal asymptote. When you switch back to the original function $ x = \pi/2$ is now a vertical asymptote. $$\left|\lim_{x \to \pi/2 } f(x)\right| = +\infty, $$ approaching from left.


1

Yes: it tells you about the rate of change of the curvature of a plane curve, which is given by the formula $$ \kappa = \frac{y''}{(1+y'^2)^{3/2}} $$ The derivative of this is $$ \kappa' = \frac{y'''}{(1+y'^2)^{3/2}} - \frac{3y''^2}{2(1+y'^2)^{5/2}}. $$ If you work in more than two dimensions, the torsion of a curve involves the third derivative: this tells ...


1

For 1 see the asymptotic $f_1(x) \sim \ln(x^2) = 2\ln |x|$ and note that $1+x^2 \ge 1$ so $f_1(x) \ge 0$ and opens up like $\ln$ to both sides (resembles a smooth bucket) Similarly, the shape of $f_2$ is like the shape of $\frac1{1+x^2}$, somewhat like a bell with its only maximum at $0$ with value $4$. For 3 just try some algebraic manipulations: ...


1

change $x$ by $x_2-x_1$ changes the value of $y$ by an integer so just add $x_2-x_1$ to $x_2$ to get another integer. There are therefore an infinite number of integral points.


1

It is the amplitude, it will make the upper & lower bounds of the function on the $y$-axis higher and lower. Normally $\cos x$ and $\sin x$ are between $-1$ and $1$ on the $y$-axis. By taking $4 \cos x$ and $4 \sin x$, your values will be between $-4$ and $4$. You could say that it "streches out" your function along the $y$-axis. Everything gets out of ...


1

The first derivative is used to detect critical points. Find where the first derivative is zero or undefined. These critical points can be local maxima, local minima, and inflection points. Now that you have a list of critical points, you want to know which of them are maxima, which are minima, and which are critical points. For this you need(*) the ...


1

For simplicity, I talk about functions $\mathbb{R} \to \mathbb{R}$. An extremum means either a maximum or a minimum. A relative extremum is a local extremum, they are synonyms. Some older texts seem to prefer to use "relative" rather than "local", say Apostol's calculus. That the derivative of a function $f$ at a point $x$ equals $0$ is a necessary ...


1

Your idea is close, though, I presume you mean $z=\ldots$, rather than $x=\ldots$. Note that a level set of your function has the form $f(x,y,z)=k$ for some real number $k$. For this particular function, that is $$9x^2-4y^2-36z = k$$ or $$z = \frac{9x^2-4y^2}{36}-\frac{k}{36}.$$ Presumably, you know that each of these is a hyperbolic paraboloid; the ...


1

You are confusing degrees with radians. $\pi/6$ is ~0.523, it is not 30. edit to try and make this more helpful. From the question, as well as later comments, it is clear that you expect $cos(30 f(x,y))$ to be the same function as $cos(\frac {\pi} {6} f(x,y))$. Trigonometric functions such as $cos$ are defined to take an angle as an input, and when no ...


1

Use some version of $$\lfloor \sin(a \cdot t+b)+c \rfloor$$ Where $\lfloor x \rfloor$ is the floor function and a,b, and c are constants. You can integrate this piecewise to find the area under the graph. By the way, your graph is referred to as a square wave.


1

I suggest using a digitizing software to convert the graph into numbers. Once you have the graph as a data file, you can use whatever fitting algorithm you wish (usually a least square estimator provides good results). For the digitizing software, I use im2graph. im2graph is free and available for Linux and Windows. Converting graphs to data requires ...


1

Assuming that $$y = \frac{A_{i}}{ln[x]^{3}} + B_{i}$$ is the right model to use, the general approach you used is very correct for me (at least in its principle). For each value of $z$, you adjusted the $A_z$ and $B_z$ and plotting the values of these coefficients as a function of $z$ you observed regular trends and you decided to fit them using a cubic ...



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