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1

$$ 2 \frac{\sin(x)}{x} + \frac{\sin(x/2)}{x/2} e^{-i x/2} $$ since we know $$ \sin x = \frac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i} $$ thus $$ \frac{2}{x}\frac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i} + \frac{2}{x}\frac{\mathrm{e}^{i\frac{x}{2}}-\mathrm{e}^{-i\frac{x}{2}}}{2i}e^{-i x/2} $$ which is equal to $$ ...


1

Why not just forget about the different variations of the formula (e.g. point slope, 2 points, slope intercept, etc. ) and just memorize this one. $$\color{Tomato}{m=\frac{\Delta y}{\Delta x}}$$ If the slope ($m$) and a single point ($x_0, y_0$) are known, then this forumla becomes $$\begin{align} m&=\frac{\Delta y}{\Delta x}\\ ...


2

You almost have it with $y-4=(3x/4)+(6/4)$. Just get the 4 to the right hand side $y=\frac34x+\frac32+4$. So $$y=\frac34x+\frac{11}2$$


0

the graph of $y = f(x) = x^3 - 4x = x(x+2)(x-2)$ is odd and $y\ge 0$ on $[-2,0]\cup [2, \infty).$ the graph of $y = f(x)$ has a local maximum at $(-\frac2{\sqrt 3}, \frac{16}{3\sqrt 3}).$ the domain of the graph of $y = \sqrt{f(x)}$ is $[-2,0]\cup [2, \infty).$ the graph of $y = \sqrt{f(x)}$ looks like a semicircle in $-2 \le x \le 0$ and like a half of a ...


1

No, sketching those $3$ curves is unlikely to help. Instead, sketch the radicand polynomial $g(x) = x(x + 2)(x - 2)$. Recall that the original function $f(x) = \sqrt{g(x)}$ is defined only where $g(x)$ is on or above the $x$-axis. Can you use this to figure out the domain of $f$? Now to transform $g$ into $f$, focus only on the parts of $g$ that lie above ...


0

With MatLab : for k=1:1999 x(k)=-5+k*5/1999; y(k)=x(k)-1; z(k)=x(k); w(k)=x(k)+1; t(k)=x(k)*exp(1/(x(k))); end x(2000)=0; y(2000)=-1; z(2000)=0; w(2000)=1; t(2000)=0; for k=2001:4000 x(k)=(k-2000)/400; y(k)=x(k)-1; z(k)=x(k); w(k)=x(k)+1; t(k)=x(k)*exp(-1/(x(k))); end plot(x,y,x,z,x,w,x,t) The result is ...


2

The second derivative represents the inflexion of a curve at a given point. In the picture you have given, it shows the concavity of the graph of function. When the concavity is turned up, the second derivative is positive; otherwise it is negative. When it is zero, the graph of function inflects at that point; that is, the graph has an S-shaped form at the ...


0

I won't say it is (or isn't) the best choice, but I will say it's possible, though maybe a bit challenging, with Sage: This image is from this link. I can confirm that Sage plays nicely with OS X (better than with Windows, I believe), although I've never tried to animate anything; it may have quite a learning curve. You can download and run everything on ...


1

Like any graph, to find $f(x)$ you go to the value $x$ on the $x$ axis and then vertically to find $f(x)$. In the first graph, if I asked you for $f(1.5)$, you would have not trouble reporting that $f(1.5)=1$. If I asked for $f(1)$ and the dots were not there, you wouldn't know whether it is $0$ or $1$. The filled dot is part of the graph and the open one ...


0

Too long for a comment. Something that I wish someone had told me when I started trying to learn calc on my own is how important it is to learn the relationships between position, velocity, and acceleration, and how the areas/tangent lines relate them all. Although these concepts don't inherently involve calculus, they help you understand the uses of the ...


0

Hint: For something like $\lim_{x\to 0^{+}}\arctan(\ln(x))$, think about what happens on the inside first: $\ln(x)\to-\infty$. So then ask, what is $\lim_{u\to-\infty}\arctan(u)$? For this, you can either think about what $\arctan(x)$ means... or if you just know what the graph looks like (it has two horizontal asymptotes).


1

Just a sketch for $x>0$: this transcendental equation cannot be solved explicitly. Nevertheless, by putting it in the form $\tan x^2 = \frac 1 {2x^2}$ one notices that the graph of the left-hand side (LHS) is made of several disjoint branches, each one increasing continuously from $0$ to $\infty$, while the right-hand side (RHS) is decreasing from ...


1

We can do a few simplifications: Substitute $z=x^2$ so $x = \pm\sqrt z$. Then $$\begin{align*} 2\cos x^2 - 4x^2 \sin x^2 & = 0\\ \Leftrightarrow \cos z - 2z\sin z & = 0 & z \ge 0 \end{align*} $$ Now since $z=0$ is no solution and $\cos z = 0 \Rightarrow \sin z \ne 0$ is also no solution, we can divide by $\cos z$ and get $$1 = 2z\tan z$$ so $$z = ...


0

At $x \approx 4$, $y \approx 224(x-4)$, at $x \approx -3, y \approx -126(x+3)$, and at $x \approx 0, y \approx -24x^2$. Also $ y \to + \infty$ as $x \to \pm \infty$, and the graph is continuous. I hope you can piece the parts together.


0

Hint: the y-intercept of a function is the point at which the function crosses the y-axis (x value is $0$). The roots, a.k.a. the x-intercepts, are the points at which the function crosses the x-axis (y value is 0).


0

It will tend towards a (scaled) Gaussian distribution function; see the Central Limit Theorem. If we scale $\chi$ so that $\int\chi(x)dx=1$ then we get the following sequence of functions (for the first 10)


0

It seems like a logistic equation would model this perfectly. $$\frac{dP}{dx}=k(P-P_L)(P-P_U)$$ where $P_L$ and $P_U$ are our lower and upper bounds (by letting $P_L=-P_U$ we are given a slope of 1 at x=0, this makes things nicer) solving this DE and using your intitial conditions, we get: $$P(x)=\frac{100(e^{0.022x}-1)}{e^{0.022x}+1}$$ Obviously, we ...


0

It seems that you look at function such as $$y(x)=100(1-e^{-bx})$$ which goes through the origin and will never touch $100$. So, your requirements give $$80=100(1-e^{-100b})$$ This gives $$b=\frac{\log (5)}{100}$$


0

To expand on @jgon's comment, the function $$100(1-\exp((0.01\ln 0.2)x))$$ would work.


0

It looks like you know all of the equations you need to solve this problem. I also see that you know that the slope of the asymptote line of a hyperbola is the ratio $\dfrac{b}{a}$ for a simple hyperbola of the form $$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$$ So, given $y=\dfrac{1}{2}x+10$, we can see that the ratio $\dfrac{b}{a}=\dfrac{1}{2}$. Furthermore, ...


0

Let me explain how this works. First ill explain what does, what is going on is that if x is less than a x-a will always be negative so the square root of x-a when it is negative would be undefined so it can't be graphed.multiplying the square root of x-a by just makes x, x again after it finds out if it is undefined or not. The plus b is just so the line ...


0

If a finite number (the first few) of terms of a sequence are given and if that sequence is:- (1) in Arithmetic Progression; or (2) in Geometric Progression; or (3) in some nicely behaved pattern (like a Harmonic Progression or a Fibonacci Sequence ), then answer is YES – a formula for the general term can be found. Otherwise is NO, because the values ...


0

Try $x^2+y^2-z^2=-1$, $z>0$ (one component of a two sheeted hyperboloid) or $x^2+y^2-z^2=1$ intersected with $x= z-1$.


2

On can observe that $y=2(x-2^k)+1$ in $2^k\leq x<2^{k+1}$ or $k\leq \frac{\ln(x)}{\ln(2)}<k+1$ hense $k=\left \lfloor\frac{\ln(x)}{\ln(2)}\right \rfloor$ So, the function is : $$y=2\left(x-2^{\left \lfloor\frac{\ln(x)}{\ln(2)}\right \rfloor}\right)+1$$ where $\lfloor \, \rfloor$ is the symbol of the floor function.


3

You might observe that $f(2^n)=1$ and $f(2^n+k)=2k+1$ if $0 \le k \lt 2^n$, though you might want to prove it. So consider whether you can define $k=x-2^{\lfloor \log_2 x \rfloor}$ and so $$f(x)=2x-2^{1+\lfloor \log_2 x \rfloor}+1.$$


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 ...


0

All circular paraboloids are elliptical paraboloids but not all elliptical paraboloids are circular paraboloids. More precisely, an elliptical paraboloid in a surface which has parabolic cross sections in 2 orthogonal directions and 1 elliptical cross section in the other orthogonal direction. (An elliptical paraboloid) Because a circle is just a special ...


0

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: ...


0

$$\arg(-5-5\sqrt3\,i)=\arctan\frac{-5\sqrt3}{-5}=\arctan\sqrt3=\frac\pi3\pm k\pi\;,\;\;k\in\Bbb Z$$ Since $\;-5-5\sqrt3\,i\;$ is in the third quadrant, it must be its argument is $\;\dfrac\pi3+\pi=\dfrac{4\pi}3\;$ (of course, this is defined only up to multiples of $\;2\pi i\;$ )


1

The centers of the circles should be $(c, 0)$, which you can see if you rewrite your equation as follows: $$ (x-c)^2 + y^2 = x^2-2xc + c^2 + y^2 = (x^2-2xc + y^2) + c^2 = c^2 $$


0

Well, its simple to visualize the loci. First, think of $z$ as a point in the Argand plane, such that $z=x+iy$. Then fit it in the equation you have. For the given equation, $$|z-(i+1)|=|1+i|$$ $$|(x-1)+i(y-1)|=|1+i|$$ Now take the magnitudes and equate. You would get something like $$(x-1)^2+(y-1)^2=1^2+1^2=2$$ This loci would be a circle. Any complex point ...


3

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 $r$.


0

Simplify the equation using $z=x+iy$ to get $$(x-1)^2+(y-1)^2=2$$ Try to find this equation yourself using the definitions. So, the locus is a circle centered at $(1,1)$ and with radius $\sqrt{2}$.


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 ...


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.


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 ...


0

This should do the trick function mse Kc = 0.865; n = 2.08; dc = 3*1e-6; d = linspace(0,10,100); % divide the interval (0,10) into 100 steps Td = 1-exp(-Kc*(d/dc).^n); plot(d,Td); end Note, that the plot looks quite boring. Which is due to the fact, that $d/dc$ is very large, and then $\exp(-Kc(d/dc)^n)$ is almost zero, if $d>0$.


0

To solve this question for a general $n$, define a new variable $u = x^n$. Then $x = u^{1/n}$. Now plug this into $f(x)$ to obtain $y = au^{2/n} + bu^{1/n} + c + du^{-1/n}$ You can then see that there is no value of $n$ that makes this a linear function of $u$ over the entire $u$-axis. However, consider each of the proposed values of $n$. $n=2$ $y = ...


0

This is due to some bad software and/or asking software to do something it wasn't meant to do. Suppose $y=\gcd(x/y,xy)$. Then $y$ must divide $x$, else $x/y$ is not an integer, and the gcd would not be defined. Say $x=yk$ for some integer $k$. Then $x/y$ = $k$, and $\gcd(x/y,xy)=\gcd(k,y^2 k) = k$. So $k=y$, i.e., $x=y^2$. In fact, $y=\gcd(x/y,xy)$ ...


0

I will assume that by $x^n$ as abscissa you mean that on the $y$-axis a distance of one unit corresponds to an increase of the $y$ value by one whereas on the $x$-axis an increase of one unit corresponds to an increase of the $x$ value that depends on how far from the origin you are, in such a way that at distance $k$ units form the origin the $x$-value is ...


0

The graph could have been displayed better. From the speed-vs-time graph we can see that there is a linear relationship, namely $$s=s_0+at,$$ where $s$ is the speed, $s_0$ is the initial speed (the speed at tinme $t=0$), and $a$ is the acceleration. If we integrate this with respect to time we get $$x=\int(s_0+at) \ dt=\left(s_0x+\frac{1}{2}at^2\right)+c.$$ ...


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 $\mathbb R$, ...


0

It is a circle around the origin having radius 5 if you draw it as 2D points $(x(t), y(t)) = (5 \cos t, 5 \sin t)$. Check it here.


0

For instance, $\lim_{x\rightarrow \frac{\pi}{2}^-}\sec x=+\infty$ In particular, the solutions of $\sec x=\frac{\pi}{2}+k\pi,\ k\in\mathbb{N}$ accumulate on the left of $x_0=\frac{\pi}{2}$. But every time $\sec x=\frac{\pi}{2}+k\pi,\ k\in\mathbb{N}$, $\tan\sec x$ has a vertical asymptote. Which gives you the graphical effect you see on the left of $x_0$. ...


0

Here's what I've got: X = (Ax+Bx)/2 plus or minus [(1/2)*sqrt(3)*sqrt((Bx-Ax)^2+(By-Ay)^2)]/[sqrt((-(Bx-Ax)/(By-Ay)+1)] Then solve for y using any random formula such as the distance formula (since you have x) Where (X,Y) = unknowns, and (Ax, Ay) is one vertex, and (Bx, By) is another. Note the above equation may be wrong, but how I can tell you how I got ...


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

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 ...


0

It's the solid with a parabola as a top and a flat, isosceles triangle as a base.


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.


0

you can use the geometric interpretation of the absolute value function $|a-b|$ as the distance between the points represented $a$ and $b$ on the number line. that is rewrite $y = |x+2| + |x-3|$ as $y=|x-(-2)| + |x-3|.$ now interpret $y$ as the sum of the distances between the point $x$ to the points $-2$ and $3.$ with that, it becomes clear now that if ...



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