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0

$$x \to 0^\pm \implies \frac{1}{x} \to \pm\infty \implies \tan^{-1}(\frac{1}{x}) \to \pm \frac{\pi}{2}$$ Alternatively, to emphesize the fail of $y$ "at" $0$: $$ x = \frac{1}{\tan(y)} \to 0 \implies \tan(y) \to \pm \infty \implies y \to \pm \frac{\pi}{2}$$


1

Graphing calculators are usually unable to deal with discontinuities, they just keep on plotting and connecting points, so you have to know this "feature" and guess there is something going wrong, or you can see it directly by inspecting the function, since $$\arctan x \underset{x\to+\infty}{\longrightarrow}+\frac{\pi}2$$ $$\arctan x ...


4

WolframAlpha clearly states below the graph that the domain is $\mathbb{R} \setminus \{0\}$. It also states that the range is $(-\pi/2,0) \cup (0,\pi/2) $. Perhaps the vertical line is to emphasize the jump discontinuity (non-removable). It would be nearly impossible to see any graph that stops short of the y-axis. (I.e. where would we put the points ...


0

The curve you are trying to find an equation for also seems to pass through the point $(-5,15)$. Given three points, it's always possible to find a parabola that passes through them, that is, an equation of the form $y=ax^2+bx+c$. To find the coefficients $a$, $b$, and $c$, you need to plug in the $x$ and $y$ coordinates of the three points, $(5,17)$, ...


1

There is no vertical line joining these two "y-intercepts" There is a non removable discontinuity taking place on the y-axis. With the limit definition of the derivative you can find that both tangents at the "y-intercetps" have slope -1 But obviously the derivative as a function does not exist because the given function is and remains discontinuous


3

The original picture tells you the curve has the form $y=a+c\log_b(dx)$ for some constants $a$, $b$, $c$, and $d$, and that it passes through the points $(-27,17)$, $(-9,13)$, and $(-3,9)$. Since $27$, $9$, and $3$ are powers of $3$, it makes sense to try $b=3$ for the base of the logarithm (as you did). If you let $d=-1$ (it has to be negative because ...


1

I don't think that you need to resort to an implicit plotting method for this (which can involve solving an implicit representation repeatedly to try and find the border on the fly, and which can result in a rendering with irregular edges or a need to mess with grid resolution). It seems to me that you have an explicit formulation already, and using such ...


0

This is how I would attempt this Ok first for simplicity lets call: the line made by joining points $(-17,0)$ and $(-3,9)$ as Line A and the line made by joining points $(-12,0)$ and $(-3,9)$ as Line B and the left curve and right curve as shown in the diagram. Now to find the line of A and B is pretty simple, Il do for one (Line A) to demonstrate ...


0

You have an equation for the left curve: $f(x)=a+c\log_b(dx)$. This curve has to pass through three points: $(-27,17),(-9,13)$ and $(-3,9)$. That means that for each point $P$ of coordinates $(x_P,y_P)$ you must have $f(x_P)=y_P$. Try to find the coefficients $a,b,c$ and $d$ letting the logarithm pass through the three points.


3

At a point where a function is not $C_2$ (i.e. not twice differentiable), there will be a discontinuity in the curvature of its graph. In other words, the curvature will jump abruptly from one value to another. Designers can see these jumps, if they are large enough, but many people can't. You can try some experiments with circular arcs that join ...


0

Maybe thinking about the increase in steepness in the positive and negative $y$ direction will help (darker lines being steeper). Alternatively you can somewhat see that the graph is $z=y^2$ so at each $z=c$ you have the two lines $y=\pm\sqrt{c}$. For example, for $z=9$, the two lines are $y=3, y=-3$. You can see that the lines get farther away from the ...


0

yeah that's a log-log plot. Is it enough for you ?


1

Your answer for (a) is correct. Putting that into standard form, we get $$\frac {x^2}2 - \frac {y^2}2 = 1$$ We therefore see that the graph is a hyperbola centered at the origin with branches to the left and the right. (You do know enough analytical geometry to see that, I hope!) At least, your graph is part of that hyperbola. You need to find which part. ...


0

Use the reflection formula: $$\frac{x'-x}a=\frac{y'-y}b=\frac{-2(ax+by+c)}{a^2+b^2}\quad\text{for line }ax+by+c=0$$ Case A: Line is $y-4=0$: $$\frac{x'-x}0=\frac{y'-y}1=\frac{-2(y-4)}{1}$$ where a convention is to take $0/0$ which is a result of the derivation. $$\implies x'=x,y'=8-y$$ So curve becomes: $$y'=8-e^{x'}\equiv y=8-e^x$$ Case B: Line is ...


0

My suggestions are 1. time( $10^{-1} s $ ) 2. t($10^{-1} s $ ) 3. t/$10^{1} s^{-1} $ but what i think is as long as it express the basic idea of that axis to anyone who reads ur graph it will be fine


1

Since we only know how to reflect about the $x$ or $y$ axis, we're going to have to reduce the given problem to a simpler problem. Notice that reflecting about the line $y = 4$ is equivalent to: Translating $4$ units down. Reflecting about the $x$-axis. Translating $4$ units up. So we get: \begin{align*} \boxed{y = e^x} ...


0

Would I calculate the utility for (a,b) , then set U(x,y) equal to that utility, then plot that level curve? This is a good idea. Suppose $U(x,y)=x^{1/2}\cdot y^{1/2}$ and $(a,b)=(36,100)$. Thus $U(36,100)=6 \cdot 10=60$ $60=x^{1/2}\cdot y^{1/2} \Rightarrow y=\frac{3600}{x}$


1

Main features: intercepts, critical points, inflections, asymptotes, concave up/down, increasing/decreasing, symmetry.


1

This is the Gamma Function, the continuous version of the factorial. http://en.wikipedia.org/wiki/Factorial#The_Gamma_and_Pi_functions


1

$$A(t) = 35e^{-0.17t}$$ when power of $e$ becomes $0$, $A(t)=35$ when $t$ progresses in positive direction, power of $e$ keeps on increasing in negative direction and hence value of $\left[e^{(\text{negative power})}\right]$ keeps on diminishing, but it will never become zero! (?) when $t$ progresses in negative direction, power of $e$ keeps on increasing ...


3

I suspect the scale of your calculator when you tried to graph the function needs to be extended (zoom out). $$A(t) = 35 e^{-0.17t} = \frac {35}{e^{0.17 t}}$$ Note that as $t \to \infty$, rather quickly, $A(t) \to 0$. In contrast, as $t$ gets smaller (as we move to the left along the "t" axis, $A(t)$ grows exponentially, and as $t \to -\infty$, $A(t) ...


2

This may be an unsatisfying answer, but: A function $f(x)$ is twice differentiable if neither the graph of $f(x)$ nor the graph of $f'(x)$ has sharp bends in the region of interest. If I were given $$f(x) = \left\{ \begin{array}{l l} x^2 & \quad \text{if $x \geq 0$}\\ -x^2 & \quad \text{if $x < 0$} \end{array} \right.,$$ I ...


1

In the specification of the function $f$, we've restricted the domain $x \geq 1$, which by construction must be the image of the inverse. We can see that of the two branches (the "two functions to graph") only one takes values $y \geq 1$, namely, $y = 1 + \sqrt{x - 5}$, so the inverse of the function must be given by just that half. Another way of seeing ...


1

I do not know if it is valuable to answer this question at this moment, but it is indeed a valuable question in terms of how to use (and think in terms of) GNU Octave. GNU Octave is based on vector and matrices. We have a function $f$ f = @(x) ( (1) .* and((0 < x),(x <= 1)) + (-1) .* and((1 <x),(x<=2))); which, in terms of GNU Octave, is a ...


0

Your graphing calculator sees $1.4142135$ as $\sqrt{2}$ altough it's just approximation. Keep in mind that your graphing calculator might not have enough capabilities. Even an online calculator such as Wolfram Alpha or Desmos shows a right line. You are right in this case. This is just another example of why you shouldn't always rely on your graphing ...


0

Let's consider two methods for graphing a function: method #1 is to plug in numbers and plot points, for which you can plunk down a few dollars and purchase a cheap graphing calculator; method #2 is to apply algebra and calculus and training and deep understanding and intuition, for which you can pay the tuition of a few years of mathematical training. Which ...


1

Type in RegionPlot[x^2+y^2<=2y,{x,-1,1},{y,0,2}] link or RegionPlot[(x^2+y^2)^2<=2*2^2(x^2-y^2),{x,-3,3},{y,-3,3}] (using $a=2$) link to WolframAlpha or Wolfram Cloud.


3

For the second, try polar coordinates: $x^2 + y^2 = r^2$ while $x^2 - y^2 = r^2 (\cos^2(\theta) - \sin^2(\theta)) = r^2 \cos(2\theta)$. So the equation becomes $$ (r/a)^2 \le 2 \cos(2\theta)$$ Note that the right side is positive for $-\pi/4 < \theta < \pi/4$ and for $3\pi/4 < \theta < 5\pi/4$, negative for $\pi/4 < \theta < 3\pi/4$ and ...


3

For $(x^2 +y^2)^2 \leq 2a^2(x^2 - y^2)$ I would immediately go to polar coordinates, and rewrite as $$r^4\le 2a^2(r^2\cos^2\theta-r^2\sin^2\theta)=2a^2r^2\cos(2\theta).$$ Remembering that $r=0$ is the origin, we can almost rewrite the inequality as $r^2\le 2a^2\cos^2(2\theta)$. The polar curve is easy to draw, by just imagining $\theta$ growing, and ...


5

Here are some methods: If possible, express y explicitly in terms of x (or vice versa) on the line of equality. For instance in the first example write: $$ x = \sqrt{y(2-y)} $$ Plot this function (not forgetting the negative square root). It's a circle. Picture each side as a two dimensional surface and imagine where they intersect (and where one is higher ...


4

$x^2+y^2 \leq 2y \Rightarrow x^2+y^2 - 2y \leq 0 \Rightarrow x^2 + (y-1)^2 -1 \leq 0 \Rightarrow $ $\Rightarrow x^2 + (y-1)^2 \leq 1$ We have the interior and boundary of a circle with center in (0,1)


0

I have no plotting software at hand to show right now, but a quick answer for now: if you replace $y$ with $Z$ and $x$ with $\sqrt{X^2+Y^2}$ in your equation, you will have an implicit equation of a surface of revolution, that is really your curve rotated around the $y$ axis (but I rename axes, since I prefer the vertical to be $Z$ instead of $y$). To see ...


2

The slope measure form 0 to 1 is basically an angle (lets call it $\alpha$), the other one is the slope, lets call it $m$. The relation is $$ m = \tan\left( \alpha \cdot \frac{\pi}{2} \right)$$ if you are calculating with radians or $$ m = \tan\left( \alpha \cdot 90° \right)$$ if you are calculating in degrees.


0

Replace $x=\frac{2}{y}$ in equation $x^3+2x^2-4=0$ and get $$ \left(\frac{2}{y}\right)^3+2\left(\frac{2}{y}\right)^2-4=0 $$ and after simplifications $$ y^3-2y-2=0 $$ Now use the Cardamo's formula for cubic equation $y^3+py+q=0$ $$ y=\sqrt[3\,]{-\frac{p}{2}+\sqrt{\left(-\frac{p}{2}\right)^2+\left(-\frac{q}{3}\right)^3}} + ...


1

$$\cos4\alpha=-\sin5\alpha=\cos\left(90^\circ+5\alpha\right)$$ $$\iff4\alpha=360^\circ m\pm\left(90^\circ+5\alpha\right)$$ where $m$ is any integer Consider '+', '-' sign one by one. For the second one use $$\sin2y=\frac{2\tan y}{1+\tan^2y}$$


0

$$f(x)=x^3+2x^2-4$$ $$f(0)=-4,f(1)=-1,f(2)=12$$ Since $f(1)f(2)<0$ there is one root in between. Now this is messy: $f(1.5)=3.375+4.5-4>0$ So root is in between $1$ and $1.5$ $f(1.25)=1.078>0,f(1.125)=-0.0449<0$ So approximate value of $x\approx1.1(8)$ satisfies your equation, with exact value at $x=1.13039543476\cdots$ seems close, or you ...


0

Marginal value tells you how much value the production of an additional widget gives you. Since the Value function you described seems to "increase greatly" at first, I take this to mean that the value function increases at an increasing rate. Thus, each additional unit gives you more additional value than the previous unit. Then as the value function ...


0

The formula for slope is $$\begin{array}{llllr}m=\frac{\Delta y}{\Delta x}&&&&(1)\end{array}$$ To find the equation of a line passing through 2 points the formula is $$\begin{array}{llllr}m=\frac{\Delta y}{\Delta x}&&&&(2)\end{array}$$ For example, if we wanted to find the equation of a line passing through points $(-2, 11)$ ...


0

If (x1,y1) and (x2,y2)be two points in the plane.Then the equation of the line joining these two points is $(y-y1) (x1-x2)=(x-x1)(y1-y2)$. From this equation you will get slope intercept form.


2

HINT: Let the slope intercept from be $$y=mx+c$$ Set $(-8,1);(-8,7)$ to find two linear simultaneous equation in $m,c$


0

The complex multiplication by geometric interpretation means that the lenght are multiply each other and the arguments are addition each other and here the argument means the angle with the positive part of the real axis. For nonnegative integer exponents with trigonometric form you can get by induction, that $$C^n = (a+bi)^n = |C|^n\left(\cos(nt)+i ...


0

Every complex number can be expressed as $z=re^{i\theta}$, $r\ge0,\;\theta\in[0,2\pi[$. Hence $z^{n}=r^ne^{in\theta}$. Think on your own at the meaning of $r$ and $\theta$ and what the results of raise to the $n$-th power for $r$ and multiplication by $\theta$ for $n$ are.


4

$$x^2 - y^2 + 2y = 1\iff x^2 = y^2 -2y+1 = (y-1)^2 \iff y-1 = \pm x \iff y = 1\pm x$$ Thus the two intersecting lines are given by $y = 1+x$ and $y = 1-x$. The point of intersection is given by $(0, 1)$. One bisector is the y-axis: the line $x=0$. The other bisector is given by $y = 1$. Can you take it from here?


0

Hint: The slope of the line passing through the pair of points with coordinates $(x_1,y_1)$ and $(x_2,y_2)$ is $$m=\dfrac{\Delta y}{\Delta x}=\dfrac{y_2-y_1}{x_2-x_1}$$ In this context, it will be: $m=\dfrac{-5-1}{-4--2}=3$.


2

Hint: First lets calculate all the points that form the parallelogram. Let us assume that $n\neq 0 \neq m$ The lines $y=mx$ and $y=nx$ intersect at $(0,0)$. The intersection of $y = mx+1$ and is at $(0,1)$ Then the intersection of $y=mx+1$ and $y=nx$ is at $(1/(n-m),n/(n-m)$ and because of the 'symmetry of the equations the intersection of $y=nx+1$ and ...


0

We may suppose that $a\not =0$. The line $x+y=|a|$ passes through two points $(0,|a|),(|a|,0)$. And the line $ax-y=1\iff y=ax-1$ passes through the point $(0,-1)$ with the slope $a$. Since the two lines intersect in the first quadrant, the slope $a$ of the second line has to be larger than $a_0$ such that $y=a_0x-1$ passes through the point $(|a_0|,0)$. ...


3

$x \longmapsto -\tan \left[\frac{\pi}{2} \left(\frac{2 x}{R}-1\right)\right]$


2

It is not a log scale. Every time you go a certain distance upward, you add $20$. On a log scale, every time you go a certain distance upward, you multiply by a particular number. When you go from $100$ to $200$, you multiply by $2$. When you go that same distance upward again, you'd multiply by $2$ again, getting $400$. Then the same distance again and ...


3

You stumbled upon a classic counterexample to the "theorem" $f\colon \mathbb{R} \to \mathbb{R}$ has an horizontal asymptote if and only if $f'(x) \to 0$ as $x\to \infty$. Now, your function $f(x) = \log(1+x^2)$ is a counterexample to the "$\Leftarrow$" implication, since $\lim_{x\to \infty} f(x) = + \infty$. This can be shown by many means; but let's try ...


1

Short answer: Logarithms don't have horizontal asymptotes. Long answer: the derivative becomes smaller as $x\to\infty$, but it has no horizontal asymptote. The asymptote you thought you found, is not an asymptote since $\ln(1+x^2)=14.25$ has a solution (check it yourself). A way to think of it is: logarithms and exponential functions are inverses of each ...



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