# Tag Info

3

Firstly, let's take a look at the surface described by the equation $z^2 = x^2 + y^2$. Since $z^2$ is always non - negative for every $z \in \mathbb R$ and $x^2 + y^2$ is non - negative as well for every $(x,y) \in \mathbb R^2$, there is no restriction for $x,y$ (see below). If we look carefully, $x^2 + y^2= r^2$ reminds us the equation of a circle centered ...

3

The local minima is at $x=\frac{3\pi}{2}$ This is very obvious from the graph of $f(x)=\sin{x}$ On a second look at the graph below, I believe $x=\frac{\pi}{4}$ is also a local minimum. This is because it is lesser than all other values within its locality. Thus we have two local minima: $x=\frac{\pi}{4}, \frac{3\pi}{2}$

2

Hint: In a neighborhood of 6 $f(x)$ is continuous and we have $f(x)>2$ , so : $$\lim_{x \rightarrow 6}f(f(x))= \lim _{f(x) \rightarrow 2^+} f(f(x))=\lim _{y \rightarrow 2^+} f(y)$$ and you can see this limit from the graph.

2

a) You are correct. b) The tiding is coming in, since the height of it is rising as time goes on. c) The domain should include $0$ and the range should be $\{y \: \:|\:5 \le y \le 30, y \in \mathbb{R} \}$

2

There is no discontinuity at the origin.

2

You can change the definition of your function $f$ to $\quad \quad f(x,y)=\tanh(\sqrt{(a\, b)^2}-\sqrt{(b\, x)^2 + (a\, y)^2}\,)$ To visualize the resulting elliptical contours of $f$ in Mathematica, you can use the following code: With[{a = 2, b = 3}, Module[{f, c}, c = Max[a, b] + 1; f[x_, y_] := Tanh[Sqrt[(a b)^2] - Sqrt[(b x)^2 + (a y)^2]]; ...

2

Hint: Graph what $\sin ( 2x / (1+x^2))$, you'll see the only way you can have a line passing through $(0,0)$ and 2 other points of the graph is if the line has slope smaller than 0 and bigger than $-2$ Since $$\frac{2x}{1+x^2} = \sin(-px) \implies 2x \approx -px \quad \text{around zero}$$

2

If we define the ramp function $r$ as $$r(t)= \begin{cases} t, & t\ge 0\\\\ 0, & t<0 \end{cases}$$ then the function $f$ plotted in the post can be represented as $$f(t)=r(t)-r(t-2)$$ Note that if one introduces (i.e., adds) a step function, the resulting plot would exhibit a jump discontinuity. Inasmuch as the plot exhibits no jump, then ...

1

This will be a 2-dimensional graph, since the position vector r has 2 components. Draw an x,y plane. At time t, the x position is $\cos(\omega t)$ and the y position is $\sin(3\omega t)$.

1

Both are same, one is rotated by $45^0$ with respect to the other. $x = (x_1+ y_1)/2, y =(x_1 - y_1)/2$ gets from one form to the other.

1

$$f(x) = \begin{cases}1 & \text{ if } x\leq 0 \\ 0&\text{ if }x>0\end{cases}$$ is one such function. Also, for any $\alpha > 0$, the function $$\frac 1{(x + 1)^\alpha}$$ satisfies your conditions. Furthermore, take any function $f(x)$ which is not identically equal to $0$ and is defined on $(a, \infty)$ for some $a\in\mathbb R$. Also, ...

1

We can implement this as a translation of the line to the origin, a reflection about the line through the origin, and then a translation back in the same direction. In particular: suppose we want to reflect a point $(x_0,y_0)$ across this line. First translate it to the point $(x_1,y_1) = (x_0,y_0 - b)$. Then, reflect $(x_1,y_1)$ across the line $y = ax$ ...

1

a) One possible interpretation is that the student makes an initial deposit of USD 450 into his account, on the same day he gets his paycheck. Out of each successive paycheck he deposits USD 100 into the account. The $r$ is the number of paychecks he has gotten after the initial deposit. b) The slope of the graph is 100 and represents the deposit amount for ...

1

For, discontinuity at $x=0$, let's check both left hand & right hand limits as follows $$LHL=\lim_{x\to 0^{-}}\frac{2x^2}{x-3}$$ $$=\lim_{h\to 0}\frac{2(0-h)^2}{(0-h)-3}$$ $$=\lim_{h\to 0}\frac{2h^2}{-h-3}$$ $$=\lim_{h\to 0}\frac{-2h^2}{h+3}=\frac{-2(0)^2}{0+3}=0$$ Similarly, $$RHL=\lim_{x\to 0^{+}}\frac{2x^2}{x-3}$$ $$=\lim_{h\to ... 1 x^y-y^x=0 has a general solution in terms of the Lambert W function. x^y-y^x=a\neq0, however, has no such general solution. If you wish, you may write {\color{red}y}=\sqrt[\Large x]{x^{\color{red}y}-a} , and then repeatedly iterate the expression with regard to y. 1 "Horizontal asymptotes" is a geometric statement of the limits of the following two limits:$$\lim_{x \to +\infty} \frac{1 - 2^x}{1 + 2^x} \text{ and } \lim_{x \to -\infty}\frac{1 - 2^x}{1 + 2^x}.in which \begin{align*} & \lim_{x \to +\infty} \frac{1 - 2^x}{1 + 2^x} = \lim_{x \to +\infty}\frac{\frac{1}{2^x} - 1}{\frac{1}{2^x} + 1} = \frac{0 - 1}{0 + ... 1 Hint : Two lines are parallel if and only if they have the same slope. If the line is given in the form y=mx+b, the slope is m. 1 Hint The equation defining the surface can be written in the formf(x^2 + y^2, z) = 0,$$and so the surface is symmetric about the z-axis. In particular, it is the surface of revolution generated by the intersection of the surface with the half-plane$$\{(x, y, z) : x \geq 0, y = 0\}, and substitution of the conditions x \geq 0, y = 0 in the ... 1 If the graph is more rounded than a parabola at a point where it touches the x-axis, then its square-root would be rounded there too, having zero gradient. If it is less rounded, then its square-root would have a cusp there with infinite gradient on both sides. If it is equally rounded, then its square-root would have a straight corner there. The problem ... 1 To scale a function f\left(x\right) down horizontally (in the x direction), what you do is replace x with nx, i.e. f\left(nx\right), where it will be scaled down by a factor of n. Vertically, simply scale it down by taking \frac{1}{n}f\left(x\right), where n is the factor by which it will be scaled down. Simply combine these two methods to ... 1 Why not use a definition...? According to Wikipedia Maxima and minima — Definition: If the domain X is a metric space then f is said to have a local (or relative) [...] minimum point at x^∗ if [there exists some ε > 0 such that] f(x^∗) ≤ f(x) for all x in X within distance ε of x^∗ Of course \pi/4 fits the definition criteria: ... 1 Edit: Starting from the initial point x=\frac{\pi}{4}, we see that f(x)=\sin x clearly has a maxima at x=\frac{\pi}{2} & x=\frac{\pi}{4} is in the given domain \left[\frac{\pi}{4}, \frac{7\pi}{4}\right] of the function & it is minimum in the locality of x in \left[\frac{\pi}{4}, \frac{\pi}{2}\right] Hence x=\frac{\pi}{4} is also ... 1 In the first place: Don't flip the plot, because it mixes up the names of the variables. When studying f and f^{-1} at the same time you see the graph of f^{-1} in the original f-plot when you tilt your head 90^\circ sideways. For functions f:\>{\mathbb C}\to{\mathbb C} and their inverses f^{-1} there is an analogous mischievous symmetry; ... 1 Notice that, at the given instant: x=1 cm and \begin{align*} \frac{d^2x}{dt^2}&=\frac{d}{dt}\left(\frac{dx}{dt}\right)\\ &=\frac{d}{dt}\left(\frac{v}{\sqrt{1+8x^2}}\right)\qquad\text{where }v=||\mathbf{v}||\\ &=\frac{\sqrt{1+8x^2}\dfrac{dv}{dt}-\frac{8x}{\sqrt{1+8x^2}}\frac{dx}{dt}v}{1+8x^2} \\ ... 1 As I mentioned in my comment above, you can figure out \dfrac{d^2x}{dt^2} from your equation in |\mathbf{v}|, given that you know that the change of speed is 3cm/s/s. \begin{eqnarray} |\mathbf{v}| &=& (\sqrt{1+8x^2})\dfrac{dx}{dt}\\ \dfrac{d|\mathbf{v}|}{dt} &=& (\sqrt{1+8^2})\dfrac{d^2x}{dt^2} + ... 1 It is very simple to plot graph of a straight line, by finding out the points of intersection of the given line with the coordinate axes as follows \color{red}{\text{Intersection point with the x-axis}}: Setting y=0 in the equation of the line: 6x-3y-18=0 we get6x-0-18=0x=\frac{18}{6}=3$$Hence the point of intersection with the x-axis is ... 1 we now that the domain of the function is:$$ax+b\gt 0\Rightarrow x\gt\frac{-b}{a}\text{so the domain is:}(\frac{-b}{a},+\infty)f'(x)=\frac{a}{ax+b}$$in the domain of the function since we have x\gt\frac{-b}{a}\Rightarrow ax+b>0  the sign of f'(x)=\frac{a}{ax+b} will be dependent to the sign of a so: if a\gt 0\Rightarrow f'(x)\gt 0 and ... 1 I suspect that what you're really trying to do is find a and b that best fit the equation y = b (a-x)^{8/25}, and further that "best fit" is in the least squares sense, i.e. you want to minimize$$ \sum_i (y_i - b(a-x_i)^{8/25})^2 where $(x_i, y_i)$ are your data points. This is a non-linear least squares problem, so it is not easy, but methods are ...

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