# Tag Info

## Hot answers tagged graphing-functions

4

When testing for minimums and maximums, we check critical points, and included in these critical points are endpoints of intervals, in the case of a function defined on such an interval. In your function, $y' = 5 \neq 0$ for all $x$. That leaves us with only the endpoints of the interval on which $y$ is defined as possible candidates for extrema. Since ...

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In this example, every two maxima have a unique distance, but the average distance stays constant: $$f(x)=\cos(g_\epsilon (x)), \qquad g_\epsilon (x)=x+\epsilon \sin(\sqrt{2}\cdot x), \qquad \epsilon=0.4$$ Edit: To understand the idea, first draw the unperturbed inner function $g_0(x)=x$ on a sheet of paper. Whenever this diagonal line crosses the ...

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I explain the "horizontal shift" this way: when we graph $\ y = f(x+h) \ ,$ we are composing $\ f(x) \$ on the function $\ x + h \ .$ This is to say that we are first adding $\ h \$ to $\ x \ ,$ evaluating the function $\ f \$ at $\ x + h \ ,$ and then plotting the result at $\ x \ .$ This has the effect of reading off values of the ...

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I assume you know what the graph of $$y=e^x$$ looks like. The graph of $$y=b^x$$ is similar, only the slope of the tangent line at $x=0$ is $\log b$ instead of 1. The graph of $$y=b^{x-h}$$ is similar to that of $y=b^x$, only pushed $h$ units to the right. The graph of $$y=ab^{x-h}$$ is similar to that of $y=b^{x-h}$, just that the distances to the ...

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You will get a very different graph from the one you have drawn for intensity against distance if you plot intensity against $\cfrac 1{\text{distance}^2}$ What you seem to be trying to do is to use the numbers in your final column on the $x-$ axis, which will then run up to a maximum value of $\cfrac 1{25}=0.04$, so you might choose a scale which goes from ...

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Here's your surface: I generated it with the following Mathematica code. If you can make sense of the code, particularly the ParametricPlot3D command, then you can probably see how the $y=-x$ affects the surface and what your bounds of integration will be after you set up the surface integral as an iterated integral. bounds = ContourPlot3D[{y == 0, y ...

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As in other post, I think you should do it as follows: [> restart; with(plots): f := (a, b, c)-> a*x+b*(y-6)+c*(z-6): d := solve(f(12, -6, 6) = 0, z); w1 := plot3d(d, x = -15 .. 15, y = -15 .. 15, color = green); w2 := pointplot3d({[0, 6, 6], [1, 3, 1], [1, 9, 7]}, axes = boxed, color = black, filled = true); w3 := arrow(<,>(12, -6, ...

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I assume $K>1$. If you only want to know $a, b, c$, you can diagonalize the matrix $$A=\pmatrix{K & 1 & 1\\1 & K & 1\\ 1 & 1 & K}$$ the eigenvalues are $K-1, K-1$ and $K+2$. So $a, b, c$ would be $$\frac{1}{\sqrt{K-1}}, \frac{1}{\sqrt{K-1}}, \frac{1}{\sqrt{K+2}}$$

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A dot plot is just a bar chart that uses dots to represent individual quanta. So if you wanted to plot the number of pets per household, you might have 10 households with 0 pets, 20 with 1 pet, 12 with 2 pets, etc. Over zero, you'd draw 10 dots; over one, you'd draw 20 dots, etc. A scatter plot puts a point representing a single realization of a tuple of ...

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If by $\left[\frac{1}{x}\right]$ you mean "the integer part of $x$", then the plot is correct. Note that if $x < -1$, then $\left|\frac{1}{x}\right| < 1$, so $\left[\frac{1}{x}\right]$ has integer part $0$. Edit: This answer is based on the following convention: http://mathworld.wolfram.com/IntegerPart.html

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Do I understand it correctly that you are looking for a function $f$, where $f(1)=1$ and $f(x)=0$ for all $x \neq 1$, and which is defined for all real $x$? This function is formally written down as $$f(x)= \begin{cases} 1, & x=1 \\ 0, & x\in \mathbb{R}\setminus \{1\}\end{cases}.$$ If you are looking for a function which is only defined at $x=1$, ...

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The data set provided by William has a very low scatter and the points are numerous (500) and well distributed. As a consequence, the method of "regression with integral equation" gives a very accurate result (attachment : points in black, fitted curve in red)

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Google is plotting the graph incorrectly. There is no asymptote. (The function cannot tend to infinity, since it is bounded between $-\pi/2$ and $\pi/2$.) In fact, since $f'(x)=1/(1+x^2)$ for $x<-1$ and for $x>-1$, it follows that $f(x)=\arctan x + C_1$ for $x<-1$ and $f(x)=\arctan x+C_2$ for $x>-1$, but $C_1$ may not be the same as $C_2$. If ...

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Consider the data $$\{0,4,5,5,5,6,6,6,6,7,20\}.$$ The median is $6$, the first quartile is $5$, and the third quartile is $6$. So the IQR is $1$ and it easily follows that $\{0\}$ is a lower outlier and $\{20\}$ is an upper outlier. What you need to take into account is that the box shows you where 50% of the data lies, so if this is particularly narrow, ...

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For 1, the graph $g(x)$ near $0$ is helpful: If a function has a tangent at any point, there must exist the derivative at the point [why?]. So, only if $g'(0)$ exists, then $g(x)$ has a tangent at the origin. The graph shows that such functions really behave oddly near $0$, so its important to deal with them carefully. So, computing $g'(0)$ by ...

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The instantaneous period (what you call the "trough spacing") at $x$ is the reciprocal of the cyclic frequency at $x$. The frequency of $y = \cos( f(x) )$ is defined to be $f'(x)/(2\pi)$, so all you need to do is choose $f(x)$ such that $f'(x)$ is not a constant. A common example would be $$y = \cos( \pi \gamma x^2),$$ such that $f'(x) = 2\pi \gamma x$. ...

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