# Tag Info

## New answers tagged graphing-functions

0

Consider $f(x)=max(x, x^2)$.It is continuous but not differentiable at x=1.

1

Consider the Thomae function; $f:[0,1]\to\mathbb R$ where $$f(x)=\begin{cases} \frac1q,& x\in\mathbb Q,\ x=\frac pq, \ p,q\in\mathbb Z, \ p\not\lvert q, \ q>0\\ 0,& x\notin\mathbb Q. \end{cases}$$ Then $f$ is continuous on $[0,1]\setminus\mathbb Q$ but not continuous on $[0,1]\cap\mathbb Q$. See this MSE question for a proof: Proving Thomae's ...

0

To reduce the “multiple” sine wave to a “single” one, multiply with a boolean value such as $x^2<\pi^2$, for instance. Just type sin(x) (x^2 < pi^2) into the input bar, and then hit enter. To transform a horizontal wave into a vertical one, use Rotate[...].

3

It's not that hard. You should just use the summation formula for sines: $$\sin (x + y) = \sin (x)\cos (y) + \cos (x)\sin (y)$$ This is how it works \eqalign{ & \sin (x) + \cos (x) = \sqrt 2 \left( {{1 \over {\sqrt 2 }}\cos (x) + {1 \over {\sqrt 2 }}\sin (x)} \right) \cr & ... 0 A vertical wave is not a function of x. You might try x = \cos y, which is just a normal wave, except vertically. 1 Here I just say how to do it with Maple for other softwares maybe one other will tell you, also for Mathematica you can ask the Mathematica section of stackexchange. As an example I'll compute the NullSpace of the following matrix;\begin{bmatrix}1 & 1 & 1\\ 0 & 2 & -1\\ 2 & 0 & 3\end{bmatrix}$$with(LinearAlgebra): M := ... 1 The graph is correct, although the currently-attached statement f'(x) = 3 at x \le 1 is incorrect. It would be true to say that f'(x) = 3x^2 at x < 1, for which the limit of f'(x) approaches 3 as x approaches 1 from the left. 1 Here is a graph generated with MATHEMATICA Well, I have had installed the program in a PC. There are some sites where you can graph functions, but I'm not sure if you can get graphs with the same quality: Wolfram Alpha Desmos Geogebra I hope they will be useful to you. 1 Yes. To the degree of detail. For x<= 1 the graph will resemble the graph of y = x^3, and for x > 1 the graph will resemble the graph y = x. As at x = 1, both x^3 and x equal 1, the graph with "connect". 1 Pick two points on the graph with the same x-coordinate, multiply their y values together and you'll have the y-coordinate of the product function. Do this for every point and you'll have the new graph. Obviously it's impractical to actually do that for every point so your final graph will be an estimate. For example, if the two y-values are large than ... 1 Expand in terms of complex exponentials.$$\sin^4 x + \cos^4 x = \left( \frac{e^{ix} - e^{-ix}}{2i} \right)^4 + \left( \frac{e^{ix} + e^{-ix}}{2} \right)^4$$Notice that i^4 = +1, so we get$$\sin^4 x + \cos^4 x = \frac{1}{16} \left( 2e^{4ix} + 2 e^{-4ix} + 12 \right)$$where we use the relation (a+b)^4 = a^4 + 4 a^3 b + 6 a^2 b^2 + 4 ab^3 + b^4. The ... 1 If you want to express in functions of higher frequencies like this$$\sum_{k=0}^N \sin(kx) + \cos(kx)$$Then you can use the Fourier transform together with convolution theorem. This will work out for any sum of powers of cos and sin, even \sin^{666}(x). 3 Note that a^2 + b^2 = (a+b)^2 - 2ab$$(\sin^2 x)^2 + (\cos^2 x)^2 = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x\cos^2 x =(\sin^2 x + \cos^2 x)^2 - 2(\sin x\cos x)^2 = \\ 1 -\frac{ \sin^2 2x}{2}$$Note the following results:$$ \sin^2 x + \cos^2 x = 1 \sin x \cos x = \frac{\sin 2x}{2}6 \begin{align} \sin^4 x +\cos^4 x&=\sin^4 x +2\sin^2x\cos^2 x+\cos^4 x - 2\sin^2x\cos^2 x\\ &=(\sin^2x+\cos^2 x)^2-2\sin^2x\cos^2 x\\ &=1^2-\frac{1}{2}(2\sin x\cos x)^2\\ &=1-\frac{1}{2}\sin^2 (2x)\\ &=1-\frac{1}{2}\left(\frac{1-\cos 4x}{2}\right)\\ &=\frac{3}{4}+\frac{1}{4}\cos 4x \end{align} 2 Let\displaystyle y=\sin^4 x+\cos^4 x = \left(\sin^2 x+\cos^2 x\right)-2\sin^2 x\cdot \cos^2 x = 1-\frac{1}{2}\left(2\sin x\cdot \cos x\right)^2$$Now Using$$\bullet\; \sin 2A = 2\sin A\cos A$$So we get$$\displaystyle y=1-\frac{1}{2}\sin^2 2x$$1 It might be easier to understand if we write the scaling and translation in two steps. To that end, suppose the f(n) is a given function. Then for two numbers a and b, we define g(n) as the function given by$$g(n)=f(an)$$and h(n) as the function given by$$\begin{align} h(n)&=g(n+b/a)\\\\ &=f\left(a(n+b/a)\right) \tag 1\\\\ ...

2

If you write $x[4n+1] = x\left[4\left(n+\frac{1}{4}\right)\right]$, you would see more clearly which is the correct sequence of operations. To better understand this, let $y_1[n] = \mathcal{T}_1\{x[n]\} = x[4n]$ be the output of a system that downsamples its input by a factor of 4, and $y_2[n] = \mathcal{T}_2\{x[n]\}=x[n+1]$ be the output of a system that ...

0

If you think of the both sides of that equation as functions of $x$, you're looking for a value $x$ with $$f(x) = g(x)$$ where $f$ is a sum and $g$ is a constant function. The sum happens to also be $$F'(x)$$ where \begin{align} F(x) &= \sum_{n=1}^\infty x^{n+1}\\ &= \sum_{n=2}^\infty x^{n}\\ &= -1-x + \sum_{n=0}^\infty x^{n}\\ &= ...

2

If the discriminant of a cubic is $\Delta$ and if $\Delta \lt 0$ the equation has one real root and two nonreal complex roots. There is no difficulty with the example you have used. Read further in https://en.wikipedia.org/wiki/Discriminant#Cubic.

0

The statements in the answer you already have are all correct. Here is a more explicit way to solve the necessary equations, assuming the coordinates of $P$ are $(p_x,p_y)$ and both coordinates are positive. The apex of the desired function lies on the intersection of the line $y=\frac{p_y}{p_x} x$ and the circle $x^2 + y^2 = 1$, so it is the point $$A = ... 2 Let t the triple root and consider the coefficients r,s of the quadratic factor so one has the polynomial (x-t)^3(x^2+rx+s). The equality of corresponding coefficients gives$$r-3t=0s-3tr+3t^2=0-3ts-t^3+3t^r=-10a^33t^2s-t^3r=b^4-t^3s=c^5$$Solving easily this system we have a parameterization of a,b,c in function of the triple ... 3 Note: Although your answers contain some true aspects, your reasoning is not completely satisfying. Most of them is due to the incomplete specification of the function f. It's absolutely crucial to be fully aware about the domain and codomain of a function. Otherwise no precise analysis of a problem is possible. You may want to check the example in the ... 5 Let$$f(x) = x^5 - 10a^3x^2 + b^4x + c^5 = (x - m)^3g(x)$$where m is the repeated root, and g(x) is some second-order polynomial. Then, differentiating and substituting x = m,$$5x^4 - 20a^3x + b^4 = (x - m)^3g'(x) + 3(x - m)^2g(x)5m^4 - 20a^3m + b^4 = 0$$Differentiating another time, and similarly substituting x = m,$$20x^3 - 20a^3 = ...

2

The following statements certainly have standard definitions: 1) "$f$ is continuous at $x$" ; 2) "$f$ is continuous". For the function $f(x) = x^{-1/3}$, it's certainly true that "$f$ is continuous" and that "$f$ is not continuous at $0$". However, what about the statement, "$f$ is continuous on $[a,b]$"? Perhaps there is some confusion or disagreement ...

1

You're right: It only makes sense to say that a function $f$ is continuous in $x$ if $x$ is in the domain of $f$, i.e. if $f$ is defined in $x$. Let me add that for other examples it could be possible that $f$ can be extended to a continuous function $$\tilde{f}(x) = \begin{cases} f(x), &x\neq 0,\\ c, & x=0,\end{cases}$$ with some value ...

0

If you have an n-dimensional function, you can observe the behavior of two dimensions at a time using contour lines (see https://en.wikipedia.org/wiki/Contour_line). By looking at several two-dimensional slices (aka contour graphs or contour plots) of your n-dimensional space, you may be able to integrate in your mind what the n-dimensional function really ...

3

In modern mathematics, the definition of continuity discards points that fall outside the domain of definition. In other words, continuity must be checked only at point of the domain. Since your function is defined on $\mathbb{R} \setminus \{0\}$, it is continuous at every point of its domain of definition.

1

The function $\dfrac1x$ is the well-known equilateral hyperbola, that has an horizontal and a vertical asymptote. $\dfrac1{x+a}$ is the same, horizontally shifted to the left by $a$. $-\dfrac1x$ is the same, mirrored around the horizontal axis. $\dfrac2x$ is the same, scaled by two vertically. You can draw an approximate sketch by overlapping three ...

0

Yes, Google and Wolfram are awesome, but just so you know: when you see terms like $\frac{1}{x+1}$ you have to think to yourself that $x\neq-1$ or else you will be dividing by zero. So in this case, there are three values of $x$ where $f(x)$ "blows up," so to speak.

1

You can google it: See here or you can use wolfram alpha.

-3

we say $x^{-1/3}$ is only defined for $x>0$. I would say (1) if is continuous for $x>0$ (in $x=0$ only right continuous) (2) f is not bounded for $x>0$ and we have $\lim_{x\to 0+}x^{-1/3}=+\infty$. And what is $A$?

1

HINT: Notice when a given function say $y=f(x)$ is vertically shrunk (scaled down/scaled up) by a factor $k$ its transformed equation is given as $$y_1=kf(x)$$ Now, the function $y_1=kf(x)$ is reflected about the x-axis, the transformed equation is given as $$y_2=-kf(x)$$

1

You plot number of locations vs ratings to see the normal distribution. Let the X-axis have ratings chosen at certain resolution, say, 0.5. The Y-axis has number of locations that have the rating fall in the window you have chosen. eg: between 3.5 and 4 you have 3 locations. Since the data is too small dont expect to see a perfect bell curve.

0

In Maple, Pi, the well-known constant, is spelled Pi, not pi.

1

Since this is a homework, and the solution would take time to develop, I post this text as a Hint: If you want to create a picture composed of several equations, the general idea is to divide the picture to parts where each part could be represented with a curve in a given range. Better yet to begin with symmetrical parts and then figure out the curve(s) ...

1

Assuming that you hold the origin fixed while compressing, it does. If you compress first, you’ll get $x[4n+1]$. If you shift first, you’ll get $x[4(n+1)]=x[4n+4]$.

0

The graph of $f$ is all the points $(x, y)$ where $y = f(x)$. So to find the value of $f(-1)$, look for $-1$ on the $x$-axis. Follow it straight up (or down for some functions, but not for this one) until you arrive at the graph of $f$. Then head across horizontally until you get to the $y$-axis to find the value of $y = f(-1)$. That tells you what $f(-1)$ ...

0

Thank you for taking the time to address my question. My goal is to have the function rewritten with respect to the polar system so that r values are mapped to theta values, but to shift the curve as if the Pole were at {px, py} instead of {0, 0}. Although I suppose I could have a process where the function were rewitten as if the Pole were at {r_p, ...

0

An attempt to solve the problem for any number of points on the number line while always being in 2-D space: Label the points $P_1,P_2,...,P_n$ progressively from left to right; Let $P_{i,j,...}$ represent the mean of points $P_i, P_j,...$ Find $P_{1,2}$ Find $P_{1,2,3}$ by dividing the line segment joining $P_{1,2},P_3$ in the ratio $1:2$ (Note ...

2

The centroid of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is the point whose cartesian components are the means of the individual $x$-coordinates and $y$-coordinates: $$(x_C, y_C) = \frac{1}{3} (x_1 + x_2 + x_3, y_1 + y_2 + y_3)$$ The centroid of a triangle can also be easily constructed by bisecting each side of the triangle and then ...

1

You could separate the variables: $$x^2 - \ln x = y^2 + \ln y.$$ It's easy then to show that $y\to\infty$ as $x\to\infty$. Take the ratio of the two sides: $$\frac{y^2 + \ln y}{x^2 - \ln x} = 1.$$ For large positive $x$, you have large positive $y$ and $y^2/x^2 \approx 1$, so $y/x \approx 1$. You can even write \begin{align} \frac{y^2}{x^2} & = ...

1

You can establish that $$\frac{dy}{dx}=\frac{2x-\frac 1x}{2y+\frac 1y}\rightarrow 1$$ as $x,y\rightarrow +\infty$ Hence for large $x,y>0$, the graph is approximated by the line $y=x$

1

Your function that maps $x$ to $y$ is $$f(x)=2x+x^2$$ Such that $f(1)=3$, $f(2)=8$, $f(3)=15$ and $f(4)=24$. The nature of your graph is quadratic as $$f(x)=x(2+x)$$ it goes through the origin $(0,0)$ and intersects the axis at $x=-2$ and $x=0$ it has a minimum point at $(-1,1)$. Hope this helps.

3

Symmetry in line $y=-x$ maps point $(x,y)$ to $(-y,-x)$ (and vice versa), so you indeed can simply replace $x$ by $-y$, $y$ by $-x$ and check whether you get the same equality.

0

Booklet plot is a plot of $2 (\sin(x /2 - \pi/4) - 1),\, -2 \pi <x <2 \pi.$

0

We rewrite the equation as $e^{\cot^2\theta}=2\cos^22\theta-\sin^2\theta+4\sin\theta-4$. The equation has a period of 2π. L.H.S.≥1, then R.H.S.≥1, i.e. $2\cos^22\theta-\sin^2\theta+4\sin\theta-4≥1$, then we have $\cos4\theta≥(sin\theta-2)^2$. Since $\cos4\theta$∈[-1,1] and $(sin\theta-2)^2$∈[1,9] for θ∈[0,2π], θ must be π/2. Therefore we can see that the ...

1

It's easily checkable that $n\pi$ is not a solution. LHS and RHS have periods of $\pi,2\pi$ respectively so we only need to worry about the interval $(0,2\pi)$. If $\xi$ is a solution of the equation in $(0,2\pi)$, $\xi+2n\pi$ is also a solution in $(2n\pi,2(n+1)\pi)$. So at least one solution in $(0,2\pi)$ guarantees at least $5$ solutions in $[0,10\pi]$ ...

1

Let $$y=\sum_{x=0}^{100}\left[(100-x)(x^2-x)\right] = \sum^{100}_{x=0}\left\{101x^2-x^3-100x\right\}$$ So we get $$y = 101\sum^{100}_{x=0}x^2-\sum^{100}_{x=0}x^3-100\sum^{100}_{x=0}x$$ Now Using $\displaystyle \bullet \sum^{n}_{x=1}x = \frac{n(n+1)}{2}$ Using $\displaystyle \bullet \sum^{n}_{x=1}x^2 = \frac{n(n+1)(2n+1)}{6}$ Using $\displaystyle ... 2 You already know that there is a bijection$(c,d) \to (-1,1)$and it thus suffices to find a bijection$f \colon (-1,1) \to \mathbb R$. The following will do: For all$n \in \mathbb N_0$we let$f \restriction_{[1- 2^{-n},1-2^{-n-1}]}$be the linear function from$(1- 2^{-n} ; 2^n)$to$(1-2^{-n-1} ; 2^{n+1})$and likewise$f \restriction_{[-1+ ...

3

Are you allowed to use fractions? In that case, you could use $$\frac{1}{c-x} + \frac{1}{d-x}$$ Of course, you would have to do some work to prove that it is monotonous on $(c, d)$, but that shouldn't be too hard (you are allowed to use differentiation, are you not?)

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