# Tag Info

0

First off, as Svetoslav comments, the graph is not a parabola since the expression for the function is cubic. You are correct that $f(5) = 0$ and $f(6)=0$. But there is another zero. Dividing $x^3-18x^2 + 107x-210$ by $x-5$ or $x-6$ will yield a quadratic that you can factor to find the third root. That gives one more root of the polynomial, or ...

4

max{x,3x} + min{x,3x} = 4x max{x,3x} - min{x,3x} = |2x| ---------------------------- 2max{x,3x} = 4x + 2|x| max{x,3x} = 2x + |x| As a bonus, min{x,3x} = 2x - |x|

9

I will probably put more emphasis on those transformation properties of the $\max(\cdot)$ function which is reasonably "obvious" and easy to remember/use. For example, $\max(a+b,a+c) = a + \max(b,c)$. $\max(ab,ac) = a \max(b,c)$ when $a > 0$. $\max(a,-a) = |a|$. This may help a student to get familiar with such "tools" for attacking similar problems. ...

0

You need to do some analysis.I recommend take the following points. From basic function f and g: See when are f and g zero Find the max and min value of the f and g (example : for sin(x) +1 and -1) Plot the envelopes of the shape in the enlarged size to get an idea of the graph. OR Calculate the roots of function(sum) if possible. Analyse the value at ...

1

Since $h(x)=(f+g)(x):=f(x)+g(x)$ for every $x$ in the domain, the graph is the one that you obtain summing the two functions pointwise. That is, at $x=x_0$ will correspond the point $h(x_0)=f(x_0)+g(x_0)$. Edited after seeing the comment about discontinuities: if one of the functions $f$ and $g$ has a discontinuity, remember that the domain of $f+g$ is ...

0

You can think about the graph of h(x) pointwise, adding the heights of the two graphs f(x) and g(x) at each point x. For example, if f(1) = 2 and g(2) = 3. h(1) = f(1) + g(1) = 2 + 3 = 5.

25

You can use the fact that, $\max\{x,y\}=\frac{x+y+|x-y|}{2}$. Therefore, $\max\{x,3x\}=\frac{x+3x+|x-3x|}{2} = \frac{4x+|-2x|}{2} =2x+|x|$.

1

Note that $f(-x) = -ax + b|x|+ c$, thus breaking into symmetrical and anti symmetrical combinations one obtains $$f(x) - f(-x) = 2ax\\ f(x) + f(-x) = 2b|x| + 2c$$ Thus $$a = \frac{f(1) - f(-1)}{2} = \frac{3 + 1}{2} = 2\\ c = f(0) = 0\\ b = \frac{f(1) + f(-1)}{2} - c = \frac{3 - 1}{2} = 1$$ Second variation of the same idea: $$f(-x) = \max(-x, -3x) = ... 2 I like the question! It is not immediately obvious to me, however, that it suffices to match your form to the given function at three points. Perhaps that is the case, but if so it requires a separate argument. Alternatively, having discovered the final form you could verify it directly. I would address the original problem this way: Your function, f, ... 1 You're on the right track, but you're having trouble with the endpoints. The points (-3, -2) and (4, -2) are on the graph in a), which means that your domain should actually be x \in [-3, 4] in interval notation to indicate that x = -3 and x = 4 are included. This would be reflected in the set-builder notation by using inequalities with "or equal ... 0 If you move down one and to the left three, then \Delta y = -1 and \Delta x = -3. Therefore,$$\frac{\Delta y}{\Delta x} = \frac{-1}{-3} = \frac{1}{3} \neq -\frac{1}{3} = m$$We cannot obtain a negative slope if both \Delta y and \Delta x have the same sign. To obtain a negative slope, \Delta y and \Delta x must have opposite signs. ... 1 Very easy to construct, \sin goes from -1 to +1 so you just shift and stretch:$$f(x)=\text{min}+\frac12(\text{max}-\text{min})(1+\sin (kx+\phi))$$or, if you prefer$$f(x)=\text{average} + \text{amplitude}\cdot \sin (kx+\phi)$$where average=(max+min)/2 and amplitude = (max-min)/2. 2 Desmos,Wolfram Alpha,gnuplot ,Geogebra,Microsoft Mathematics,Matlab,FooPlot,GraphSketch,MAFA Function Plotter. 1 I assume that f(60)=f(0) (It doesn't really look so in your graph). h =  highest, l = lowest, m =  highest point's x-value, p =  period (60 in your graph). Your function is:$$f(x) = \frac{h-l}{2}cos \left( \frac{2\pi}{p}(x-m) \right) + \frac{h+l}{2}$$For h=0.9, l=0.6, m=0, p=60 it looks like this: Of course, there are many many other ... 1 Regard your data points as simple support points through which an elastic thin beam has to pass. The beam deforms to a curve that minimizes the energy required for bending it. You might be interested in that amount of energy. It can be used to measure the beam curve's deviation from a straight line. The simplest theory for thin elastic beams is Bernoulli's ... 2 A simple approach would be to calculate the slope of the interpolating line between every successive pair of points, then take the standard deviation of the slopes. 1 To find the shape of graph is also to plot it. You have got the basic approach. I repeat the essential steps: Find as many pairs of points$$ (x, f(x)) ... \tag {1}$$as possible. Plot these points on x- and y- axis parallel lines respectively. This fixes the domain. Solve  f'(x) = 0  and mark x values with short vertical lines on x-axis. Calculate ... 3 That looks pretty good, though I would add three more items in your "Original function" section. produces the y-intercept of the graph by finding f(0) produces the x-intercepts of the graph by solving f(x)=0 gives the domain of the graph by finding where f(x) is undefined That then covers what I teach in my calculus class. There are always more ... 2 To prove that of dartboard rule, just take two points not in the lines, in adjacent regions. The sign of each parenthesis is the same for two, except for a single parenthesis. This means that one point is in the region and the other not. The second rule is obvious since the repeated parenthesis does not change the sign of the product. If you want to use ... 0 Nothing wrong with the Maple output. The coefficients you provide for the quadratic are miniscule, so for practical purposes, on your indicated range, the function is very close to:$$PN(Q)=-3.3\cdot 10^{-4}Q+45$$which is a line of very small slope. To see the quadratic nature of the original, versus the linear approximation, try increasing the desired ... 0 The quadratic term is much smaller than the linear term. The closer you look at any curve, by narrowing the range of x, the straighter it will look. Instead, try larger ranges of x, you will start to see the curve. For the very narrow range, Maple doesn't store numbers between 44.99999999999997 and 44.99999999999998. This is called round-off error. ... 1 For sure, graphing the function is a good idea since it will show where are the roots ... if they exist ! Let us consider for more generality function$$f(x)=10^{x-a}-x$$Its derivative is$$f'(x)=10^{x-a}\,\log (10) -1$$the point where it cancels is given by$$x_*=a-\frac{\log (\log (10))}{\log (10)}$$At this point$$f(x_*)=-a+\frac{\log (\log ...

0

For $x>0$, $$|x|=x,\ |x|'=1,\ |x|''=0.$$ For $x<0$, $$|x|=-x,\ |x|'=-1,\ |x|''=0.$$ For $x=0$, neither $|x|'$ nor $|x|''$ are defined. You can summarize with $$\text{for }x\ne0,\ |x|'=\text{sign}(x),\ |x|''=0.$$ Let us solve the exercise: For $x^2-9\ne0$, $$f'(x)=2x\text{ sign}(x^2-9)$$ has a zero at $x=0$, and changes sign at $x=\pm3$. ...

1

We have $$f(x)= \begin{cases} x2-9,&\text{for}\,\,|x|\ge 3\\\\ 9-x^2,&\text{for}\,\,|x|\le 3& \end{cases}$$ Thus, the derivative $f'$ is $$f'(x)= \begin{cases} 2x,&\text{for}\,\,|x|\ge 3\\\\ -2x,\text{for}\,\,|x|\le 3& \end{cases}$$ Both the derivative from the left and right are zero at $x=0$ and so this is a local extremum. To ...

-1

No. The simplest example I can think of? $$f(x) = x$$ Then $G(x) = \frac{x^2}{2} + C$, which is not guaranteed to be always $0$.

0

As I commented, another counterexample is given by $f(x)=e^x-e^{-x}$, whose antiderivative is $F(x)=e^x+e^{-x}+C$. Note that $f(0)=e^0-e^0=0$, and $F(0)=e^0+e^0+C$, so $F(0)=0\iff C=-2$. Variations of the coefficients of the exponentials or the exponents would force you to choose a different value for $C$: $f(x)=2e^x-e^{-x}$, with antiderivative ...

2

No, this is not true in general. Let $F(x)$ be a anti-derivative of $f(x)$ then $$\int f(x) \, \mathrm{d}x = F(x) + k$$ now, even if $F(0) = 0$ there's still the $+ k$ constant to contend with. An example would be $f(x) = x\exp x$ then we have $$F(x) = \int x\exp x \, \mathrm{d}x = e^x (x-1) + k$$ Now, even if $k = 0$, then $F(0) = -1$ whilst $f(0) = 0$.

2

No. Take $f(x) = xe^{-x^2}$. Clearly, $f(0) = 0$. The integral function of $f(x)$ is: $$G(x) = \int_{-\infty}^{x} te^{-t^2} dt = \left.-\frac{1}{2}e^{-t^2}\right|_{t=-\infty}^{t=x} = -\frac{1}{2}e^{-x^2}$$ In this case, $G(0) = -\frac{1}{2} \neq 0.$

1

I assume by "hypercube" you mean a $d$-dimensional cube, that is, the graph with $2^d$ vertices indexed by length-$d$ bit strings, and edges joining two vertices precisely when their bit strings differ in exactly one bit. If that's the case, then finding shortest paths in this graph is very simple: if two vertices' bit strings differ in $k$ spots, then just ...

0

It seems what you want is left_p+image_width/2 to be the center point for both the image and the text. If this is so, the left edge of the text starts at left_p+image_width/2-text_width/2.

0

Note that if left_p=200 then your left margin is 200 and text is 300 so if you want to be on the page (which is 500 wide), right_p must be 0. If you use that, then your formula gives $$(500-(200+0)-300)/2+200 = 200$$ as expected (i.e. start the text directly on the margin)

1

I'll do it for the $z$ axis to give you the idea; it's also possible for the other axes, but a bit more complicated. The $z$ axis case is simpler both because, as noted in the comments, the integration over $\phi$ yields $2\pi$, and because in this case we have simply $\vec r\cdot\hat k=r\cos\theta$. Thus after integrating over $\phi$, we're left with $$... 1 You can just do y=10 \sin (x) + \sin (10x) You can play with the parameters to make it like you want. Here is a link to Alpha 2 Here is a color plot (aka. scalar field) of$$x^2+y^2+{y\over x}$$White is around zero, gray is positive and red is negative. The thick line is 0 and the thin line is 1, your original curve. It's easy to see that by taking the inverse, the thin line will be preserved, but you will get a discontinuity at the thick line, because one side will go to ... 1 rotationally symmetric about the origin; your version does not allow x=0 but the curve becomes a smooth variety if (0,0) is included The smooth implicit function is$$ x^3 + x y^2 + y - x = 0 $$with gradient$$ \left\langle 3 x^2 + y^2 - 1, 2xy + 1 \right\rangle $$For large |y|, solving the quadratic formula in y shows xy \approx -1. Indeed, ... 1 A simple code is x = zeros(1,2000); % Initialize x-vector v = rand(1,2000); % Create noise v for n = 3:2000 % Main loop x(n) = 0.6530*x(n-1) - 0.7001*x(n-2)+ v(n); end plot(x) Although, you have to adjust exactly what you mean by white noise. I just added some uniform noise between 0 and 1. 0 Okay, i got it. hyp:=x^2+(y/2)^2-z^2=5: vlak1:=z+1/2*x+3=0: vlak2:=z+1/2*x-3=0: I defined them explicit this way: f(x,y):=solve(vlak1,z); g(x,y):=solve(vlak2,z); h(x,y):=solve(hyp,z)[1]; So now it was possible to solve them, to obtain the intersection solve(g(x,y)=h(x,y),y); 0 The derivative of your quartic is$$4(45x^3 - 108x^2 + 73x - 14) = 4(3x-2)(3x-1)(5x-7)$$So the stationary points are at x = \frac{1}{3}, \frac{2}{3}, \frac{7}{5}, does this help you draw the graph? In addition to this, we have the second derivative as$$540x^2-864x + 292$$that will allow you to check whether your stationary points are minimums or ... 1 Yes, the dark red curve is at least approximately exponential. You can check this by looking at how far along the horizontal you have to go to double the vertical; it's the same distance each time. More information about what you're analyzing would help guide further suggestions. 2 What you're looking for is called "exponential regression" which is a process which fits your data to the graph of an exponential function, which does in fact, look like what you have as a picture. Furthermore, the process of exponential regression gives a number called a correlation coefficient which is between 0 and 1 which tells you just how well the data ... 2$$x - {(\cos(x) + i\sin(x))^{ix}} = 0\Longleftrightarrowx - {(e^{ix})^{ix}} = 0\Longleftrightarrowx - e^{ixix} = 0\Longleftrightarrowx - e^{(ix)^2} = 0\Longleftrightarrowx - e^{i^2x^2} = 0\Longleftrightarrowx - e^{-x^2} = 0\Longleftrightarrow$$(because x is real we can write)$$e^{x^2}x= 1\Longleftrightarrowe^{x^2}= ...

1

We have $f(x)=x-(\cos x+i\sin x)^{ix}=x-e^{-x^2}$. Plot this function and find that $f(x)=0$ when $x\approx. 0.65291862487151$.

3

Hint: If $\hspace{0.25cm}\displaystyle\prod_{k=1}^n a_k = 0\hspace{0.25cm}$ then for some $K$, $a_K = 0$. Now let $a_k = (x^2+y^2 - k^2)$... what can you conclude?

4

Hint: $$\frac{2x^2+20}{x^2+5}=2+\frac{10}{x^2+5}$$ and $$0<\frac{10}{x^2+5}\le \frac{10}{0+5}= 2$$

1

It seems that the lines between 1 and 3 are 1.5 and 2. Including 1.5 gives a more equal spacing to the grid, since the largest gap is between 2 and 1. Also, observe that the distance between 1.5 and 2 is the same as the distance between 3 and 4 (since they have the same ration and we have a logarithmic scale). There is also a horizontal line between 100 and ...

1

Edit: the last answer didn't work for it was symmetrical $$f(x) = \begin{cases}g(x)^2 & \text{ if } g(x)>0\\ \frac{g(x)^2}{9}& \text{ if } g(x) \leq 0\end{cases}$$ where $g(x) =2( x-[x])-1,5$

6

When I was your age, I discovered a way to do this, although it's a bit artificial. Suppose that you want a function which takes on the values of a function $f(x)$ when $x<a$ and $g(x)$ when $x>a$. Then $h(x)$, the function we are looking for, can be given by ...

5

How about this? $$y=\frac{x-|x|}{2}+\sin\left(\frac{x+|x|}{2}\right)$$

3

This is a fairly technical question, and therefore will admit a technical answer. If $x,y$ are restricted to be real (or rational) numbers, then $\sqrt{-x}$ is undefined for $x>0$. Once we have an undefined quantity, we cannot proceed further, even multiplying it by zero. Hence with this restriction Desmos is correct. However, if $x,y$ are allowed to ...

2

Either order is fine. The issue with Desmos is that it is restricting the domain (artificially) to non-positive numbers because $\sqrt{-x}$ is complex for other numbers. In reality, having complex numbers is fine, and indeed, the next step, which is to multiply by zero makes the number real again. Wolfram alpha's plot is more correct, while Desmos' plot is ...

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