# Tag Info

26

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

11

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. ...

6

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|

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

4

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

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?

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 ...

3

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, $f$, ...

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 ...

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, $G(0) = -\frac{1}{2} \neq 0.$

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 ...

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

Desmos,Wolfram Alpha,gnuplot ,Geogebra,Microsoft Mathematics,Matlab,FooPlot,GraphSketch,MAFA Function Plotter.

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 ...

2

$$x - {(\cos(x) + i\sin(x))^{ix}} = 0\Longleftrightarrow$$ $$x - {(e^{ix})^{ix}} = 0\Longleftrightarrow$$ $$x - e^{ixix} = 0\Longleftrightarrow$$ $$x - e^{(ix)^2} = 0\Longleftrightarrow$$ $$x - e^{i^2x^2} = 0\Longleftrightarrow$$ $$x - e^{-x^2} = 0\Longleftrightarrow$$ (because $x$ is real we can write) $$e^{x^2}x= 1\Longleftrightarrow$$ $$e^{x^2}= ... 2 A simple approach would be to calculate the slope of the interpolating line between every successive pair of points, then take the standard deviation of the slopes. 1 Regard your data points as simple support points through which an elastic thin beam has to pass. The beam deforms to a curve that minimizes the energy required for bending it. You might be interested in that amount of energy. It can be used to measure the beam curve's deviation from a straight line. The simplest theory for thin elastic beams is Bernoulli's ... 1 You're on the right track, but you're having trouble with the endpoints. The points (-3, -2) and (4, -2) are on the graph in a), which means that your domain should actually be x \in [-3, 4] in interval notation to indicate that x = -3 and x = 4 are included. This would be reflected in the set-builder notation by using inequalities with "or equal ... 1 We have f(x)=x-(\cos x+i\sin x)^{ix}=x-e^{-x^2}. Plot this function and find that f(x)=0 when x\approx. 0.65291862487151. 1 rotationally symmetric about the origin; your version does not allow x=0 but the curve becomes a smooth variety if (0,0) is included The smooth implicit function is$$ x^3 + x y^2 + y - x = 0 $$with gradient$$ \left\langle 3 x^2 + y^2 - 1, 2xy + 1 \right\rangle $$For large |y|, solving the quadratic formula in y shows xy \approx -1. Indeed, ... 1 First off, as Svetoslav comments, the graph is not a parabola since the expression for the function is cubic. You are correct that f(5) = 0 and f(6)=0. But there is another zero. Dividing x^3-18x^2 + 107x-210 by x-5 or x-6 will yield a quadratic that you can factor to find the third root. That gives one more root of the polynomial, or ... 1 You can just do y=10 \sin (x) + \sin (10x) You can play with the parameters to make it like you want. Here is a link to Alpha 1 To find the shape of graph is also to plot it. You have got the basic approach. I repeat the essential steps: Find as many pairs of points$$ (x, f(x)) ... \tag {1}$$as possible. Plot these points on x- and y- axis parallel lines respectively. This fixes the domain. Solve  f'(x) = 0  and mark x values with short vertical lines on x-axis. Calculate ... 1 Edit: the last answer didn't work for it was symmetrical$$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}$$where g(x) =2( x-[x])-1,5 1 Very easy to construct, \sin goes from -1 to +1 so you just shift and stretch:$$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)$$where average=(max+min)/2 and amplitude = (max-min)/2. 1 Yes, the dark red curve is at least approximately exponential. You can check this by looking at how far along the horizontal you have to go to double the vertical; it's the same distance each time. More information about what you're analyzing would help guide further suggestions. 1 I assume that f(60)=f(0) (It doesn't really look so in your graph). h =  highest, l = lowest, m =  highest point's x-value, p =  period (60 in your graph). Your function is:$$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 ...

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