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There is a pretty natural relationship with physics, specifically with the notion of changing reference frames. If you add a symmetry, say the length of the lines, then Einstein Relativity is all about this. Step into a valid reference frame and take the world-line of a rigid body, call this the receiving function. The world line of any other particle in ...

2

Let's call $f(x)=0.04x^{1.7}$. The function is one-to-one and therefore has an inverse. To find the equation of the inverse, solve the equation $$x=0.04y^{1.7}$$ for $y$. I will leave that part up to you ☺

2

Let us go with $y = \alpha \log (x+\beta)$. Let us say that you want the axis intersections to be $(a, 0)$ and $(0, b)$. Then: $0 = \alpha \log(a + \beta) \Rightarrow a + \beta = 1 \Rightarrow \beta = 1-a$. Similarly: $b = \alpha \log \beta \Rightarrow \alpha = \frac{b}{\log (1-a)}$. So, for your logarithmic function to intersect the axes at $(a,0)$ ...

2

What you wrote according to order of operations is $-1^x=-1$. You meant $(-1)^x$ probably. See: http://m.wolframalpha.com/input/?i=%28-1%29%5Ex&x=0&y=0

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Here's a general answer: rotation of the graph of a function $f$ around a point $(a,b)$ is but the symmetry of the graph w.r.t. this point, hence it is $$y=g(x)\stackrel{\text{def}}{=}2b-f(2a-x).$$ To see this, denote $x,y=f(x)$ a point on the original graph, and $(X,Y=g(X)$ athe symmetric point. By definition, we have $X=2a-x$, $Y=2b-y$ (writing the ...

2

If the formula for a function can be expressed as a ratio of two polynomials, then the function can change from being positive to negative or vice versa only at a vertical asymptote or at a place where it crosses the $x$-axis. In your example, the asymptotes occur at $x = 1$ and $x = 3$. The zeros occur at $x = 0$ and $x = -2$ (look at the numerator). ...

2

If you want to plot the data below in a linear fashion, you could write $$\frac{C_v} T=A + B\, T^2$$ So define $y=\frac{C_v} T$ and $x=T^2$ to have $y=A+Bx$. But you must take care that the results would not be the same fitting the first and second models (except in the very ideal case where the data would be perfectly exact). For the second model, this ...

2

The question you ask is a standard linear regression problem. Your function $C_v=AT+BT^3$ is indeed not linear in $T$, but it is linear in $A$ and $B$, which is all that matters. A transformation that maps your data to a straight line exists, but you will not find it without knowing $A$ and $B$, and you do not need to find it to learn the values of $A$ and $... 2 Are you seeing the inflexion point now!? :) For a better quality of the image click here or on the image itself. Analytically, if you solve for$f^{''}(x)=0$then you will see that this equation has only one real root namely$x=2\sqrt[3]{2}+\sqrt[3]4$. 1 To reiterate what Eff said, a function is more than a formula; it requires a domain. The domain of$f$and the domain of$g$are different, so they are not the same function. doesn't the cancellation of$(x-1)$allow us to calculate the “true” value of$f(1)$? I think that your notion of “true value” is exactly what the concept of limit is supposed ... 1 The first function is not defined when$x = 1$, whereas the second is. The cancelling is only possible assuming that$x \neq 1$; when$x = 1$the denominator is zero and hence the expression is undefined. Since the domains are different, the functions are different. However, the functions are obviously equal when restricted to the domain that excludes$x = ...

1

Let's first address the question of finding solutions to the equation $$\Pr[X \le x] = x, \quad 0 \le x \le 1,$$ for $X \sim \operatorname{NormalDistribution}(\mu,\sigma^2)$, or equivalently, $$\Phi\left(\frac{x - \mu}{\sigma}\right) = x$$ where $\Phi$ is the CDF of the standard normal distribution. This of course has no closed form solution for $x$, but ...

1

Two ways to detect the inflection point, not mentioned yet: (1) Blow up: Instead of plotting $f(x)$, plot $20\cdot f(x)$. You'll see immediately that there's an inflection point. (2) Look closer: From your graph it is for sure that $f'$ is increasing up to approximately up to $x=4$. Now have a close look: the average slope between $4$ and $5$ is about $... 1 You may not have seen it because of the scale you have chosen on your graph. Note that to the right of the minimum, the graph is increasing but tending towards the horizontal asymptote, so there must be an inflection to the right of the minimum point. 1 Not terribly sophisticated, but something like the following may be a nice visual aid when comparing the change of any two candidates: https://public.tableau.com/s/sites/default/files/rank11.png 1 The only piece of information that you need to distinguish between the points$(1,3)$and$(3,1)$is a component in the direction of$u_1$. In particular: the centroid of our data is$\mu = (2,2)$(or right around there, anyway). We can deduce whether a signal$s = (s_1,s_2)$should be$(1,3)$or$(3,1)$by stating that a signal is$(3,1)$if$(s - \mu)^...

1

My students used to have trouble with this idea too, and it came down to this: you don't add the x-values of the point, you just add the y-values. That's really all you need to know, but if you want to think more deeply, this goes back to how functions are designed. All that the x-value tells us is where we are looking on the graph. The x-axis itself ...

1

This is how points on elliptic curves are added. You can google "addition on elliptic curves" to get many explanations. This may also help: http://www.math.vt.edu/people/brown/class_homepages/elliptic_curve_addition.pdf

1

Let $c\in(0,2]$. The equation $f(x,y)=c$ is: $$\sqrt{4-x^2-y^2}=c.$$ Squaring both sides: $$4-x^2-y^2=c^2.$$ Rearranging $$x^2+y^2=4-c^2.$$ This is the circle of radius $\sqrt{4-c^2}$ centred at the origin.

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The graph of the function is shown below: It is clear that if we rotate the part $\{(x,f(x)):x \geq 1\}$ we get the portion of the graph $\{(x,f(x)): x \leq 1\}$. Hence $g = f$. This can also be seen as follows: If we rotate the point $P(x,f(x))$ on the curve by $180^\circ$ about $(1,-1)$, then we obtain the point $(2-x, -2-f(x))$. Note that $f(2-x)+f(x) = ... 1 From geography, we know that Variation in daylight time: at equator, and latitudes near equator, is minimum. The daytime is approximately constant throughout the year. So most likely, none of the graphs correspond to this. is maximum near poles, so most likely, plot C corresponds to a place near arctic circle (not near antarctic circle, since graph C is ... 1 OK, I think I got it... Prerequisites I will be calling the graphs by letters A-D. (i.e. A corresponds to Graph A). I will be calling the cardinal directions N, E, S, W. (i.e. N corresponds to north). Solution A is given to be N. D is A flipped horizontally. It is fairly obvious that D is S. The rest of the solution is where I was stumbling. ... 1 You can use the value at t=8 to calculate the value of r.$7.58=48\cdot e^{-r\cdot 8}$Soving for r.$\frac{7.58}{48}=e^{-r\cdot 8}ln\left(\frac{7.58}{48} \right)=-r\cdot 8\frac{ln\left(\frac{7.58}{48} \right)}{8}=-rr=-\frac{ln\left(\frac{7.58}{48} \right)}{8}r=\frac{ln\left(\frac{48}{7.58} \right)}{8}\approx 0.230711$I checked it for ... 1 OK, so I think the point of this problem is to get you to understand some different kinds of behaviour that drug concentrations might have in the body. In fact these are also important forms of behaviour for all sorts of situations involving variables changing in time, such as population growth, nuclear decay, disease spread, and so on. Let's look at the ... 1 If you mean a "straight line", then it's because a function with such a graph is always linear (so, not non-linear). Linear functions are precisely those whose graphs are (nonvertical) lines. I'm assuming we're talking about real-valued functions of a single real variable (functions$\mathbb R\to\mathbb R$). 1 A simple way a computer can make a graph of a function$f$on an interval$[a,b]$is to divide$[a,b]$up in$n$ascending points$x_1,\dots x_n$with$x_1=a$and$x_n=b$and then draw straight lines between$\left(x_i,f(x_i)\right)$and$\left(x_{i+1},f(x_{i+1})\right)$. The number$n$and the ways the partition$x_1,\dots x_n$is chosen depends on the ... 1 The result you're thinking of does not work for$\ln x$since it takes nonreal values at$x<0$. However, note that$\mathfrak{R}(\ln x)$is even on$\mathbb{R}\setminus\{0\}$(where$\mathfrak{R}z$is the real part of$z$) 1 Showing the limits at infinity is straightforward. Simply consider the related rational function $$g(x) = \frac{3-x}{1+x^2}$$ which, by definition of the floor function, satisfies $$f(x) \le g(x) < f(x) + 1$$ for all$x\$, and for which $$\lim_{x \to -\infty} g(x) = \lim_{x \to -\infty} \frac{\frac{3}{x^2} - \frac{1}{x}}{\frac{1}{x^2} + 1} = 0,$$ and ...

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