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0

A statement like $g(x) = 0.20(a)$ is a bit of a mis-match: The notation $g(x)$ tells us the function should depend on $x$, but $0.20(a)$ (and your remarks) suggest that you're considering $0.20(a)$ to be a function of some variable $a$ (the amount of a bill, which is quite smart). So, you have two options: Use $x$ instead of $a$, so that $g(x) = 0.20x$. ...


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Let $x$ be the number of people. The cost of renting a room can be represented as a function of the number of people: $f(x) = 975 + 39.95x$. Let $y$ be the amount of a bill. The cost after a 20% discount can be represented as a function of the amount of a bill: $g(y) = .8y$. A 20% discount on the cost of renting a room can be represented by a composition ...


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Let's look at a few different ways to do the first one, then maybe you'll see how to do the second: Let $f(x) = \sqrt{x^2 + 6}$ and $g(x) = x$. Then $$f(g(x)) = f(x) = \sqrt{(x)^2+6} = h(x)$$ That one sorta feels like it's cheating, though, doesn't it? Let's find a not so trivial way to do it. How about $f(x) = \sqrt{x+6}$ and $g(x) = x^2$. Then ...


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Well, you can always take $g(x)=x$, and $f(x) = h(x)$. Then, you have $f(g(x)) = h(g(x)) = x$ and you are done.


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$f(g(x)) = f(x^2) = \sqrt{16-(x^2)^2} = ....$Can you continue?


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Here is a simple rule: the domain of all polynomial functions(a sum of positive powers of x) is all Real numbers unless you are told otherwise. Additionally the below also holds for polynomials: They are continuous on R They are differentiable on R So when you see functions such as below, rejoice! f(x)=-2x^2+x-7 f(x)=x^3/3-3x^2+0.755 f(x)=4x^4+3x-55


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You have the right idea, but you just have to do one thing at a time. Here, $g(f(x))$ $= g(2x^2 + x)$ $= (2x^2 + x)^2 + 1$ Just think of $g$ as a machine that takes its input, and outputs the input squared plus one. Now, assuming that the domains of $f$ and $g$ are both $\mathbb{R}$, see if there are any real numbers $x$ for which the above expression ...


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$g(f(x)) = g(2x^2+x) = (2x^2+x)^2+1 = (2x^2)^2 + 2(2x^2)x + x^2 + 1 = ....$ can you take it from here?


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http://www.padowan.dk/download/ I use this graph calculator. It has a feature: Point series and trendlines You can create series of points with different markers, colors and size. Data for a point series can be imported from other programs, e.g. Microsoft Excel. It is possible to create a line of best fit from the data in a point series, either from ...


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You're absolutely right about $f(x)$: no asymptotes there. For $g(x)$, given your tags, I'm not sure with you've been exposed to the concept of a limit. If you have, simply taking $\lim_{x\to\infty} g(x)$ and you will find $g(x)$ has a horizontal asymptote $y=1$. If you, are unfamiliar with this concept, just remember that $g(x)$ is an exponential decay, so ...


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Well, I would start by looking for the horizontal asymptotes. For $f$, there is no horizontal asymptote, since the quadratic inside grows larger and larger as $x$ grows and as $x$ decreases. For $g$, the same argument as for $f$. For vertical asymptotes of $f$, we need to look at where it is unbounded. The natural is unbounded at 0, so $x^2+6x+9 = 0$. So ...


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no horizontal asymptotes exist. $log(x)$ has domain $x>0$ and an asymptote at $x=0$ so $f(x)$ will have an asymptote where $x^2+6x+9=0$ $g(x)$ has domain $-1\le x \le +1$ and range $0 \le y \le 1$ but no assymptotes. $h(x) = f(g(x))$ also has domain $-1\le x \le +1$ it has no asymptotes because $x^2+6x+9>0$ whenever $0 \le x \le 1$


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Alternatively, Find the point where your curve will intersect the axes. That can be done by equating y=0 for getting x-axis point $\sqrt{x+1}-3=0$ $x+1=9$ $x=8$ Now, put x=0, you will get y=-2 So, 2 points are definitely $(8,0)$ and $(0,-2)$. Now you need just 2 more points to plot the graph. Lets, put x=3, because x+1=4 which is a square of 2. That ...


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To obtain integer values for $y = \sqrt{x + 1} - 3$, $x + 1$ must be a perfect square. The first five perfect squares are $0, 1, 4, 9, 16$. Thus, we must set $x + 1$ equal to $0, 1, 4, 9, 16$. If $x + 1 = 0$, then $x = -1$. If $x = -1$, then $y = \sqrt{-1 + 1} - 3 = \sqrt{0} - 3 = -3$. Hence, the point $(-1, -3)$ is on the graph. If $x < -1$, then ...


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Let $$x:=t-{5\over2}\qquad\left(-{7\over2}\leq t\leq{17\over2}\right)\ .$$ We have to find the range of $$g(t):=\left(t-{3\over2}\right)\left(t-{1\over2}\right)\left(t+{1\over2}\right)\left(t+{3\over2}\right)=\left(t^2-{9\over4}\right)\left(t^2-{1\over4}\right)$$ when $t$ ranges in the given interval. Since $t^2$ then assumes values between $0$ and ...


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Hint $$(x+1)(x+2)(x+3)(x+4)+5=[(x+1)(x+4)][(x+2)(x+3)]+5=(x^2+5x+4)(x^2+5x+6)+5 =(x^2+5x+5)^2+4$$


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$\bf{My\; Solution::}$ Let $\displaystyle \left(x+\frac{1}{2}\right) = t\;,$ and $\displaystyle -\frac{11}{2}\leq t \leq \frac{13}{2}$. Then expression convert into $$\displaystyle f(t) = \left(t-\frac{3}{2}\right)\cdot \left(t-\frac{1}{2}\right)\cdot \left(t+\frac{1}{2}\right)\cdot \left(t+\frac{3}{2}\right)=\left(t^2-\frac{9}{4}\right)\cdot ...


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Indeed Cameron, thanks. I was not using pen and paper. Doing it in my head, but I still think the derivative will factor out by grouping terms as below (hopefully right this time :)). f'(x) = (x+1)(x+2)(x+3)+(x+2)(x+3)(x+4)+(x+1)(x+3)(x+4)+(x+1)(x+2)(x+4) = (x+2)(x+3)(2x+5)+(x+1)(2x+5)(x+4) = (2x+5)[(x+3)(x+2)+(x+1)(x+4)] The derivative nicely ...


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First case: From Newton's Second Law, you have: $$F=ma=mg-mkv\Longrightarrow\dfrac{dv}{dt}=g-kv$$ We can perform an integration using separation of variables: $$\int_0^v\dfrac{dv}{g-kv}=\int_0^t dt\Longleftrightarrow -\dfrac{1}{k}\ln{|g-kv|}\Bigg\vert_0^v=t\Bigg\vert_0^t$$ Rearranging: $$v(t)=\dfrac{g}{k}\left(1-e^{-tk}\right)$$ And you can see that: ...


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First of all by newtons second law we have choosing the coordinate axis as down positive and up as negative we have the following equation for the first one. $F = mg - mkv = ma \rightarrow g - kv = a \rightarrow g - kv = dv/dt$ Integrate and we will get the graph second case we do something similar. Now for the second equation we have $F = mg - mkv^2 = ma$ ...


12

$$\left(\frac{2}{5}\right)^x+\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x=1$$ LHS is strictly decreasing for all $x\in\Bbb R$, since $(2/5)^x, (3/5)^x, (4/5)^x$ are. RHS is constant. $\lim_{x\to -\infty}\text{LHS}=+\infty$ and $\lim_{x\to +\infty}\text{LHS}=0$, so LHS crosses $1$. Exactly one solution exists.


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There is no ambiguity on the function definition. It is defined as $x-\pi$ on $(0,\pi]$ and even with period $2\pi$, so that it is $-x-\pi$ on $(-\pi,0]$. Anyway, the Fourier series is to be computed on the range $(-2\pi,2\pi]$, there will be no $2\pi$ harmonic.


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Before deleting it, can you please tell me why this would be the graph ?


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The Fourier series converges to a certain extension of the given function. cosine series on $[0,L]$: even periodic extension sine series on $[0,L]$: odd periodic extension full Fourier series (sines and cosines on $[-L,L]$): periodic extension. To obtain the required extension, Define $f(x)=f(-x)$ on $[-L,0)$ and proceed to 3. Define $f(x)=-f(-x)$ ...


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Such a $c$ does exist and $c=0$ Here is a graph of the function (red) and its tangent at $x=0$ (blue):


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Let us prove that this can only happen when $f''(c) = 0$: Since $f$ is continuous and $\frac{f(a) - f(b)}{a - b} \ne f'(c)$ we have either $\frac{f(a) - f(b)}{a - b} \ge f'(c)$ for all $a,b$ OR $\frac{f(a) - f(b)}{a - b} \le f'(c)$ for all $a,b$ (because $f$ is continuous and hence it takes all intermediate values). So WLOG let us assume that $\frac{f(a) - ...


3

Your statement is ambiguous; I shall interpret it as saying that $\exists c$ such that $f'(c) \neq \frac {f(b)-f(a)} {b-a}$ for $\forall a, b$ with $a \neq b$. Well, in this case there is no such $c$, by the following argument: note that $f'(x)=3c^2+3$ and $\frac {f(b)-f(a)} {b-a} = a^2 + ab +b^2 +3$. Equating the two expressions, you claim that there is a ...


1

If $f$ is twice differentiable and $f''(c)\ne 0$ then we can find two distinct $a,b$ with $f'(c)=\frac{f(b)-f(a)}{b-a}$ - why? Assume wlog. that $f''(c)>0$. Then for sufficiently small positive $h$ we have $\frac{f'(c+h)-f'(c)}{h}>0$ and $\frac{f'(c-h)-f'(c)}{h}<0$, hence $f'(c+h)>f'(c)>f'(c-h)$. Again, for sufficiently small $\eta$, we have ...


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The terms diagonal and anti-diagonal are descriptive and "culturally apt". The first is standard in geometry and topology (the diagonal embedding of a topological space $X$ is the inclusion $X \hookrightarrow X \times X$ defined by $x \mapsto (x, x)$), and I'm almost positive I've seen the second in connection with the normal bundle of a diagonal embedding ...


2

Joining to other answers, this is a "sketch". One could consider continuous functions as their graphs. Then the ball is a set of functions which graphs in this banded pipe (as @copper.hat has noted). In case of closed ball graphs could touch boundary of the pipe.


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The closed ball $\bar{B}(\sin,1)$ consists of continuous $f$ such that $|f(x)-\sin x| \le 1$ for all $x$. To visualise, form the curves $x \mapsto \sin x \pm 1$, then the closed ball is all the continuous functions whose graph lies inside the 'envelope' formed by these curves. Or, imagine putting a pipe of radius one around the sine curve, then all ...


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The basic idea is that you need to resort to the definition of a closed ball. In a metric space with distance function $d(x, y)$, the closed ball centered at $x_0$ with radius $r$ is given by $$\overline{B_r}(x_0) = \{x : d(x, x_0) \leq r\}$$ (note that in my comment to the question, I had mistakenly used "$=$", when I meant "$\leq$"; it's all about the ...


3

You were able to solve the first question by figuring that $$x^2 + 3x + 5 = (x^2 + 3x - 2) + 7$$ And so the equation is equivalent to $$x^2 + 3x + 5 = -7$$ so that the solutions are the points where your parabola cuts the line $y=-7$ (and not $y=-9$, but I assume that was just a typo or an arithmetical error). Now observe that ...


0

Yes, this is essentially correct; as usual, one must take care when taking square roots as one does here. Substituting as in the question gives $$r^2 \sin \theta \cos \theta = 4,$$ which is equivalent to $$r = \pm \sqrt{\frac{4}{\sin \theta \cos \theta}}.$$ In our case, however, we have (in polar coordinates) that \begin{align} \left(\sqrt{\frac{4}{\sin ...


2

Yes.   You have $xy = 4$ in cartesian coordinates, so in polar coordinates that is indeed: $$r^2\sin(\theta)\cos(\theta) =4$$ You can leave it at that, or rearrange to suit.   I recommend using $ 2\sin(\theta)\cos(\theta)=\sin(2\theta)$ . $$r^2\,\sin(2\theta) = 8$$ $$r = +2\sqrt{2\csc(2\theta)}$$


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$\displaystyle\frac{1}{1+e^{-x}}$ is bounded while any translation or scaling of $e^{-x}$ is not.


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Wolfram Mathematica has a great help with many examples. Also it does have its own community on StackExchange http://mathematica.stackexchange.com/ . You'd better address questions regarding Mathematica there G1 = CompleteGraph[{7, 2}]; G2 = EdgeDelete[G1, {1 <-> 8, 2 <-> 8}]; GraphPlot[G1] GraphPlot[G2]


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For graphs of the form $y=f(x)sinx$, draw the graphs of $y=f(x)$ and $y=-f(x)$. Then draw the graph of $y=sinx$ with the usual roots but with the amplitude increasing or decreasing according to the values of $f$. The graphs should touch whenever $sinx$ reaches its maximum and minimum.


2

Evaluate some points of $f$ for intuition. Note that $f(x\pm \pi k) = 0$. Now look at the first and second derivative for increasing/decreasing and for concavity. The general graph will be of an increasing, unbounded oscillation as $e^x$ is increasing without bound and $\sin x$ oscillates.


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Note that if you look at the transition from $q_1$ with the symbol $0$, then it can go to $q_1$ as well by using the empty transition after passing through $q_0$ first. The DFA you've written down doesn't accept "00", which is accepted by the NFA above. You also need, as mentioned in the other answer, to make the start state $\{q_0, q_1\}$, since that ...


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If lambda is the empty string then that isn't correct. Remember the start state is the epsilon closure of the start state - in other words the start state should be the union of q0 and q1 *note I havent done NFA to DFA in awhile so correct me if im wrong.


1

$$ 2 \frac{\sin(x)}{x} + \frac{\sin(x/2)}{x/2} e^{-i x/2} $$ since we know $$ \sin x = \frac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i} $$ thus $$ \frac{2}{x}\frac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i} + \frac{2}{x}\frac{\mathrm{e}^{i\frac{x}{2}}-\mathrm{e}^{-i\frac{x}{2}}}{2i}e^{-i x/2} $$ which is equal to $$ ...


1

Why not just forget about the different variations of the formula (e.g. point slope, 2 points, slope intercept, etc. ) and just memorize this one. $$\color{Tomato}{m=\frac{\Delta y}{\Delta x}}$$ If the slope ($m$) and a single point ($x_0, y_0$) are known, then this forumla becomes $$\begin{align} m&=\frac{\Delta y}{\Delta x}\\ ...


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You almost have it with $y-4=(3x/4)+(6/4)$. Just get the 4 to the right hand side $y=\frac34x+\frac32+4$. So $$y=\frac34x+\frac{11}2$$


0

the graph of $y = f(x) = x^3 - 4x = x(x+2)(x-2)$ is odd and $y\ge 0$ on $[-2,0]\cup [2, \infty).$ the graph of $y = f(x)$ has a local maximum at $(-\frac2{\sqrt 3}, \frac{16}{3\sqrt 3}).$ the domain of the graph of $y = \sqrt{f(x)}$ is $[-2,0]\cup [2, \infty).$ the graph of $y = \sqrt{f(x)}$ looks like a semicircle in $-2 \le x \le 0$ and like a half of a ...


1

No, sketching those $3$ curves is unlikely to help. Instead, sketch the radicand polynomial $g(x) = x(x + 2)(x - 2)$. Recall that the original function $f(x) = \sqrt{g(x)}$ is defined only where $g(x)$ is on or above the $x$-axis. Can you use this to figure out the domain of $f$? Now to transform $g$ into $f$, focus only on the parts of $g$ that lie above ...


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With MatLab : for k=1:1999 x(k)=-5+k*5/1999; y(k)=x(k)-1; z(k)=x(k); w(k)=x(k)+1; t(k)=x(k)*exp(1/(x(k))); end x(2000)=0; y(2000)=-1; z(2000)=0; w(2000)=1; t(2000)=0; for k=2001:4000 x(k)=(k-2000)/400; y(k)=x(k)-1; z(k)=x(k); w(k)=x(k)+1; t(k)=x(k)*exp(-1/(x(k))); end plot(x,y,x,z,x,w,x,t) The result is ...


2

The second derivative represents the inflexion of a curve at a given point. In the picture you have given, it shows the concavity of the graph of function. When the concavity is turned up, the second derivative is positive; otherwise it is negative. When it is zero, the graph of function inflects at that point; that is, the graph has an S-shaped form at the ...


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I won't say it is (or isn't) the best choice, but I will say it's possible, though maybe a bit challenging, with Sage: This image is from this link. I can confirm that Sage plays nicely with OS X (better than with Windows, I believe), although I've never tried to animate anything; it may have quite a learning curve. You can download and run everything on ...



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