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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.


4

Think of a critical point as a "good candidate" for a point at which a local extremum could occur. To come up with a sensible way of formalizing this, think about common places where local extrema occur: for, say, $f(x)=x^2$, it's where $f'(x)=0$, but for $g(x)=|x|$, it's where $g'(x)$ is undefined (i.e. at $x=0$). We also want a critical point of a function ...


4

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 ...


3

Where does the function $|x|$ attain it's minimum? Where does $\sqrt[3]{x}$ change concavity?


3

To better grasp the behavior of complex functions, in the first place you must master the behavior of the classical ones, like the powers ($x^d$, for all $d\in\mathbb R$), the polynomials, the exponential, the trigonometric functions and their inverses. You must know by heart their range and their domain, their behavior at $\pm\infty$, their asymptotes and ...


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 ...


3

Your equation is equivalent to $$y-(-6)=\frac 1{x-(-5)}$$ We know that the graph of $y=\frac 1x$ is a right hyperbola, centered at the origin with vertical asymptote $x=0$ and horizontal asymptote $y=0$, with two branches, one above and to the right of the origin and one below and to the left of the origin. One vertex of the hyperbola is $(1,1)$, and the ...


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 ...


2

y = a * sin( x * 2 * pi / w ) ?


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)$ . $$r^2\,\sin(2\theta) = 8$$ $$r = +2\sqrt{2\csc(2\theta)}$$


2

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


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.


2

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 $$y=\frac34x+\frac{11}2$$


2

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

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 ...


2

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.


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 ...


1

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


1

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.


1

If I well understand you want a stright line that pass thorough the pont $P=(x_m,y_m)=(50,1)$ and a point $P'=(x_M,y_M)=(x_M,1.5)$ with $x_M>50$. This line has equation: $$ y-y_m=\dfrac{y_M-y_m}{x_M-x_m}(x-x_m) $$ So for every value of $x_M$ you have a differnt stringt line: $$ y-1=\dfrac{1.5-1}{x_M-50}(x-5) $$


1

Like any graph, to find $f(x)$ you go to the value $x$ on the $x$ axis and then vertically to find $f(x)$. In the first graph, if I asked you for $f(1.5)$, you would have not trouble reporting that $f(1.5)=1$. If I asked for $f(1)$ and the dots were not there, you wouldn't know whether it is $0$ or $1$. The filled dot is part of the graph and the open one ...


1

$\bf{My\; Solution::}$ 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 ...


1

I will give you some hints. We can graph a parabola with the function $$f(x) = ax^2+bx+c $$ where $a,b,c$ are parameters that we need to figure out. The height of the parabola is then $f(x)$ at the position $x$ on the ground. If we place our coordinate system in a way such that the origin is on the ground in the center of the parabola, we know that $$f(0)=52 ...


1

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]


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}\\ ...


1

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 ...


1

Just a sketch for $x>0$: this transcendental equation cannot be solved explicitly. Nevertheless, by putting it in the form $\tan x^2 = \frac 1 {2x^2}$ one notices that the graph of the left-hand side (LHS) is made of several disjoint branches, each one increasing continuously from $0$ to $\infty$, while the right-hand side (RHS) is decreasing from ...


1

We can do a few simplifications: Substitute $z=x^2$ so $x = \pm\sqrt z$. Then $$\begin{align*} 2\cos x^2 - 4x^2 \sin x^2 & = 0\\ \Leftrightarrow \cos z - 2z\sin z & = 0 & z \ge 0 \end{align*} $$ Now since $z=0$ is no solution and $\cos z = 0 \Rightarrow \sin z \ne 0$ is also no solution, we can divide by $\cos z$ and get $$1 = 2z\tan z$$ so $$z = ...



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