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$$\left(\frac{2}{5}\right)^x+\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x=1$$ 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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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 ...

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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$. Equating the two expressions, you claim that there is a ... 3 Where does the function$|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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