# Tag Info

7

Neither is correct. With $z = x+iy$, the inequality is just $x < 2$, representing a half-plane to the left of the vertical line $x = 2$:

5

$\sin^2x=\dfrac{1-\cos2x}{2}$, $\sin2x=2\sin x\cos x \to\\ f(x)=2\sqrt{\dfrac{3}{2}-\dfrac{1}{2}(\sin2x+2\cos2x)}=2\sqrt{\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}\sin(2x+\phi)}.$ Now you can plot by hand.

4

For all $x \in \mathbb{R}$, $$e^x \geq 1 + x$$ Taking log on both sides we get, $$\ln (1 + x) \leq x, \forall x > -1$$ Substituting $x = \frac{1}{k}, k \notin [0, -1]$, we get, $$\displaystyle{\ln \left(1 + \frac{1}{k}\right) \leq \frac{1}{k}}$$ Substituting $x = \frac{-1}{k + 1}, k \notin [0, -1]$, we get, $$\ln \left(\frac{k}{k + 1}\right) \leq ... 4 Let g(x)=f(f(x)), then g'(x)=f'(x)f'(f(x))\geq0 since f(x) is monotonic function. Note that x=1 is the solution of g(x)=f(f(x))=\frac{1}{x^2-2x+2}=\frac{1}{(x-1)^2+1} Now, since \frac{1}{(x-1)^2+1} is strictly decresing function on interval [1,\infty), and g(x) is nondecreasing on that interval, the solution x=1 is unique. Hence, N=1. 4 We see$$3y-2=-4\sqrt{1-y^2}$$so$$(3y-2)^2=16-16y^2$$therefore$$9y^2-12y+4=16-16y^2$$rearranging gives$$25y^2-12y-12=0$$And so the solutions are$$y_0,y_1=\frac{12}{50}\pm\frac{1}{50}\sqrt{144+1200}$$and simplifying yields$$y_0,y_1=\frac{6\pm4\sqrt{21}}{25}$$Checking these solutions will give us the unique solution$$y_0=\frac{6-4\sqrt{21}}{25}$$3 \sin(\pi/2-X)=\cos(X) hence \sin(\pi/2-2x)=\cos(2x) which you should be able to draw. You can't be getting \sin(2x) since \sin(2*0)=0\neq\sin(\pi/2-2*0)=1 3 Point z=re^{i\theta} is given in polar coordinates. It means that you have a radius r and an angle \theta. You can always go back to cartesian coordinates using the formula:$$re^{i\theta}=r\cdot \cos(\theta)+r\cdot i\sin(\theta)$$There's a nice representations of both coordinate systems at Wikimedia: 3 Note that the two planes represent the same thing, with points drawn in different methods. This is expected since re^{i \theta} and a+bi both just represent complex numbers. A point on the plane can be identified by its real and imaginary coordinates - this is the case of a + bi since we identify it as the point (a, b). A point on the plane can ... 3 A string of length n is a finite sequence s_1\cdots s_n with s_i\in\{1,\ldots,m\} for each i. Writing s(i)=s_i you get a map s\colon\{1,\ldots,s\}\to\{1,\ldots,m\}. 3 "If (x,y) represents a point on the graph of y = 2x + 1" means you get a point whose coordinates are$$(x,2x+1)$$"which of the following could be a portion of the graph of the set of points (x,y^2)" is equal to "which of the following could be a portion of the graph of the set of points (x,(2x+1)^2 = 4x^2 + 4x+1)" Hence you get the equation$$y = ...

2

... Figured out as I was rereading my own question: -1 is the function's middle line, which shifts the whole graph down by -1. The inequality I started with has no solutions because, indeed, the function never evaluates to a y greater than zero. I was greatly anxious, and as I just can't handle these kinds of situations, mistakes like this I guess may ...

2

The elements of $[n]$ are naturally ordered, and this order gives a sequence $[s_{1}, \ldots, s_{n}]$, where each of the $s_{i} \in [m]$. In other words, you have $n$ symbols from $[m]$ laid out in a specified order. It is sometimes convenient to think of this functionally, so the $i$th character of the string is given by $s(i) = s_{i}$. As a programmer, ...

2

Hint: You can use a function like $$y=ax-b|\sin (cx+d)|$$ adjusting the constants $a,b,c,d$.

2

Set $f:[k,k+1]\to\mathbb{R}$ given by $f(x)=\log(x)$ Then, $f$ is continuous in $[k,k+1]$ and differentiable in $(k,k+1)$. Thus, there is $\xi\in(k,k+1)$ such that $\frac{f(k+1)-f(k)}{k+1-k}=f^\prime(\xi)$. That is $\log(k+1)-\log(x)=\frac{1}{\xi}$ for some $\xi\in(k,k+1)$. Then ...

2

It's easier to start studying the function under the square root, $$g(x)=8\sin^2(x)+4\cos^2(x)-8\sin(x)\cos(x).$$ This is an homogenous trigonometric polynomial of the second degree, and we have some hope of simplifying it by the double angle formulas. Indeed, $$g(x)=8\frac{1-\cos(2x)}2+4\frac{1+\cos(2x)}2-8\frac{\sin(2x)}2=6-2\cos(2x)-4\sin(2x).$$ We ...

2

If $a>0$ is the same in the two functions, the correct graph is this. Note that $y=x^2-a$ is a parabola that has negative values $y<0$ in the interval $x \in(-\sqrt{a},\sqrt{a})$. For the squared equation $y=(x^2-a)^2$ we can have only positive values, so the graph become positive in the interval $(-\sqrt{a},\sqrt{a})$, but we have always $y=0$ for ...

2

Note that $\forall k \in \mathbb{N}$ the points such that $f(x) = k$ are only those in $\Big[\frac{k(k+1)}{2};\frac{k(k+3)}{2}\Big]$. Now, let $a = \frac{k(k+1)}{2}, b = \frac{k(k+3)}{2}$: to find the inverse you have to solve for $k$ $$\begin{cases} a = \frac{k(k+1)}{2} \\ b = \frac{k(k+3)}{2} \end{cases}$$ Simple algebra caluli lead to $$\begin{cases} k = ... 2 To answer your first question: No there is no error in your solution. Now to answer: "What does \{z\in \mathbb C\mid z^z\in \mathbb R\} look like?" Well, it does not look very pretty. If we write \arg(z)=\theta, then \Re(z)=r\cos \theta. Rewriting what you found, this means that$$z^z\in \mathbb R\iff\frac{\theta}{\pi}r\cos\theta\in\mathbb Z\iff ...

2

Thinking about it the way that you are saying, you can write $\sin(\pi/2-2x)=\sin(-(2x-\pi/2))=-\sin(2x-\pi/2)=-\sin(2(x-\pi/4))$, where in only the second to last step we have used a trig identity. So you shrink by a factor of $2$, shift to the right by $\pi/4$, and reflect through the $x$ axis. This winds up putting a maximum at $0$, which you might ...

2

One canonical example of an equation of one variable that can't be solved without graphing is $$x^2=e^x.$$ Try it for yourself if you'd like.

2

Hint: $$y-x = c(x+y) \Rightarrow y = \bigg(\frac{1+c}{1-c}\bigg)x$$ And $d$ is the only graph with a positive gradient.

2

If you want to express it as a function, then you do so by defining the function piecewise: $y = f(x) = \left \{ \begin{array}{ll} 12x & x \lt 12 \\ 6x + 12 & x \ge 12 \end{array} \right.$ Alternatively, you can write it using something like the Heaviside step function, but that's just doing exactly the same thing but hiding the piecewise nature of ...

2

Notice, the equation of the tangent at $(3, 36)$ to the parabola: $y=4x^2$ $$y-36=\left(\frac{dy}{dx}\right)_{(3, 36)}(x-3)$$ $$y-36=24(x-3)\ \ or \ \ \ y=24x-36$$ The above tangent line intersects the x-axis at $\left(\frac 32, 0\right)$ Draw a diagram, drop a perpendicular from the point of tangency $(3, 36)$ to the x-axis at the point $(3, 0)$ & ...

2

if you rotate the [graph of a] function by $90$ degrees positive turn (counter-clockwise) and then flip it through the $y$-axis you get the [graph of the] inverse That is correct. I wouldn't express it as "the same plot but rotated" because there is also that "flip", or reflection, that you did after rotating it.

1

Well...the easiest way to find the function $f$ rotated 90 degrees is simply to first find $f^{-1}$, and then to set $g(x) = f^{-1}(-x)$. But that kind of defeats the purpose, since you were trying to rotate as an intermediate step in finding the inverse. To rotate it 90 degrees directly, you could convert it into polar form as a function $r(\theta)$. Then ...

1

I suspect the symbol was meant to be $\infty$ (infinity) but was printed incorrectly. If so, then the domain is $\{x:x\ge -5\}$ and the range is $\{y:y\ge 0\}$.

1

$$\text{cis}\left(x^2\right)=\text{cis}(x)\Longleftrightarrow$$ $$\cos\left(x^2\right)+\sin\left(x^2\right)i=\cos(x)+\sin(x)i\Longleftrightarrow$$ $$e^{ix^2}=e^{ix}\Longleftrightarrow$$ $$\ln\left(e^{ix^2}\right)=\ln\left(e^{ix}\right)\Longleftrightarrow$$ $$ix^2=ix\Longleftrightarrow$$ $$x^2=x\Longleftrightarrow$$ Notice, if $a,b\in\mathbb{R}$ and ...

1

how do I know which one is y and which one is x? You can use any consistent set of variable names. You can name your axis whatever you want as long as you are consistent and don't make an error. Here I assume that the horizontal access is "a" and the vertical access is "b" - To draw the constraints I did this: Note: I am only showing how I draw the ...

1

This is a typical example of the loss of significance occurring in numerical calculations. Let's focus on the numerator. When $x=0.005$, the terms being subtracted, $\sin \tan x$ and $\tan \sin x$, are of size $\sim 0.005$ themselves. In double precision, their representation incurs errors of size $0.005\cdot 2^{-53}\approx 5\cdot 10^{-19}$. The Taylor ...

1

Hint: If $f(x)$ is monotonic, then $f(f(x))$ is (possibly non-strictly) increasing, so… Full solution: Clearly if $f$ is increasing, so is $f\circ f$. Furthermore, if $f$ is decreasing, then $x \leq y$ implies $f(x) \geq f(y)$, which in turn implies $f(f(x)) \leq f(f(y))$. Either way, $f \circ f$ is increasing. Since $f$ is $1$ at $x=1$, it follows ...

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