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Starting with the red graph: This is obviously a sine wave. A sine wave is $$A\sin(\omega x+\varphi),$$ where $A$ is the amplitude, $f$ the angular frequency and $\varphi$ the phase. We can easily see that $A=1$ and $\varphi=0$. The angular frequency is defined as $\frac{2\pi}{T}$, where $T$ is the lenght of a period, which in this case is $\pi$. So the red ...
$$f(x)=\arccos\left(\ln{\frac{x+1}{x-3}}\right)$$ domain of $\arccos$ is $[-1,1]$ thus we must have $$-1\le \ln{\frac{x+1}{x-3}}\le 1;\;\frac{x+1}{x-3}> 0$$ $$x\leq -\frac{3+e}{e-1}\lor x\geq \frac{3 e+1}{e-1}$$ $$\lim_{x\to\infty}\arccos\left(\ln \left(\frac{x+1}{x-3}\right)\right)=\frac{\pi}{2}$$ Horizontal asymptote $y=\frac{\pi}{2}$ $$\lim_{x\to \... 1 You could just study it as a function: see where it equals zero, take the derivatives and study their signs finding local maxima/minima an then study it's behaviour towards infinity. This should give you all the informations about the graph. 1 You have a function$$f(x, y, z) = x^2 z^3 + \frac{9}{80} y^2 z^3 - (x^2 + \frac{9}{4} y^2 + z^2 - 1)^3$$This describes an implicit surface f(x, y, z) = 0. There are a number of applications that can plot implicit surfaces (or more generally, isosurfaces f(x, y, z) = c). In this particular case,$$\begin{aligned} f(x,y,z) & = 0 \\ x^2 z^3 + \frac{9}...