$$x^4-18x^2+56$$
$$\begin{aligned}
y &= f(x) \\
\mathbf{X} &\longrightarrow \mathbf{Y} \\
\mathbf{X} &= (-\infty, +\infty) \\
\mathbf{X} &= \mathbb{R} \\
\end{aligned}$$
Given $x^*$ a finite boundary of $f(x)$,
A horizontal asymptote is present at
$$\begin{aligned}
x_a &\in \{\} \text{ because } x^* \in \{\} \\
&& \\
y_a &\in \{\} \text{ as found in "Limits at Boundaries"} \\
\end{aligned}$$
$$\begin{aligned}
&\lim_{x \to -\infty} x^4-18x^2+56 \\
\sim &\lim_{x \to -\infty} x^4 \\
= &+\infty \\
&& \\
&\lim_{x \to +\infty} x^4-18x^2+56 \\
\sim &\lim_{x \to +\infty} x^4 \\
= & +\infty \\
\end{aligned}$$
$$\begin{aligned}
f(x) &= (x^2-4)(x^2-14) \\
&= (x+\sqrt{14})(x-\sqrt{14})(x+2)(x-2) \\
x_0 &\in \{-\sqrt{14}, -2, 2, \sqrt{14}\} \\
&\approx \{-3.74, -2, 2, 3.74\} \\
&& \\
y_0 &= f(0) = 56 \\
\end{aligned}$$
$$\begin{aligned}
f'(x) &= 4x^3-36x \\
&= 4x(x^2-9) \\
&= 4x(x-3)(x+3) \\
x'_0 &\in \{-3, 0, 3\} \\
\end{aligned}$$
$$\begin{aligned}
f'(x) < 0, && x &\in (-\infty,-3) && (-)\\
f'(x) > 0, && x &\in (-3,0) && (+) \\
f'(x) < 0, && x &\in (0,3) && (-) \\
f'(x) > 0, && x &\in (3, +\infty) && (+) \\
\end{aligned}$$
$$\begin{aligned}
f''(x) &= 12x^2 - 36\\
&= 12(x^2-3) \\
&= 12(x-\sqrt{3})(x+\sqrt{3}) \\
x''_0 &\in \{-\sqrt{3}, \sqrt{3}\} \\
&\approx \{-1.73, 1.73\} \\
\end{aligned}$$
$$\begin{aligned}
f''(x) > 0, && x &\in (-\infty,-\sqrt{3}) && (\cup)\\
f''(x) < 0, && x &\in (-\sqrt{3},\sqrt{3}) && (\cap) \\
f''(x) > 0, && x &\in (\sqrt{3}, +\infty) && (\cup) \\
\end{aligned}$$
$$\begin{aligned}
x'_0 &\in \{-3, 0, 3\} && \\
f''(-3) &= 12(-3)^2 - 36 > 0 &\Rightarrow min \\
f''( 0) &= 12( 0)^2 - 36 < 0 &\Rightarrow max \\
f''( 3) &= 12( 3)^2 - 36 > 0 &\Rightarrow min \\
\end{aligned}$$
$$\begin{aligned}
x''_0 &\in \{-\sqrt{3}, \sqrt{3}\} \\
\end{aligned}$$
$$\frac{2x-x^2}{x^2-2x+1}$$
$$\begin{aligned}
f(x) &= \frac{g(x)}{h(x)} \\
h(x) &= x^2-2x+1 \\
&= (x-1)^2 \\
x^* &\in \{1\} \\
&& \\
\mathbf{X} &= (-\infty,1) \cup (1,+\infty) \\
\mathbf{X} &= \mathbb{R} \backslash \{1\} \\
\end{aligned}$$
$$\begin{aligned}
x^* &\in \{1\} \\
&\lim_{x \to 1} \frac{2x-x^2}{x^2-2x+1} = \frac{1}{0} \to \infty \\
&& \\
x_a &\in \{1\} \\
&& \\
y_a &\in \{-1\} \text{ as found in "Limits at Boundaries"} \\
\end{aligned}$$
$$\begin{aligned}
&\lim_{x \to -\infty} \frac{2x-x^2}{x^2-2x+1} \\
= &\lim_{x \to -\infty} \frac{\frac{2}{x}-1}{1-\frac{2}{x}+\frac{1}{x^2}} \\
= &-1
&& \\
&\lim_{x \to +\infty} \frac{2x-x^2}{x^2-2x+1} \\
= &\lim_{x \to +\infty} \frac{\frac{2}{x}-1}{1-\frac{2}{x}+\frac{1}{x^2}} \\
= &-1
\end{aligned}$$
$$\begin{aligned}
g(x) &= 2x-x^2 \\
&= x(2-x) \\
x_0 &\in \{0,2\} \\
&& \\
y_0 &= f(0) \\
&= \frac{2(0)-(0)^2}{(0)^2-2(0)+1} \\
&= 0
\end{aligned}$$
$$\begin{aligned}
f'(x) &= \frac{(2-2x)(x^2-2x+1) - (2x-x^2)(2x-2)}{(x^2-2x+1)^2} \\
&= \frac{-2 ((x-1)(x^2-2x+1) + (x-1)(2x-x^2))}{(x-1)^4} \\
&= \frac{-2 (x^2-2x+1 + 2x-x^2)}{(x-1)^3} \\
&=-\frac{2}{(x-1)^3} = -2(x-1)^{-3} \\
&\qquad x'_0 \in \{\} \qquad x'^* \in \{1\} \\
\end{aligned}$$
$$\begin{aligned}
f'(x) > 0, && x &\in (-\infty,1) && (+) \\
f'(x) < 0, && x &\in (1,+\infty) && (-) \\
\end{aligned}$$
$$\begin{aligned}
f''(x) &= 6(x-1)^{-4} \\
x''_0 &\in \{\} \qquad x''^* \in \{1\} \\
\end{aligned}$$
$$\begin{aligned}
f''(x) > 0, && x &\in (-\infty,1) && (\cup) \\
f''(x) > 0, && x &\in (1,+\infty) && (\cup) \\
\end{aligned}$$
$$\begin{aligned}
x'_0 &\in \{\} \\
\end{aligned}$$
$$\begin{aligned}
x''_0 &\in \{\} \\
\end{aligned}$$
$$\frac{2x^2+3x+1}{x^2-1}$$
$$\begin{aligned}
f(x) &= \frac{g(x)}{h(x)} \\
h(x) &= x^2-1 \\
&= (x+1)(x-1) \\
x^* &\in \{-1,1\} \\
&& \\
\mathbf{X} &= \mathbb{R} \backslash \{-1,1\} \\
\end{aligned}$$
$$\begin{aligned}
x^* &\in \{-1,1\} \\
&\lim_{x \to -1} \frac{2x^2+3x+1}{x^2-1} \\
= &\lim_{x \to -1} \frac{(x+1)(2x+1)}{(x+1)(x-1)} \\
= &\lim_{x \to -1} \frac{2x+1}{x-1} = \frac{1}{2}\\
&& \\
&\lim_{x \to 1} \frac{2x^2+3x+1}{x^2-1} = \frac{6}{0} \to \infty \\
&& \\
x_a &\in \{1\} \\
&& \\
y_a &\in \{2\} \text{ as found in "Limits at Boundaries"} \\
\end{aligned}$$
$$\begin{aligned}
&\lim_{x \to -\infty} \frac{2x^2+3x+1}{x^2-1} \\
= &\lim_{x \to -\infty} \frac{2+\frac{3}{x}+\frac{1}{x^2}}{1-\frac{1}{x^2}} \\
= &2 \\
&& \\
&\lim_{x \to +\infty} \frac{2x^2+3x+1}{x^2-1} \\
= &\lim_{x \to -\infty} \frac{2+\frac{3}{x}+\frac{1}{x^2}}{1-\frac{1}{x^2}} \\
= &2 \\
\end{aligned}$$
$$\begin{aligned}
g(x) &= 2x^2+3x+1 \\
&= (x+1)(2x+1) \\
x_0 &\in \{-\frac{1}{2}\} \\
&& \\
y_0 &= f(0) \\
&= \frac{2(0)^2+3(0)+1}{(0)^2-1} \\
&= -1 \\
\end{aligned}$$
$$\begin{aligned}
f'(x) &= \frac{(4x+3)(x^2-1)-(2x^2+3x+1)(2x)}{(x^2-1)^2} \\
&= \frac{4x^3-4x+3x^2-3-4x^3-6x^2-2x}{(x+1)^2(x-1)^2} \\
&= \frac{-3x^2-6x-3}{(x+1)^2(x-1)^2} \\
&= \frac{-3(x+1)^2}{(x+1)^2(x-1)^2} \\
&= \frac{-3}{(x-1)^2} = -3(x-1)^{-2} \\
&\qquad x'_0 \in \{\} \qquad x'^* \in \{1\} \\
\end{aligned}$$
$$\begin{aligned}
f'(x) < 0, && x &\in (-\infty,1) && (-) \\
f'(x) < 0, && x &\in (1,+\infty) && (-) \\
\end{aligned}$$
$$\begin{aligned}
f''(x) &= 6(x-1)^{-3} \\
x''_0 &\in \{\} \qquad x''^* \in \{1\} \\
\end{aligned}$$
$$\begin{aligned}
f''(x) < 0, && x &\in (-\infty,1) && (\cap) \\
f''(x) > 0, && x &\in (1,+\infty) && (\cup) \\
\end{aligned}$$
$$\begin{aligned}
x'_0 &\in \{\} \\
\end{aligned}$$
$$\begin{aligned}
x''_0 &\in \{\} \\
\end{aligned}$$
$$\sqrt{x^2-2x-5}$$
$$\begin{aligned}
f(x) &= \sqrt{g(x)}, \quad g(x) \geq 0 \\
g(x) &= x^2-2x-5 = x^2-2x+1-6 \\
&= (x-1+\sqrt{6})(x-1-\sqrt{6}) \\
g''(x) &= 2 > 0 \quad (\cup) \\
&\therefore g(x) > 0, x \in (-\infty, 1-\sqrt{6}) \cup (1+\sqrt{6}, +\infty) \\
&& \\
\mathbf{X} &= (-\infty, 1-\sqrt{6}) \cup (1+\sqrt{6}, +\infty) \\
& \approx (-\infty, -1.45) \cup (3.45, +\infty) \\
\end{aligned}$$
$$\begin{aligned}
&\lim_{x \to 1 \pm \sqrt{6}} \sqrt{x^2-2x-5} = 0 \\
\\
x_a &\in \{\} \\
&& \\
y_a &\in \{\} \text{ as found in "Limits at Boundaries"} \\
\end{aligned}$$
$$\begin{aligned}
&\lim_{x \to -\infty} \sqrt{x^2-2x-5} \\
\sim &\lim_{x \to -\infty} \sqrt{x^2} \\
= &+\infty \\
&& \\
&\lim_{x \to +\infty} \sqrt{x^2-2x-5} \\
\sim &\lim_{x \to +\infty} \sqrt{x^2} \\
= &+\infty \\
\end{aligned}$$
$$\begin{aligned}
\sqrt{g_0} &= 0, \quad g_0 = 0 \\
g(x) &= x^2-2x-5 = 0 \\
&= (x-1+\sqrt{6})(x-1-\sqrt{6}) = 0 \\
x_0 &\in \{1-\sqrt{6}, 1+\sqrt{6}\} \\
&& \\
y_0 &= f(0) \\
0 &\not \in \mathbf{X} \\
&\therefore \text{ no intercept } y_0 \\
\end{aligned}$$
$$\begin{aligned}
f'(x) &= \frac{1}{2} (x^2-2x-5)^{-\frac{1}{2}} (2x-2) \\
&= (x-1)(x^2-2x-5)^{-\frac{1}{2}} \\
&\qquad x'_0 \in \{\} \qquad x'^* \in \{1-\sqrt{6}, 1+\sqrt{6}\} \\
\end{aligned}$$
$$\begin{aligned}
f'(x) < 0, && x &\in (-\infty, 1-\sqrt{6}) && (-) \\
f'(x) > 0, && x &\in (1+\sqrt{6}, +\infty) && (+) \\
\end{aligned}$$
$$\begin{aligned}
f''(x) &= (1)(x^2-2x-5)^{-\frac{1}{2}} + (x-1)(-\frac{1}{2})(x^2-2x-5)^{-\frac{3}{2}}(2x-2) \\
&= (x^2-2x-5)^{-\frac{1}{2}} -(x-1)^2(x^2-2x-5)^{-\frac{3}{2}} \\
&= \frac{x^2-2x-5}{(x^2-2x-5)^{\frac{3}{2}}} -\frac{x^2-2x+1}{(x^2-2x-5)^{\frac{3}{2}}} \\
&= -\frac{6}{(x^2-2x-5)^{\frac{3}{2}}} \\
&\qquad x'_0 \in \{\} \qquad x'^* \in \{1-\sqrt{6}, 1+\sqrt{6}\} \\
\end{aligned}$$
$$\begin{aligned}
f''(x) < 0, && x &\in (-\infty, 1-\sqrt{6}) && (\cap) \\
f''(x) < 0, && x &\in (1+\sqrt{6}, +\infty) && (\cap) \\
\end{aligned}$$
$$\begin{aligned}
x'_0 &\in \{\} \\
\end{aligned}$$
$$\begin{aligned}
x''_0 &\in \{\} \\
\end{aligned}$$
$$ln(4x+3)$$
$$\begin{aligned}
f(x) &= ln(g(x)), \quad g(x) > 0 \\
g(x) &= 4x+3 > 0 \\
x &> -\frac{3}{4} \\
&& \\
\mathbf{X} &= (-\frac{3}{4}, +\infty) \\
\end{aligned}$$
$$\begin{aligned}
x_a &\in \{-\frac{3}{4}\} \text{ as found in "Limits at Boundaries"} \\
&& \\
y_a &\in \{\} \text{ as found in "Limits at Boundaries"} \\
\end{aligned}$$
$$\begin{aligned}
&\lim_{x \to -\frac{3}{4}} ln(4x+3) \\
= &-\infty \\
&& \\
&\lim_{x \to +\infty} ln(4x+3) \\
= &+\infty \\
\end{aligned}$$
$$\begin{aligned}
ln(g_0) &= 0, \quad g_0 = 1 \\
g(x) &= 4x + 3 = 1 \\
x_0 &\in \{-\frac{1}{2}\} \\
&& \\
y_0 &= f(0) \\
&= ln(4(0)+3) \\
&= ln(3) \approx 1.098 \\
\end{aligned}$$
$$\begin{aligned}
f'(x) &= \frac{4}{4x+3} = 4(4x+3)^{-1} \\
&\qquad x'_0 \in \{\} \qquad x'^* \in \{-\frac{3}{4}\} \\
\end{aligned}$$
$$\begin{aligned}
f'(x) > 0, && x &\in (-\frac{3}{4}, +\infty) && (+) \\
\end{aligned}$$
$$\begin{aligned}
f''(x) &= -4(4x+3)^{-2}(4) \\
&= -16(4x+3)^{-2} \\
&\qquad x''_0 \in \{\} \qquad x''^* \in \{-\frac{3}{4}\} \\
\end{aligned}$$
$$\begin{aligned}
f''(x) < 0, && x &\in (-\frac{3}{4}, +\infty) && (\cap) \\
\end{aligned}$$
$$\begin{aligned}
x'_0 &\in \{\} \\
\end{aligned}$$
$$\begin{aligned}
x''_0 &\in \{\} \\
\end{aligned}$$
$$e^{2x}-3e^x+2$$
$$\begin{aligned}
f(x) &= g^2(x)-3g(x)+2 \\
g(x) &= e^x \\
\mathbf{X} &= \mathbb{R} \\
\end{aligned}$$
$$\begin{aligned}
x_a &\in \{\} \text{ because } x^* \in \{\} \\
&& \\
y_a &\in \{2\} \text{ as found in "Limits at Boundaries"} \\
\end{aligned}$$
$$\begin{aligned}
&\lim_{x \to -\infty} e^{2x}-3e^x+2 \\
= &(0)-(0)+2 \\
= &2 \\
&& \\
&\lim_{x \to +\infty} e^{2x}-3e^x+2 \\
\sim &\lim_{x \to -\infty} e^{2x} \\
= &+\infty \\
\end{aligned}$$
$$\begin{aligned}
f(x) &= (e^x)^2-3e^x+2 \\
&= (e^x-2)(e^x-1) \\
x_0 &\in \{0, ln(2)\} \\
&\approx \{0, 0.693\} \\
&& \\
y_0 &= f(0) \\
&= e^{2(0)}-3e^{(0)}+2 \\
&= (1)-3(1)+2 \\
&= 0 \\
\end{aligned}$$
$$\begin{aligned}
f'(x) &= 2e^{2x}-3e^x = 0 \\
2e^{2x} &= 3e^x \\
e^x &= \frac{3}{2} \\
x'_0 &\in \{ln(\frac{3}{2})\} \\
&\approx \{0.405\} \\
\end{aligned}$$
$$\begin{aligned}
f'(x) < 0, && x &\in (-\infty, ln(\frac{3}{2})) && (-) \\
f'(x) > 0, && x &\in (ln(\frac{3}{2}), +\infty) && (+) \\
\end{aligned}$$
$$\begin{aligned}
f''(x) &= 4e^{2x}-3e^x = 0 \\
4e^{2x} &= 3e^x \\
e^x &= \frac{3}{4} \\
x''_0 &\in \{ln(\frac{3}{4})\} \\
&\approx \{-0.288\} \\
\end{aligned}$$
$$\begin{aligned}
f''(x) < 0, && x &\in (-\infty, ln(\frac{3}{4})) && (\cap) \\
f''(x) > 0, && x &\in (ln(\frac{3}{4}), +\infty) && (\cup) \\
\end{aligned}$$
$$\begin{aligned}
x'_0 &= ln(\frac{3}{2}) \\
f''(ln(\frac{3}{2})) &= 4e^{2ln(\frac{3}{2})}-3e^{ln(\frac{3}{2})} \\
&= 4e^{ln(\frac{9}{4})}-3\frac{3}{2} \\
&= 4\frac{9}{4}-\frac{9}{2} \\
&= \frac{9}{2} > 0 \Rightarrow min \\
\end{aligned}$$
$$\begin{aligned}
x''_0 &= ln(\frac{3}{4}) \\
\end{aligned}$$
$$2sin^2(x)-sin(x)-1$$
$$\begin{aligned}
f(x) &= 2g^2(x)-g(x)-1 \\
g(x) &= sin(x) \\
\mathbf{X} &= \mathbb{R} \\
\end{aligned}$$
$$\begin{aligned}
x_a &\in \{\} \text{ because } x^* \in \{\} \\
&& \\
y_a &\in \{\} \text{ as found in "Limits at Boundaries"} \\
\end{aligned}$$
$$\begin{aligned}
&\lim_{x \to -\infty} 2sin^2(x)-sin(x)-1 \\
&\qquad \lim_{x \to -\infty} sin(x) \text{ diverges }\\
&& \\
&\lim_{x \to +\infty} 2sin^2(x)-sin(x)-1 \\
&\qquad \lim_{x \to +\infty} sin(x) \text{ diverges }\\
\end{aligned}$$
$$\begin{aligned}
f(x) &= 2sin^2(x)-sin(x)-1 \\
&= (2sin(x) + 1)(sin(x) - 1) = 0 \\
sin(x_0) &= -\frac{1}{2} \Rightarrow x_0 \in \{2n\pi+arcsin(-\frac{1}{2}), 2n\pi+\pi-arcsin(-\frac{1}{2})\} \\
&= \{2n\pi-\frac{\pi}{6}, 2n\pi+\pi+\frac{\pi}{6}\} \\
&= \{\pi(2n-\frac{1}{6}), \pi(2n+\frac{7}{6})\}, \qquad n \in \mathbb{Z} \\
sin(x_0) &= 1 \Rightarrow x_0 \in \{2n\pi+arcsin(1), 2n\pi+\pi-arcsin(1)\} \\
&= \{2n\pi+\frac{\pi}{2}, 2n\pi+\pi-\frac{\pi}{2}\} \\
&= \{\pi(2n+\frac{1}{2})\} \\
\therefore x_0 &\in \{(2n-\frac{1}{6})\pi, (2n+\frac{3}{6})\pi, (2n+\frac{7}{6})\pi\} \\
&& \\
y_0 &= f(0) \\
&= 2sin^2(0)-sin(0)-1 \\
&= -1 \\
\end{aligned}$$
$$\begin{aligned}
f'(x) &= 2(2sin(x))cos(x) - cos(x) \\
&= (4sin(x) - 1) cos(x) = 0 \\
sin(x'_0) &= \frac{1}{4} \Rightarrow x'_0 \in \{2n\pi+\theta, 2n\pi+\pi-\theta\} \\
&\quad \text{ with } \theta=arcsin(\frac{1}{4}) \approx 0.25 \approx \frac{\pi}{12} \\
cos(x'_0) &= 0 \Rightarrow x'_0 \in \{2n\pi+arccos(0), 2n\pi+2\pi-arccos(0)\} \\
&= \{2n\pi+\frac{\pi}{2}, 2n\pi+2\pi-\frac{\pi}{2}\} \\
\therefore x'_0 &\in \{\pi(2n + \frac{\theta}{\pi}), \pi(2n + \frac{1}{2}), \pi(2n + 1 - \frac{\theta}{\pi}), \pi(2n + 2 - \frac{1}{2})\} \\
&\approx \{\pi(2n + \frac{1}{12}), \pi(2n + \frac{6}{12}), \pi(2n + \frac{11}{12}), \pi(2n + \frac{18}{12})\} \\
\end{aligned}$$
$$\begin{aligned}
f'(x) > 0, && x &\in (\pi(2n + \frac{1}{12}), \pi(2n + \frac{6}{12})) && (+) \\
f'(x) < 0, && x &\in (\pi(2n + \frac{6}{12}), \pi(2n + \frac{11}{12})) && (-) \\
f'(x) > 0, && x &\in (\pi(2n + \frac{11}{12}), \pi(2n + \frac{18}{12})) && (+) \\
f'(x) < 0, && x &\in (\pi(2n + \frac{18}{12}), \pi(4n + \frac{1}{12})) && (-) \\
\end{aligned}$$
$$\begin{aligned}
f''(x) &= (4cos(x))(cos(x)) + (4sin(x) - 1)(-sin(x)) \\
&= 4(cos^2(x) - sin^2(x)) + sin(x) \\
&= 4(1 - sin^2(x) - sin^2(x)) + sin(x) \\
&= -8sin^2(x) + sin(x) + 4 = 0 \Rightarrow \text{ use quadratic formula} \\
sin(x''_0) &= \frac{1-\sqrt{129}}{16} \Rightarrow x''_0 \in \{2n\pi+\alpha, 2n\pi+\pi-\alpha\} \\
&\quad \text{ with } \alpha=arcsin(\frac{1-\sqrt{129}}{16}) \approx -0.70 \approx -\frac{8\pi}{36} \\
sin(x''_0) &= \frac{1+\sqrt{129}}{16} \Rightarrow x''_0 \in \{2n\pi+\beta, 2n\pi+\pi-\beta\} \\
&\quad \text{ with } \beta=arcsin(\frac{1+\sqrt{129}}{16}) \approx 0.88 \approx \frac{10\pi}{36} \\
\therefore x''_0 &\in \{\pi(2n+\frac{\alpha}{\pi}), \pi(2n+\frac{\beta}{\pi}), \pi(2n+1-\frac{\beta}{\pi}), \pi(2n+1-\frac{\alpha}{\pi})\} \\
&\approx \{\pi(2n-\frac{8}{36}), \pi(2n+\frac{10}{36}), \pi(2n+\frac{26}{36}), \pi(2n+\frac{44}{36})\}
\end{aligned}$$
$$\begin{aligned}
f''(x) > 0, && x &\in (\pi(2n-\frac{8}{36}), \pi(2n+\frac{10}{36})) && (\cup) \\
f''(x) < 0, && x &\in (\pi(2n+\frac{10}{36}), \pi(2n+\frac{26}{36})) && (\cap) \\
f''(x) > 0, && x &\in (\pi(2n+\frac{26}{36}), \pi(2n+\frac{44}{36})) && (\cup) \\
f''(x) < 0, && x &\in (\pi(2n+\frac{44}{36}), \pi(4n-\frac{8}{36})) && (\cap) \\
\end{aligned}$$
$$\begin{aligned}
x'_0 &\approx \pi(2n + \frac{1}{12}) \\
f''(\frac{\pi}{12}) &> 0 \Rightarrow min \\
x'_0 &\approx \pi(2n + \frac{6}{12}) \\
f''(\frac{6\pi}{12}) &< 0 \Rightarrow max \\
x'_0 &\approx \pi(2n + \frac{11}{12}) \\
f''(\frac{11\pi}{12}) &> 0 \Rightarrow min \\
x'_0 &\approx \pi(2n + \frac{18}{12}) \\
f''(\frac{18\pi}{12}) &< 0 \Rightarrow max \\
\end{aligned}$$
$$\begin{aligned}
x''_0 &\approx \pi(2n-\frac{8}{36}) \\
x''_0 &\approx \pi(2n+\frac{10}{36}) \\
x''_0 &\approx \pi(2n+\frac{26}{36}) \\
x''_0 &\approx \pi(2n+\frac{44}{36}) \\
\end{aligned}$$
