$f(x) = x^3 - 4x \\g(x)= \sin(\pi x)$

$f(x)$
dominio :
$\forall x \in \mathbb{R}$

$f(-x) = -x^3 + 4x = - (x^3 - 4x) = -f(x)$
dispari

int. assi: asse y:

$x = 0 \\ f(x) = 0$

Asse x:

$y = 0 \\ x(x^2 - 4) = 0 \to x=0\\ x \pm 2$

positività:

$x^3 - 4x > 0 \\x(x-4)>0$

$-2<x<0 \cup x>2$

Comportamento all'infinito:

$lim_{x \to + \infty} x^3 - 4x = lim_{x\to \infty} x^3 (1- \no{\frac{4}{x^2}}^0) = +\infty$

E siccome dispari:

$lim_{x \to \infty} \\f(x) = -\infty$

Andamento:

$f'(x) = - 3x^2 - 4 = 0 \\f'(x) = 0 \to 3x^2 - 4 = 0 \to x^2 = \frac{4}{9} \to = x \pm \frac{2 \sqrt3}{3}$

$f'(x) = 0 \\3x^2 -4 > 0 \to x < -\frac{2 \sqrt 3}{3} \cup x > \frac {2 \sqrt3}{3}$

Cresce per

$x < \frac{ -2 \sqrt3}{3}$

In

$x= \frac{-2\sqrt3}{3}$
PUNTO DI MASSIMO RELATIVO

$f(\frac{-2\sqrt3}{3}) = (\frac{-2\sqrt3}{3})^3 - 4 (\frac{-2\sqrt3}{3}) = \frac {+16\sqrt3}{9}$

In

$x = \frac {+2\sqrt3}{3}$
MINIMO RELATIVO

$f(\frac{2\sqrt3}{3}) = \frac {-16\sqrt3}{9})$

$f''(x) = 6x \\f''(x) = 0 \to x = 0$

In

$x= 0$
Flesso a tangente obliqua

Asintoti obliqui

$lim_{x \to \infty} \to \frac{f(x)}{x} = lim_{x\to \infty} \to \frac {x(x^2-4)}{x} = +\infty$

No asintoti obliqui

$g(x) = \sin \pi x$

Dominio =

$\forall x \in \mathbb{R}$

Int. assi

$x = 0 \\y = 0$

$y = 0 \to \sin\pi x = 0 => \sin \pi x = \sin (0 + k \pi) \\ \pi x = 0 + k \pi \\ x = k$

Positività:

$\sin \pi x > 0$

$2k \pi < \pi x < \pi + 2k \pi$

$2k < x < 1 + 2k$

Crescenza/Decrescenza

$g'(x) = \pi \cos \pi x$

$g' (x) = 0 \to \pi \cos \pi x = 0 => \cos \pi x = 0 \\ \cos \pi x = \cos \frac{\pi}{2} + k \pi$

$\pi x = \frac{\pi}{2}+ k \pi \to x = \frac{1}{2} + k$

$g'(x) = 0$

$\pi \cos \pi x > 0 \\ \cos \pi x > 0$

$\cos ( - \frac{\pi}{2} + 2k \pi) < \cos \pi x < ( \frac{ \pi}{2} + 2k \pi})$

$- \frac{ \pi}{2} + 2k\pi < \pi x < \frac {\pi}{2} + 2k\pi$

$x = \frac{1}{2} + 2k$
Massimi assoluti

$g(x) = \sin \pi (\frac12 + 2k \pi) = \sin \frac {\pi}{2} + 2k \pi = 1$

$x = - \frac12 + 2k$
Minimi assoluti
$\sin \pi (- \frac12 + 2k \pi) = -1$

$g''(x) = -\pi \sin \pi x \to g''(x) = 0 \\ \pi x = k \pi \\ x = k$

$g'' (x) = 0 => - \sin \pi x > 0 \to \sin \pi x < 0 \\ \sin ( - \pi + 2k \pi) < \sin \pi x < \sin (2 k \pi) \\ -1 + 2k < x < 2k$

2)

$y = -3 \\ x^3 - 4x = - 3 \\ x^3 - 4x + 3 = 0 \\ f(1) = 1 - 4 + 3 = 0$

$\begin{array} {c|ccc|c} & 1 & 0 & - 4 & 3 \\ &&&& \\ 1 & & 1 & 1 & -3 \\ \hline \\ & 1 & 1 & -3 & 0 \end{array}$

$(x - 1) ( x^2 + x -3) = 0$

$x = 1$

$x^2 + x - 3 = 0 \to x = \frac {- 1 \sqrt{ 1 + 12}}{2} = \frac {-1 \pm \sqrt 13}{2}$

$A = (1,3) \\ B $$\frac{1 + \sqrt13}{2}; -3$$ \\ C$$\frac {1 - \sqrt13}{2}; -3$$$

$Gg : \to g'(x) = \pi \cos \pi x = 0$

$$$\frac{1}{2} + 2k, 1$$ \ \ per \ \ -3 < k < 2$

$$$-\frac{1}{2} + 2k, - 1$$ \ \ per \ \ -2 < k < 3$

3)

$\int_{0}^{2} g(x) + \int_{2}^{0} f(x) = \\ | - \frac{1}{\pi} \cos \pi x|_0^2 + |\frac{1}{4}x^4 - 2x^2|_2^0 = \\ 0 - (\frac{1}{\no{4}} \cdot \no{16}^4 - 8 ) = + 4$

4) per evitare dubbi di segno, calcoliamo il volume della piscina, dividendone la superficie.

$V= V_1 + V_2 - V_3$

$V_1 = \int_0^1 \sin \pi x \cdot h(x) dx$

$V_2 = \int_0^2 (4x - x^3) \cdot h(x) dx$

$V_3 = \int_1^2 - \sin x \cdot h(x) dx$

e dunque

$V_{TOT} = V_1 + V_2 - V_3 = \int_0^1 \sin \pi x \cdot h(x) + \int_0^2 (4x-x^3) \cdot h(x) + \int_1^2 \sin \pi x \cdot h(x) dx$

Unendo

$\int_0^1 \sin \pi x \cdot h(x) + \int_1^2 \sin \pi x \cdot h(x) dx = \int_0^2 \sin \pi x \cdot h(x) d(x)$

avremo

$V_{TOT} = \int_0^2 (\sin \pi x + 4x - x^3)(3-x) dx$

$= \int_0^2 (3-x) \sin \pi x dx + \int_0^2 (12x - 4x^2-3x^3+x^4) dx$

$= $- \frac{1}{\pi}(3-x) \cos \pi x$_0^2 - \frac{1}{\pi} \int_0^2 \cos ( \pi x) dx + $6x^2- \frac43 x^3 - \frac34x^4+ \frac{x^5}{5}$_0^2$

siccome

$\int_0^2 \cos (\pi x )dx = 0$

Allora

$- \frac{1}{\pi} + \frac{3}{\pi} +24 - \frac{32}{3} - 12 + \frac{32}{5} = \frac{2}{\pi} + \frac{116}{15}$

Il volume in m^3 sara'

$( \frac{2}{\pi} + \frac{116}{15} \) m^3$
e siccome
$1 m^3 = 1000 l$
allora i litri d'acqua saranno
$$$\frac{2}{\pi} + \frac{116}{15}$$ \cdot 10^3 l$

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