Esercizi svolti di Matematica

Esercizi per esame:

y = ^{1 - x2}/_e^x

1° DOMINIO:

D: y > 0e^x \[1 - x²] > 0f. 0 (x) = x² > 0

condizione sistema

e^x > 0x ∈ R

II° tut: ASSI:

• x = 0
• asse y: y = ^{1 - x2}/_e^x

x = 0y = ^{1 - 02}/_e⁰ = ¹/₁

(0,1)

y = 0

e^{x 1 - x2}/_e^x = 0 • e^x

1 - x² = 0x² = 1x = ±1

(-1,0)(1,0)

x = ±1 sono zero con... ass. semiasse

III° PARI in SPAZI:

f. SIMMETRIA rispetto a origine 1. Seconda simmetria riferita asse y

g(x) = ^1-x2/_e^x

g(-x) = ^1-(x)2/_e^-x = ^1-x2/e^x + g(x)+ g(x)/+g(x)

ne p.z.t. no a spazi

III° SECCHIO:

f: y = ^{1 - x2}/_e^x

n:

^{1 - x2}/e^x = 0x² = 1x = ±1

....> 0 ...

+ -

- +

Calcola area regione colorata

\( y = x^2 - x \)

area 1

area 2

\(- \int_0^1 (x^2 - x) \, dx = \text{area 1} \)

\(+ \int_1^2 (x^2 - x) \, dx = \text{area 2} \)

\(= \left[ \frac{x^3}{3} - \frac{x^2}{2} + c \right]_0^1 = \left[ \frac{x^3}{3} - \frac{x^2}{2} + c \right] = \left[ -\frac{1}{6} + c \right] = \frac{1}{6} \)

\(+ \left[ \frac{x^3}{3} - \frac{x^2}{2} + c \right]_1^2 = \left[ \frac{8}{3} - 2 + c \right] - \left[ \frac{1}{3} - \frac{1}{2} + c \right] \)

\(= \frac{8}{3} - 2 + c - \left( \frac{1}{3} - \frac{1}{2} + c \right) = \frac{8}{3} - 2 + c - \frac{1}{3} + \frac{1}{2} - c = \frac{5}{6} \)

\( \text{Area Totale} = \frac{1}{6} + \frac{5}{6} = 1 \)

D: 00 |

Studio dei limiti (Asintoti)

lim_x→+∞ ^1-x2/_ex = [ ]-∞

lim_x→∞ ^1-x2/_ex = 0

ASS orizz Dx: +∞

lim_x→-∞ ^1-x2/_ex = [∞ | 0] = ∞

m = lim_x→-∞ ^g(x)/_x + lim_x→-∞ ^1-x2/_x*ex = [-∞/0]= -∞

≠ lim_x→-∞ ^(1/x)-x

fx = -∞ ≠ non c’è as

II Limiti

• lim_x→0⁺ ((lnx)³ -(lnx)²) = [+∞+∞] F.I
• 1/x⁴ -∞

AS.: Verticale (Dx) x=0

lim_x→0⁺ ((lnx)³ -(lnx)²) = [+∞-∞] F.I.

= ±∞

lim_x→0⁺ ln(x)³ - (lnx)² = [±∞] F.I.

= 0 = m m c'è non c'è

Derivata prima di g(x) = (lnx)³ - (lnx)²

g'(x) = 3 (lnx)² - 2 (lnx)^-4

Funzione composta:

x → 2lnx, (lnx)³

1/x (1)³ (lnx)²

g'(x) = 3 (lnx)² - 2 (lnx)²

= 1/x lnx (3lnx - 2)

Def. x > x>0

Alfa:

• 1/x x→0
• Beta: 3lnx - 2 → 0, 3 (2/3), lnx ≥ 2/3

1/3e^2/3 ≈ 1.8

max/min x = 1

dominio

• → punti esclusi → lim g(x) → ∞

x→x₀

• as verticale → ∞ lim
• g(x) = l

x→∞

as orizzontale

se non c'è limitazione → obliquo

m = lim

g(x)

x→∞

q = lim (g(x) - mx)

x→∞

limiti

∞

∞

num

• → 0
• ∞ → ∞
• num

∞ = 1

es. lim

x→∞

^{2x³ - 5x⁴}/_x⁴ = [

-∞

-∞

lim → ⁵/₁ = 5

x→∞