Analisi matematica 1 - esercizi

P.163n.265

f(k) = (1-k) / (k² + 3)

1. Dominio

∀k∈R → D_f = R

2. Intersezioni con gli assi

f(0): (1-0) / (0² + 3) = 1 / 3 → I_υ = (0; 1/3)

f(k) = 0 → (1-k) / (k² + 3) = 0 → 1-k=0 ↔ k=1

3. Positività

f(k) = (1-k) / (k² + 3) > 0 ↔ (1-k) > 0 per k < 1 k² + 3 > 0 per ∀k∈R

4. Asintoti

lim (1-k) / (k²+3) k→∞

lim k (1 / k - k) / (k (k²/k)) k→∞

lim k→∞ 1 / k = 0

lim (1-k) / (k²+3) k→+∞

lim k→+∞ -1 / k

m= lim _k→∞⁺ f(k) / k = lim _k→∞ -1 / (k²) = -1 / ∞ = 0

5. Monotonia

f(k) = (1 - k) / (k² + 3)

f'(k) = -1 (k² + 3) - (1 - k) (2k) / (k² + 3)²

= (-k² - 3 - 2k + 2k²) / (k² + 3)²

= k² - 2k - 3 / (k² + 3)² > 0

φ₁,₂ = 2 < (√1 + 12) / 2

(x + 3) (x - 1)

MAX_x ∈ -1, 1/2; MIN_x ∈ -1, 1/6

MAX_xe(-1, ¹/₂)

MIN_xe(3, -¹/₆)

f(x) = ^1-x/_x²+3

Studiare la funzione

f(k) = k⁴ - 2k²

1. Dominio

X = ℝ

2. Positivita'

f(k) > 0 <=> k⁴ - 2k² > 0 <=> k² > 2 <=>

<=> k < -√2 ∧ k > √2

3. Ricerca degli Asintoti

lim_k->∞ f(k) = lim_k->∞ (k⁴ - 2k²) = +∞ = lim_k->∞ f(k)

m = lim_k->∞ (f(k)/k) = lim_k->∞ (k³ - 2k) = ∞

4. Studio della Derivata Prima

f(k) = k⁴ - 2k² => f'(k) = 4k³ - 4k > 0 <=>

<=> k³ - k > 0 <=> k³ > k <=>

<=> -1 < k < 0 ∧ k > 1

f(-1) = 1 - 2 = -1 => MIN_DEB1 = (-1, -1)

f(0) = 0 => MAX_REL = (0, 0)

f(1) = 1 - 2 = -1 => MIN_DEB2 = (1, -1)

f(h) =

^h₂/_1+h - ³/ ₂ ln (^h+1 )/_h

MIN_xT = (1; ¹/₂ + ^{3 ln 2}/ ₂)

MAX_xT = (-³/₂;-^√3 /₂;f(-³/₂;-^√3/ ₂))

= h2 + 2h/ (1+h)2 + 3 /2 h / ( h + 1 ) - ( -1 / h2 ) = h2 + 2h - 3h / (1 + h)2 2h2(1+h)2

= ^{2h2(h2 +2h) - 3h(1 +h )} / _{2h² (1 + h )² = ^{2h4 + 4h3 - 3h - 3h2 < 0 / 2h2 (1 + h )2}}

2h⁴ + 4h³ - 3h² - 3h < > 0

2h⁴ + 4h³ - 3h² - 3 / h < > 0

8)

lim_k→+∞ ^e-k/_{3^k + k + 1} = ^e-∞/_+∞ = 0

9)

lim_k→0 arctg|k|·cos(1 + log|k|)

10)

lim_k→∞ √^{k2 - 5k + 6} + k = lim_k→∞ √^{k2(1 - 5/k + 6/k2)} + k

= lim_k→∞ k√¹ + k = lim_k→∞ 2k = -∞

11)

lim_k→1 ^(k-1)2/_e^3(k-1)-1 = 0/0 = lim_k→1 ^3(k-1)/_{e^3(k-1)·6(k-1)} = 0/0

= lim_k→1 ³/_{e^{3(k-1)2/36(k-1)2 + e3(k-1)(k-1)2·6}} = ³/₆ = ¹/₃

12)

lim_k→1 ^tg(πk)/_k-1 = 0/0 = lim_k→1 ¹/_cos²(πk)

= lim_k→1 ¹/₁ = 1 = 1

13)

lim_k→π ^senk/_k-π = 0/0 = lim_k→π ^cosk/₁ = -1/1 = -1