Esercitazioni sullo Studio di funzione

2. esercizio

f(x) = (x + |x – 1|) / 4x

x – 1

(x – 1) se x ≥ 1

(1 – x) se x < 1

x ≥ 0

D(g): R – {0}

x (x – 1)/4x, se x ≥ 1

(1)

x (x – 1)/4x + x ≠ 1

(2)

x + x – 1/4x

N: |x – 1| > x – 1

0.4x > 0

lim x →+ ∞

I(g): 4x + x – 1 / 4x

x – 1 > x

x – 1 < x

2x ≥ 1/2

x < 0 ∪ x ≥ 1/2

Sequo:

4x + x – 2 > 0

x = -1 ± √17 / 8

-1 - √17 / 8

-1 + √17 / 8

lim 4x + x → x = + ∞ → cera asinto obliquo

1/4 - 1

m = 1

y = x + 1/4

g = 1/2

y = x + ⁹/₄ → asintoto obliquo 1° per x → +∞

lim_x→-1 ^4x2+x-1/_4x = +∞

g(x) = ^{(8x+1)(4x)-(4)(4x2+4x-1)}/_{32x²+4x-(16x²+9)}

= ^{16x2 -4}/_(4x)² = ^{4 (4x2 -1)}/_(4x)² = ^16x2-4/_4x²-9 = ^4x2/₉

g'(x) + g''' (4x²+1) = ^16x2-9/_16x²

g''' = ^{32 x (16x2)(16x2)-(16x2+32x)}/_(16x²) - ^{512x2 - 512x3 - 128x}/_16x³ = -128x

4x=1 → 0 → 4x²-1, x_1,2=∉R

g(x) sistema cercare + - x - + ∪ ∩

g(x) = 4x² - x + 1 / 4x

segno: 4x² - x + 1

lim_x→0 ^{4x2 - x + 1}/_4x = x

4x = ∞ → asintoto obliquo

lim_x→-∞ (x^{4 - x2 + x} - ¹) = (x⁴ - ^x24x) m = 1

lim_x→-∞ ^4x2-x+1/_4x = (4x^{6 + x + 4x} - ^{x + 1 - 4x})

= -^x + ^x(-1-x2)/_4x = -^x/₉ = -9

y = x - ⁹/₄ pv x→∞

lim_x^>0 ^{4x2 - x + 1 - ∞}/₉

lim_x→∞ g(x) = +∞

g(x) = (1)(x³-2x) - (x-3)(3x²-4x)

= x³-2x - (3x³-9x²-4x+12x)

-----------------------------------------

(x³-2x)²

= x³-2x - 3x³ + 9x²-4x²+12x

= -2x³ + 11x - 12x

--------------------------

(x³-2x)²

= -(2x³+11x-12)

(x³-2x)² > 0

x(-2x + 11x-1)

-------------------------

(x³-2x)²

(xⁿ-2x)² > 0, ∀x∈R

x > 0

2x - 11x + 12 < 0

∆ = 25, x₁₂ = 1 | 3/2

Valori interni perché >

g(x) = 0 as orizzontale

lim (x→∞) x = x|x(1 - 2/x)

lim (x→-∞) x→∞ x-3 = -∞

x = 0 as verticale

lim (x→0) x=3 = -1 | xⁿ+3 = ∞

lim (x→0) xⁿ-∞ = 1

x = 2 as verticale

g''(x) =

Wg(x) = e^2x / (e^x-1)

• Dom: e^-1 + 0, e^x + 1, e^x = 1
• x ≠ 0
• lim x→0
• e^x x (e^x -1 ) ≠ 0
• x ≠ x ∀ x ∈ ℜ
• e^x ≥ 0 e^x > 0 ∀x∈ℜ
• e^x > 0 e^x ≠ 0
• e^x = 1 ⇔ ln e = 0
• x = 0 x > 0 (x > 0)
• No simmetrie
• lim x→+∞ e^2x / (e^x-1)= -∞
• lim x→-∞ e^2x / (e^x-1)= -∞
• x=0 as verticale
• lim 2x→∞ e^2x / (e^x-1)= +∞ as orizzontale

1. 20x - e^2x-2
2. (e^x-1)
3. ⦸= 0
4. 2t - e^2x-2 ≥ 0
5. 2t + 2t - 2 ≥ 0
6. 2 - 2x + 2 = 0
7. 2 t² + 2x ≤ 0
8. e^2x (e^-2) ≥ 0
9. e^x ≥ 0 ∀x
10. x ≠ 2
11. e^x ≥ 2
12. ln ≥ ln e
13. x = ln
14. f(ln) = e^2ln / e^ln-1 = -4

g(x) = (ln(x)²-(x-1)(e^ln(x-1)) = (ln(x)-1)-(2(ln(x-1))^x

(ln(x)-1)⁴

= ln^x+1-2lnx+2lnx-2

(ln(x-1)⁴

= lnx+1-2lnx+2lnx-2

x(ln(x-1)⁴

= - e -

x(lnx-1)⁵

lnx>1

lnx>0

lnx-1 > -1 x x > 0 x

= - x ↔ e

b(x):

x > -1 -e^x e y =

.

x > 0

(1- x)

x > 0

e^k > e⁰

lim

x- > 0 1 / e^k

(x -1^x) / x - e^o

x-o as ven

f(x) - parola

e^2x-1 / 0

x > 0 1 / e^{0 * 1/ x}

s > 1

2x > 1

x > 0/

x³ x 0 > 0

e 1/2

s + x

g(e^1/2) = 0

δ = (x-1)= 1

1/e^x+1 = e^e/2 = (1 - 1/2 )