Esercitazione semplice Analisi 1

h.05.2026

8 ⋅ 6x + x⁴ - x³ + x₃² < sup> ≤ 0

x² - 8x + 24 x² + 8

(8 ⋅ 6x + 12x - x³ - x ²) (x + 1) ≤ 0

2 (x₃² + 1x - x², x + 2)

8 ⋅ 6x + 4x - x² - x³ ≤ 0

2 (x₁² , x² - x₁)

-x - 5 ≤ 0

(x₃²,x - 3) (2x - x + 5p + x ) ≤ 0

(x₄,x¹) (x - 3)

M₁

U₂ 0

D -

-

x ≤ 1 ∨ x > 3

2_x6

x₁, 12x + 12² , 2_x6

2k²x + 40 ≤ ≤ 0

(9x - 5x - h) ≤

2x - 1 + h ≤ 32 ≤ ≤ 0

4.05.2026

8 ⋅ 6x - x² - x³ + ⁵a

x³ + 8

8 ⋅ 6x ⋅ x ⋅ y - x

x³ + 8

8 ⋅ ( x³-3x²+a² ⋅ 3 - 9 )

2 8(x²+h) ⋅ x³

8 ⋅ x ⋅ 4x - x³

2(x² + h) (x - 3) (x - 3)

- x ⋅ x² - 5x + 5 < 0

(x² + h) (x - 3)

(- x² - 5) (x-1) < 0

x² + h (x - 3)

U₁ - x² > 5 -> x² <-5

U₂ x > 1

D x > 3

m u 3

M₁ - - -

U₂ - ○ ↑ ↓

D - ∅ +

x ≤ 1 ∨ x > 3

2h (x ⋅ u - x² + 3x + 16)

8 x² + h (5x-h)

12 ⋅ 12 ⋅ x² + 9x ⋅ 2

2h ⋅ x²

x ⋅ 6

2 ⋅ x ⋅ 4 + 12

2k ⋅ x²

12x ⋅ 2

40x + 32 ≤ 0

12 - 2x ⋅ 12x + 6 ⋅ 32 ≤ 0

2 x ⋅ 6 2x ⋅ 3 x ⋅ 2 ≥ 0

Δ = h ⋅ h (√ ⋅ ≤ 32)

x₁, x₂, -17 ⋅ 52

x_1,2 = -l a ⋅ h² ⋅ x ⋅ (√)

x - 1 ⋅ x ⋅ h = 52 - 12

(2 8 ꓱ 3 r)

55

3k x² - 6x + 3k - 8 = 0

Δ

= b² - 4ac > 0

= 36 - h (3k) (3k - 8) > 0

= 36 - 12k (3k - 8) > 0

= 36 - 36k² + 96k > 0

= 6 - 6k² + 16k > 0

= 3 - 2k² + 8k > 0

= 3k² + 8k + 3 ≥ 0

= 3k² - 8k - 3 ≤ 0

Δ = 6h - h (-9)

= 6h + 36 = 400

vk_1,2 = 8 ± √0 = 3

                6

= 8 ± 0 = -½ = -⅓

        6

-⅓ ≤ k ≤ 3

                  3     4

A

B

( x₁, x₂ )² > 1

x₁, x₂ = c = 3k - 8

                a                3k

^{(3k - 8)}(3k)² > 1

k ≠ 0

(3k - 8)²(> 3k)²

9k² + 64 - 48k > 9k²

- 48k > -64

k < \frac⁸

k < \frac^8k

- \frac¹ 3

^{- \frac13 ≤ k < \frac8}

            3

- \frac¹3       \frac^h3      3

72

3

x³ + 16x⁵ - 5x

x x³ + 16 x⁴ - 12x³

49 - x³ + 16 x⁶h 0

4n h x³ + 16 x⁴ 0

h x x x = 0

16 x³ (8 + x) = 0

x_i:0 V x:

√5 - x + √x = 3

5 - x ≥ 3

- x ≥ - 5

x ≤ 5

x ≥ 0

√5 - x = 3 - √x

5 - x = 9 + x - 6 √x

6√x = 9 + x - 5

6√x = h + 2x

3√x = 2 + x

9 x = h + x + h

- x² + 9x - hx - h 0

- x² + 5x - h = 0

x² - 5x + h = 0

x - x - h x + h = 0

(x - l) h (x - l) = 0

x₁ h x₂ - opp.

8x - x ≥ 8

x ≤ 8

x ≥ 2

2 ≤ x ≤ 8

8 - x = 25 + 3x - 3

25(3x - 3) = 49 + hx² + 8x

75x - 75 - 49 - hx² - 28x = 0

-hx² + 4hx - 12h = 0

hx² - 4hx + 12h = 0

Δ = h² - h(h)(12h)

x1,2 = h ± 15

203

Diam. NB = 8cm

base circolare = 2x

(CD) = 20 cm

8 - 2x =

h & other

20 = 8 + 2x + 2

32 - 8x

12 - 2x = 2

AD² = 16 - x²

AD = ^√32 - 8x

(x - 2)² = 0

AC = 20 cm

CH = 10√3 cm

CF perpendicolare AC

perimetro e area

A_ABC = ^{base ⋅ h}/₂

A_ADC = ^{AC ⋅ BC}/₂

A_ABC = ^{AB ⋅ CH}/₂

CH = 10√3

10√3 = ^{20 BC}/_AB

20 ⋅ BC = 10√3 AB

2BC = √3 AB

BC = ^√3/₂ ⋅ AB

AC² + BC² = AB²

20² + ( ^√3/₂ AB )² = AB²

400 + ³/₄ AB² = AB²

400 = AB² ( 1 - ³/₄ )

400 = ^AB²/₄

AB² = 4 ⋅ 400

( 400 kg )

AB = 40