Calcolo differenziale

f(x) = - x^2 + 3·x + 1

g(x) = 1/3·x^2 + 1/3·x + 1

0 < x < 2

f(2) = - 2^2 + 3·2 + 1=3

f(0)= - 0^2 + 3·0 + 1= 1

3 - 1 = 2

g(2) = 1/3·2^2 + 1/3·2 + 1= 3

g(0) = 1/3·0^2 + 1/3·0 + 1= 1

3 - 1 = 2

f'(x) = 3 - 2·x

g'(x) = 2·x/3 + 1/3

Th di Cauchy:

(3 - 2·x)/(2·x/3 + 1/3) = 2/2

3·(3 - 2·x) = 2·x + 1----> x = 1

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f(1) = - 1^2 + 3·1 + 1----> f = 3

[1, 3]

g(1) = 1/3·1^2 + 1/3·1 + 1-----> g = 5/3

[1, 5/3]

1^retta tangente:

y - 3 = 1·(x - 1)---> y = x + 2

2^ retta tangente:

y - 5/3 = 1·(x - 1)---> y = x + 2/3

{y = x + 2

{y = 1/3·x^2 + 1/3·x + 1

risolvo:

[x = -1 ∧ y = 1, x = 3 ∧ y = 5]

A [-1, 1]

B [3, 5]

{y = x + 2/3

{y = - x^2 + 3·x + 1

Risolvo:

[x = 2·√3/3 + 1 ∧ y = 2·√3/3 + 5/3, x = 1 - 2·√3/3 ∧ y = 5/3 - 2·√3/3]

C [2·√3/3 + 1, 2·√3/3 + 5/3]

D [1 - 2·√3/3, 5/3 - 2·√3/3]

A [-1, 1]

Determino area:

Α = 1/2·ABS((- 1·5 + 3·(2·√3/3 + 5/3) + (2·√3/3 + 1)·(5/3 - 2·√3/3) + (1 - 2·√3/3)·1) - (- 1·(5/3 - 2·√3/3) + (1 - 2·√3/3)·(2·√3/3 + 5/3) + (2·√3/3 + 1)·5 + 3·1))

Α = 8·(√3 + 3)/9