f ( x ) = { 2 x 2 + ( a + 3 b − 2 ) x + ( 2 a − b − 1 ) x < 0 1 + ln ( 1 − x + 3 x 2 ) x ≥ 0
č derivabile in x=0.
Continuitā: f − ( 0 ) = lim x → 0 − ( 2 x 2 + ( a + 3 b − 2 ) x + ( 2 a − b − 1 ) ) = 2 a − b − 1 e f + ( 0 ) = 1 + ln ( 1 ) = 1 .
Posto f − ( 0 ) = f + ( 0 ) si ricava 2 a − b − 1 = 1
Derivabilitā.
f ' ( x ) = { 12 x 2 + a + 3 b − 2 x < 0 6 x − 1 1 − x + 3 x 2 x≥0 .
f ' − ( 0 ) = lim x → 0 − ( 12 x 2 + a + 3 b − 1 ) = a + 3 b − 1 e f ' + ( 0 ) = − 1 .
Posto f ' − ( 0 ) = f ' + ( 0 ) si ricava a + 3 b − 1 = 1
Si deve risolvere il sistema lineare: { 2 a − b = 2 a + 3 b = 2 → 6 a − 3 b + a + 3 b = 12 + 2 → 7 a = 14 → a = 2
e b = 2 a − 2 = 4 − 2 = 2