f ( x ) = { a · e x 2 − 4 x≥2 1 + b x 2 + ( x − 2 ) 3 x<2
sia derivabile in x=2.
Continuitą: f − ( 2 ) = lim x → 2 − ( 1 + b x 2 + ( x − 2 ) 3 ) = 1 + 4 b e f + ( 2 ) = a
Posto f − ( 2 ) = f + ( 2 ) si ricava 1 + 4 b = a
Derivabilitą.
f ' ( x ) = { 2 a x · e x 2 − 4 x < 2 3 ( x − 2 ) 2 + 2 b x x≥2 .
f ' − ( 2 ) = lim x → 2 − ( 2 a x · e x 2 − 4 ) = 4 a e f ' + ( 2 ) = 4 b .
Posto f ' − ( 2 ) = f ' + ( 2 ) si ricava 4 a = 4 b → a = b
Si deve risolvere il sistema lineare: { 1 + 4 b = a a = b → 1 + 4 a = a → 3 a = − 1 → a = − 1 3
e b = − 1 3