ࡱ> -/,9 bjbj.FlB.000000$e TTZiZZZ^.Z.ZZ.. `Z..0.%Z%.ZEs.1  EMBED Equation.3  e  EMBED Equation.3  oppure  EMBED Equation.3  se  EMBED Equation.3  e  EMBED Equation.3  e  EMBED Equation.3  se  EMBED Equation.3  e  EMBED Equation.3  Es.2 Una possibile soluzione di questo esercizio costituita dalla dimostrazione della non esistenza della funzione richiesta tramite lapplicazione iterata del teorema di Rolle. Ricordo, infatti, che il teorema di Rolle afferma che: Dato  EMBED Equation.3  , con  EMBED Equation.3  derivabile  EMBED Equation.3  e se  EMBED Equation.3 , allora  EMBED Equation.3  tale che  EMBED Equation.3 . Pertanto si pu dimostrare che una funzione derivabile due volte e che si annulla in tre punti distinti in un intervallo chiuso e limitato debba necessariamente avere la derivata seconda che si annulla in un punto, quindi non pu essere sempre positiva, come richiesto dallesercizio. Infatti, si ha che: Se  EMBED Equation.3  derivabile due volte e tale che  EMBED Equation.3 , in modo che  EMBED Equation.3 , allora dal teorema di Rolle, applicato allintervallo  EMBED Equation.3 ,dove la funzione derivabile, considerando che la continuit della funzione in  EMBED Equation.3  implica che  EMBED Equation.3 , si ha che  EMBED Equation.3 tale che  EMBED Equation.3 . Analogamente, applicando il teorema di Rolle allintervallo  EMBED Equation.3 , si ha che EMBED Equation.3 tale che  EMBED Equation.3 . Inoltre, essendo  EMBED Equation.3  EMBED Equation.3 = EMBED Equation.3 , si ha che  EMBED Equation.3 . Si pu pertanto applicare il teorema di Rolle alla funzione  EMBED Equation.3 , che derivabile una volta e tale che  EMBED Equation.3 . Applicando tale teorema a questa funzione nellintervallo  EMBED Equation.3 , si ha che EMBED Equation.3  tale che  EMBED Equation.3 , ovvero  EMBED Equation.3 . Allora non possibile che  EMBED Equation.3  EMBED Equation.3 . Es.3 SOLUZIONE I: Una possibile dimostrazione del fatto che la funzione richiesta da questo esercizio non esiste potrebbe essere data sfruttando il teorema del limite di funzioni composte. Tale teorema afferma che: Data  EMBED Equation.3 , con  EMBED Equation.3  e  EMBED Equation.3  punto di accumulazione per  EMBED Equation.3  e data  EMBED Equation.3 , con  EMBED Equation.3  e  EMBED Equation.3  punto di accumulazione per il dominio di  EMBED Equation.3 , se si ha che  EMBED Equation.3  e  EMBED Equation.3 , dove  EMBED Equation.3 , e vale che a)  EMBED Equation.3  con  EMBED Equation.3  oppure b)  EMBED Equation.3 intorno di  EMBED Equation.3  tale che  EMBED Equation.3   EMBED Equation.3  Allora:  EMBED Equation.3  Utilizzando questo teorema si pu dimostrare lincompatibilit delle ipotesi del nostro esercizio, infatti: Se si considera  EMBED Equation.3  , per la definizione di funzione inversa si ha:  EMBED Equation.3 = EMBED Equation.3 =  EMBED Equation.3 . Ma, utilizzando il teorema del limite delle funzioni composte applicato alla funzione EMBED Equation.3 e alla sua inversa con  EMBED Equation.3  e  EMBED Equation.3  , poich dalla continuit della funzione si ha che EMBED Equation.3 , facendo valere lipotesi (a), e usando il limite dato dalle ipotesi dellesercizio, si ha che  EMBED Equation.3 = EMBED Equation.3 . Pertanto, dallunicit del limite, si dovrebbe avere che il valore della funzione in un punto sia uguale a pi infinito, che assurdo. Quindi la funzione richiesta non pu esistere. SOLUZIONE II: Alternativamente, si potrebbe affermare la non esistenza della funzione richiesta dallesercizio, trovando una contraddizione fra le ipotesi nel modo seguente: Poich  EMBED Equation.3  continua e  EMBED Equation.3  deve essere punto di accumulazione per il dominio della sua inversa,  EMBED Equation.3 con  EMBED Equation.3 . Poich  EMBED Equation.3  continua,  EMBED Equation.3  si ha che  EMBED Equation.3  un intervallo limitato, perci  EMBED Equation.3  tali che  EMBED Equation.3 , allora  EMBED Equation.3 . Poich  EMBED Equation.3  iniettiva, si ha che  EMBED Equation.3  da cui  EMBED Equation.3 . Se fosse vero che  EMBED Equation.3 , allora  EMBED Equation.3  dovrebbe esistere  EMBED Equation.3  tale che  EMBED Equation.3  EMBED Equation.3 . In particolare, se indico con  EMBED Equation.3 , si ha  EMBED Equation.3   EMBED Equation.3  e  EMBED Equation.3  in contraddizione fra loro. Es.4 Una possibile soluzione di questo esercizio pu essere costituita dalle funzioni:  EMBED Equation.3  tale che  EMBED Equation.3  e  EMBED Equation.3 , perci  EMBED Equation.3  e  EMBED Equation.3 ,  EMBED Equation.3  tale che  EMBED Equation.3  e  EMBED Equation.3 , perci  EMBED Equation.3  e  EMBED Equation.3 . Infatti, si tratta di funzioni reali a variabile reale, continue e derivabili, che si annullano in zero e in uno e tali che il vettore direzionale della retta tangente alla prima funzione, nel punto zero, sia (1,1), mentre quello della retta tangente alla seconda funzione, nello stesso punto, sia (-1, 1). Questi due vettori sono fra loro perpendicolari. In modo analogo, nel punto (1,0), la prima funzione ha retta tangente avente vettore direzionale (-1,1), mentre la seconda (1,1), fra loro perpendicolari. Infatti, la retta tangente alla prima funzione in zero la retta  EMBED Equation.3 , mentre quella tangente alla seconda funzione  EMBED Equation.3 . Nel punto (1,0), invece, la retta tangente alla prima funzione  EMBED Equation.3 , mentre quella tangente alla seconda  EMBED Equation.3 . 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