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Forum Studenti •Esercizio 2 del compito 10 gennaio 2009
Pagina 1 di 1

Esercizio 2 del compito 10 gennaio 2009

Inviato: giovedì 19 luglio 2012, 19:08
da PLA
x*arctan(e^x) = c con c= parametro reale.

Bisogna valutare quante soluzioni ha l'equazione al variare del parametro reale.
Risolvendo i limiti per x-> +/- oo, ho trovato che la funzione va ad infinito per x tendente a +infinito, mentre va a 0- per x tendente a meno infinito.
La funzione vale 0 per x=0 e la derivata prima calcolata in questo punto è maggiore di zero, quindi nel grafico avrò che proprio in zero la funzione attraverserà l'asse x per passare a valori negativi e per poi tendere nuovamente a zero per x-> -oo
Sempre considerando la derivata prima della funzione, posso facilmente dimostrare che per x>0 quest'ultima è sempre positiva. Ne deduco che non si annulli mai e che la funzione sia monotona crescente nel primo quadrante.
Per le informazioni ricavate fino ad ora sul grafico mi aspetterei di trovare un punto di minimo per un valore di x negativo.
Ma non riesco a ricavarlo in nessun modo...
Come faccio a dimostrare quindi che esistono dei valori per x<0 per cui la funzione ha almeno due soluzioni e che oltre un certo valore (non noto) la funzione non ha nessuna soluzione?

Re: Esercizio 2 del compito 10 gennaio 2009

Inviato: venerdì 18 gennaio 2013, 21:37
da ilaria11
Non c'è bisogno di trovare il punto di minimo, l'importante è dimostrare che c'è. La traccia ti chiede di determinare se esistono λ<0 per cui l'equazione ha almeno due soluzioni e se esistono λ<0 per cui l'equazione non ha soluzioni.

Re: Esercizio 2 del compito 10 gennaio 2009

Inviato: sabato 19 gennaio 2013, 9:51
da Massimo Gobbino