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Forum Studenti •Esercizio funzioni monotone
Pagina 1 di 1

Esercizio funzioni monotone

Inviato: sabato 22 settembre 2012, 15:57
da Noisemaker
[tex]\mbox { Sia }\,\,\, f:(-\infty,0)\to\mathb{R}\,\,\,\mbox{una unzione continua ed iniettiva tale che}[/tex]

[tex]\displaystyle\lim_{x \to -\infty}f(x)=-\infty.[/tex]

[tex]\mbox { Provare che } f \mbox {risulta crescente }[/tex]

Io ho seguito questo ragionamento qui:

si tratta di provare che, [tex]\forall\,\,x_1,x_2 \in (-\infty,0)[/tex] si ha che [tex]x_1<x_2 \,\,\Rightarrow\,\,f(x_1)<f(x_2)[/tex]

Allora per ipotesi la funzione è continua ed iniettiva, e ciò significa che è certamente (strettamente) monotona: infatti poichè ogni funzione strettamente monotona è iniettiva, basterebbe fare vedere che [tex]f[/tex] è invertibile (lo è per ipotesi di iniettività) implica [tex]f[/tex] strettamente monotona. Alllra essendo strettamente monotona, dobbiamo dimostrare che essa è monotona crescente; a tale scopo utilizziamo l'ipotesi del limite, cioè:

[tex]\displaystyle\lim_{x \to -\infty}f(x)=-\infty.[/tex]

questo significa, per definizione, che:

[tex]\forall \varepsilon>0,\,\, \exists\delta>0 : \,\,f(x)<-\varepsilon,\,\,\,\forall x\in (-\infty,0) : x<-\delta[/tex]

e dunque

[tex]x_1<x_2 \,\,\Rightarrow\,\,f(x_1)<f(x_2)<-\varepsilon,\quad \forall x\in (-\infty,0)[/tex]

e dunque la funzione risulta monotona(strettamente) crescente.

Re: Esercizio funzioni monotone

Inviato: sabato 22 settembre 2012, 20:49
da Massimo Gobbino
:!: :?: :!: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock: :shock:

Re: Esercizio funzioni monotone

Inviato: domenica 23 settembre 2012, 15:50
da Noisemaker

Re: Esercizio funzioni monotone

Inviato: giovedì 27 settembre 2012, 22:14
da Noisemaker
la funzione non puo decrescere per la regolarità delle funzioni monotone, cioè se [tex]f[/tex] fosse decrescente nell'intervallo [tex](-\infty,0)[/tex] allora l'estremo superiore sarebbe [tex]-\infty[/tex]

poiche


[tex]\displaystyle \sup_{]-\infty ,0[} f=\lim_{x\to -\infty} f(x) =-\infty[/tex]

ma cio è impossibile ...... perchè il [tex]\sup f=0[/tex]