Funzioni inverse
Considerando $y = g(x)$:
[*:6vnqdk3h] Se $g(x)$ è strettamente crescente, allora $g(x)<=y rarr x <= g^{-1}(y)$[/*:m:6vnqdk3h]
[*:6vnqdk3h] Se $g(x)$ è strettamente decrescente, allora $g(x)<=y rarr x >= g^{-1}(y)$[/*:m:6vnqdk3h][/list:u:6vnqdk3h]
Ho un piccolo vuoto di memoria sulle funzioni inverse. Qualcuno di voi può spiegarmi queste due implicazioni?
Risposte
Ti invito alla lettura del punto 7.4 a pg. 90-91 della di seguito indicata dispensa di Analisi Matematica I, del Prof. Marco Degiovanni insegnante presso l'Università Cattolica del Sacro Cuore (UCSC) ftp://ftp.dmf.unicatt.it/pub/users/degi ... nt/ami.pdf.
These inequalities should be strict. And, in fact, they are equivalences. I'll explain you the increasing case. The another one is analogous.
A function $f:\mathbb{R}\to\mathbb{R}$ is strictly increasing if for all $x
Let $f$ be an increasing function. Then, if $x\in\mathbb{R}$ and $y\in \text{Im}(f)$, then you have that there exists $x'\in\mathbb{R}$ such that $f(x')=y$. If you had $f(x)
Indeed, we know that $x'\ne x$ because of the fact that $f$ is STRICTLY increasing. So we have that $x'
If you repeat the argument with $f^{-1}$ in place of $f$ then (I think) you'll get the converse.
A function $f:\mathbb{R}\to\mathbb{R}$ is strictly increasing if for all $x
Let $f$ be an increasing function. Then, if $x\in\mathbb{R}$ and $y\in \text{Im}(f)$, then you have that there exists $x'\in\mathbb{R}$ such that $f(x')=y$. If you had $f(x)
Indeed, we know that $x'\ne x$ because of the fact that $f$ is STRICTLY increasing. So we have that $x'
If you repeat the argument with $f^{-1}$ in place of $f$ then (I think) you'll get the converse.
Accedi a tutti gli appunti
Tutor AI: studia meglio e in meno tempo