Funzioni inverse

[robe92]

Considerando $y = g(x)$:

• Se $g(x)$ è strettamente crescente, allora $g(x)<=y rarr x <= g^{-1}(y)$
• Se $g(x)$ è strettamente decrescente, allora $g(x)<=y rarr x >= g^{-1}(y)$

Ho un piccolo vuoto di memoria sulle funzioni inverse. Qualcuno di voi può spiegarmi queste due implicazioni?

[luca97xd]

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.

[javicemarpe]

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 $x<y$ you have that $f(x)<f(y)$. A strictly increasing function must be invertible because of the injectivity.

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 $f(x)<y$ then you would have $f(x)<f(x')$ and, as $f$ is strictly increasing, you must have $x<x'$.

Indeed, we know that $x'\ne x$ because of the fact that $f$ is STRICTLY increasing. So we have that $x'<x$ or $x<x'$. But if we had $x'<x$, then we would have $f(x')<f(x)$, which is impossible because we know that $f(x)<f(x')$. Then, the only possibility for $x'$ is to satisfy the inequality $x<x'$, i.e., $x<f^{-1}(y)$.

If you repeat the argument with $f^{-1}$ in place of $f$ then (I think) you'll get the converse.