Fie f:R-R o funcție continuă cu f(2017)=2016
Dacă f(x)f(f(x))=1 ptr orice x real
Calculați f(2015)
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Please briefly explain why you feel this question should be reported.
Va rugam explicate, pe scurt, de ce credeti ca aceasta intrebare trebuie raportata.
Motivul pentru care raportezi utilizatorul.
Cum
,f continua fie
un interval.
este tot un interval.
sau ![Rendered by QuickLaTeX.com 2017\notin f\left ( \mathbb{R} \right )](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-a011fb9a9022003489cbadcb46516678_l3.png)
![Rendered by QuickLaTeX.com 2017\in f\left ( \mathbb{R} \right )\Rightarrow \exists x^{'}\in \mathbb{R}a.i.f\left ( x^{'} \right )=2017](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-dbe9f4096db160f44e4bd85b3b6d3878_l3.png)
![Rendered by QuickLaTeX.com f\left ( x \right )f\left ( f\left ( x \right ) \right )=1,\forall x\in \mathbb{R}](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-d94e18a2f96b8e6568c5bf8cbc7430c6_l3.png)
rezulta
![Rendered by QuickLaTeX.com f\left ( x^{'} \right )f\left ( f\left ( x^{'} \right ) \right )=1](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-c8892fc7c1bfa7b7f541ef1720fe5dcc_l3.png)
![Rendered by QuickLaTeX.com 2017f\left ( 2017 \right )=1](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-0e2aa34877f4bdf524c24e4ad0822fea_l3.png)
,atunci avem :
, ceea ce este fals.
si cum
este interval ,rezulta ca:
sau
![Rendered by QuickLaTeX.com \left ( 2017,+ \infty \right )\subset f\left ( \mathbb{R} \right )](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-04c15fde6203debb2321496439996fef_l3.png)
, rezulta ca :
![Rendered by QuickLaTeX.com \left ( -\infty ,2017 \right )\subset f\left ( \mathbb{R} \right )](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-233c8109d1dd5c8125c1f371f85b4cf1_l3.png)
![Rendered by QuickLaTeX.com f\left ( x \right )f\left ( f\left ( x \right ) \right )=1,\forall x\in \mathbb{R}](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-d94e18a2f96b8e6568c5bf8cbc7430c6_l3.png)
obtinem ![Rendered by QuickLaTeX.com f\left ( 2016 \right )=\frac{1}{2016}](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-b09ceec3996fb00510ebcc3e6b4d1971_l3.png)
, adica este in imaginea functiei , rezulta ca
astfel incat :
![Rendered by QuickLaTeX.com f\left ( \lambda \right )=2015](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-a43414c46b174f84866fa8c9075dd3e5_l3.png)
![Rendered by QuickLaTeX.com x=\lambda \Rightarrow f\left ( \lambda \right )f\left ( f\left ( \lambda \right ) \right )=1\Rightarrow 2015f\left ( 2015 \right )=1\Rightarrow f\left ( 2015 \right )=\frac{1}{2015}](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-dcb8d53ce8c24701040d11bfe0f2d37e_l3.png)
Atunci
Ne punem problema daca
Presupunem ca :
Cum:
pentru
Deci :
si cum
Prin urmare :
dar cum din enunt avem :
Din
pentru
Cum
Pentru
Multumesc foarte mult
Nu inteleg de ce este important sa arat ca 2017 nu aparține lui f( x)