Fie \( n\in\mathbb{N},n\ge 2 \) si \( f:\mathbb{R}\rightarrow \mathbb{R} \) cu proprietatea:
\( f(x+y)+f(x-y)=f(nx)+nx,\forall x\in\mathbb{R} \).
Determinati \( f \).
Ecuatie functionala
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BogdanCNFB
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Daca \( y=(n-1)x => x+y=nx;\ x-y=(2-n)x. \)
Prin inlocuire: \( f(nx)+f(x(2-n))=f(nx)+nx => f(x(2-n))=nx. \)
Vom face substitutia \( x(2-n)=t => x= \frac{t}{2-n} => f(t)=\frac{nt}{2-n} \) (aici am considerat cazul cand \( n>2 \)).
Revenind la variabila x: \( f(x)=\frac{nx}{2-n} \).
Considerand celalalt caz (cand \( n=2 \)) se obtine ca \( f(0)=nx \), de unde \( f(0) \) are o infinitate de valori, ceea ce e fals.
Sper sa fie corect.
Prin inlocuire: \( f(nx)+f(x(2-n))=f(nx)+nx => f(x(2-n))=nx. \)
Vom face substitutia \( x(2-n)=t => x= \frac{t}{2-n} => f(t)=\frac{nt}{2-n} \) (aici am considerat cazul cand \( n>2 \)).
Revenind la variabila x: \( f(x)=\frac{nx}{2-n} \).
Considerand celalalt caz (cand \( n=2 \)) se obtine ca \( f(0)=nx \), de unde \( f(0) \) are o infinitate de valori, ceea ce e fals.
Sper sa fie corect.