Strategie de lucru
Fie [tex]P\in\mathbb{R}[x][/tex] cu [tex]\text{grad}(P)=n[/tex] si fie [tex]Q(x_Q,y_Q)\in\mathbb{R}^2[/tex] oarecare. Vom demonstra afirmatia prin a o reduce la o ecuatie polinomiala de gradul [tex]n[/tex], care nu poate avea mai mult de [tex]n[/tex] solutii.
Geometria problemei
Ma voi folosi, acum, de interpretarea analitica a tangentei. Fie [tex]T(x_T,y_T)\in G_f[/tex], asadar [tex]y_T=f(x_T)[/tex]. Fie Tangenta la [tex]G_f[/tex] prin [tex]T[/tex] are panta egala cu [tex]f'(x_T)[/tex], adica cu derivata functiei evaluata in punctul [tex]x=x_T[/tex] (panta este [tex]\text{tg}\varphi[/tex], unde [tex]\varphi=\angle(Ox, \text{tangenta prin }T)[/tex]). Putem sa observam, geometric (vezi desenul), ca [tex]\text{tg}\varphi=\dfrac{y_Q-y_T}{x_Q-x_Q}[/tex]. Egaland cele doua expresii ale tangentei, obtinem:
[tex]\tan\varphi=\dfrac{y_Q-y_T}{x_Q-x_T}=f'(x_T)\:\:\:\:\:(*)[/tex]
Calculul derivatei si finalizare
Cum [tex]P\in\mathbb{R}[x][/tex] cu [tex]\text{grad}(P)=n[/tex], putem defini [tex]P(X)[/tex] astfel:
[tex]P(X):=\sum_{i=0}^n a_iX^i[/tex]
Prin derivare:
[tex]P'(X)=\sum_{i=1}^n ia_iX^{i-1}[/tex]
Utilizand relatia (*):
[tex]\dfrac{y_Q-y_T}{x_Q-x_T}=\sum_{i=1}^n i\cdot a_i\cdot (x_T)^{i-1}\\\\\iff y_Q-y_T=x_Q\left(\sum_{i=1}^n i\cdot a_i\cdot (x_T)^{i-1}\right)-\sum_{i=1}^n i\cdot a_i\cdot x_T^i\\\\y_T=f(x_T)\\\\\iff x_Q\left(\sum_{i=1}^n ia_ix_T^{i-1}\right)-\sum_{i=1}^n ia_ix_T^i+\sum_{i=0}^n a_ix_T^i-y_Q=0\\\\\iff \sum_{i=0}^n\left[ x_Qia_ix_T^{i-1}+a_ix_T^i(1-i)\right]-y_Q=0[/tex]
Fie [tex]W\in\mathbb{R}[X][/tex] astfel incat [tex]W(X):=\sum_{i=0}^n\left[x_Qia_iX^{i-1}+a_iX^i(1-i)\right]-y_Q[/tex]. Este evident ca [tex]\text{grad}(W)=n[/tex], asadar ecuatia [tex]W(X)=0[/tex] are cel mult [tex]n[/tex] solutii. In concluzie, exista cel mult [tex]n[/tex] valori ale lui [tex]x_T[/tex], si deci cel mult [tex]n[/tex] puncte [tex]T\in G_f[/tex], astfel incat [tex]QT[/tex] e tangenta la [tex]G_f[/tex] in punctul [tex]T[/tex].
