Question

Determinati valorile lui a
apartine lui R pentru care punctul M(a,2a+1)€Gf,unde f:R-R este definita prin f(x)=1pe3x+2

Answer 1

[tex]f:R \rightarrow R,~f(x)=\frac{1}{3}x+2\\~ G_f = \{(x,f(x)):x \in R\}\\~\\~ M(a,2a+1) \in G_f \iff f(a) = 2a+1\\~ \frac{1}{3}a+2 = 2a+1\\~ \frac{1}{3}a-2a+1 = 0\\~ \frac{-5a}{3}+1=0\\~ \frac{-5}{3}a=-1\\~ \frac{5}{3}a=1\\~ a = \frac{3}{5}[/tex]

Am obținut punctul $M(a,2a+1)=M(\frac{3}{5},\frac{6}{5}+1)=M(\frac{3}{5},\frac{11}{5})$