Să se determine f : R -> R, f(x) = ax + b, dacă f(5)= 3, f (-10) = 0.
Rezolvare:
[tex]\it f(x)=ax+b\ \ \ \ \ (*)\\ \\ \left.\begin{aligned} \it f(5)=3\ \stackrel{(*)}{\Longrightarrow}5a+b=3 \Rightarrow b=3-5a\\ \\ \it f(-10)=0\ \stackrel{(*)}{\Longrightarrow}-10a+b=0 \Rightarrow b=10a\end{aligned}\right\}\ \Rightarrow\ 10a=3-5a \Rightarrow\\ \\ \\ 10a+5a=3 \Rightarrow 15a=3 \Rightarrow a=\dfrac{\ 3^{(3}}{15} \Rightarrow a=\dfrac{1}{5} \Rightarrow a=0,2\\ \\ b=10a=10\cdot0,2=2\\ \\ f(x)=0,2x+2[/tex]
