sa se determine functia f:R->R, f(x)=ax+b, cu a si b apartine lui R ,pentru care f(1)+f(2)+f(3)=6a+bsi f(5)=15

f(x)=ax+b

f(1)=a+b

f(2)=2a+b

f(3)=3a+b

f(1)+f(2)+f(3)=a+b+2a+b+3a+b=6a+3b

Si avem ecuatia:

6a+3b=6a+b

6a+3b-6a-b=0

2b=0

b=0

f(5)=5a+b

5a+b=15

Si avem sistemul:

[tex] \left \{ {{b=0} \atop {5a+b=15}} \right. [/tex]

[tex] \left \{ {{b=0} \atop {5a=15}} \right. [/tex]

[tex] \left \{ {{b=0} \atop {a=3}} \right. [/tex]

Finalizare:

f: R->R, f(x)=3x