Question

EX 35 VA ROG MULT!!!

Answer 1

a) f:R->R; f(x)=2x+3

g:R->R; g(x)=2+x

Gf∩Gg={A}

=> f(x)=g(x)

2x+3=2+x

2x-x=2-3 => x=-1; il inlocuim pe x

y=f(-1)=g(-1)=-2+3=2-1=1

=> A(-1; 1)

b) f:{0;1;2;3;4} ->R; f(x)=x-4

g:{0;2;4;6}->R; g(x)=-x

Gf∩Gg={B}

=> f(x)=g(x)

x-4=-x

2x=4 => x=2

y=f(2)=g(2)=2-4=-2

=> B(2; -2)

c) f:R->R; f(x)=2x-4

g:R->R; g(x)=2x+1

2x-4=2x+1

2x-2x=1+4

0=5; fals => graficele celor doua functii nu au niciun punct comun; sunt paralele

Gf || Gg; Gf∩Gg=Ф