Deoarece litera cu care se noteaza variabila nu conteaza
faci substitutia t=x
g(x)=2x-5
fsi g fiind inverse
fog(x)=gof(x)=x
gof(x)=g(f(x))=2f(x)-5=2(ax+b)-5=x
2ax-x+2b-5=0 x(2a-1)+2b-5=0 (ll
Aplici metoda identificarii
2a-1=0 =>a=1/2
2b-5=0 2b=5
b=5/2
f(x=1/28x+5/2
