f(x)=x³-12x=a
fie functia g(x)=x³-12x-a
Aplici terema Intre 2 radacini consecutive ale ecuatiei g(x)=0 se afla cel putin o radacina a derivatei
g `(x)=3x²-12
g `(x)=0 x1=-2 x2=2
Deci x1=2 se ala intre radacnile g1 si g2 si x2=2 se afla intre radacinile g2 si g3
Deci ecuatia g(x)=0 are radacinile g1, g2, g3.∀a∈R =>
Ecuatia f(x)=a admite 3 solutii reale pt ∀a
egzemplu
a=1
x³-12x=1 admite solutiille
x1∈(-∞,-2) x2∈(-2,2), x3>2
