Question

f:R-R,f(x)=(x-1)(x-2)(x-3)+2
c)Dem. ca f'(x)=0 are exact 2 radacini reale​

Answer 1

Răspuns:

f(x)=(x-1)(x-2)(x-3)+4=

(x²-3x+2)(x-3)+4

f `(x)=(x²-3x+2) `(x-3)+(x²-3x+2)(x-3) `+4 `=

(2x-3)(x-3)+(x²-3x+2)*1+0=

2x²-3x-6x+9+x²-3x+2=

3x²-12x+11

f `(x)=0

3x²-12x+11=0

Ecuatie  de gradul 2.Daca disriminantul Δ>0, ecuatia admite 2 radacini reale

Δ=(-12)²-4*3*11=

144-132=12>0

Ecuatia f `(x)=0 are 2 radacini reale ∀x∈R

Explicație pas cu pas: