Question

Demonstrați ca -5≤f(x)≤9,pentru orice x €(aparține )[-1,1]
f:R->R, f(x) =x ^3+6x+2

Answer 1

f'(x) =3x²+6≥6>0,∀x∈R⇒∀x∈[-1;1]⊂R⇒f(x) strict crescatoare pe [-1;1]
⇒min f(x) =f(-1) =-1-6+2=-7+2=-5
si
max f(x) =f(1)=1+6+2=9
⇒-5≤f(x)≤9
as simple as that!!!