Question

Fie functia f:R->R, f(x)= x+2a, x>=2 si 2x⁵+x²-2, x<2. Aratati ca ecuatia f(x)=0 are cel putin o solutie in intervalul (-1;1)​

Answer 1

$f(x) = \begin{cases}x+2a,\quad x\geq 2\\ 2x^5+x^2-2,\quad x<2\end{cases}\\ \\ \\\text{Daca }f(-1)\cdot f(1)<0,\,\,\, \text{atunci }f(x)\text{ are cel putin o solutie}\\ \text{in intervalul }(-1;1).\\ \\ f(-1)\cdot f(1) = \left[2\cdot(-1)^5+(-1)^2-2\right]\cdot (2\cdot 1^5+1^2-2) \\ =(-2+1-2)\cdot (2+1-2) = -3\cdot 1= -3 <0\quad (\checkmark)$