Explicație pas cu pas:
[tex]f(x)=(2x-1)(x+1)[/tex]
a)
[tex]f( - 1)=( - 2-1)( - 1+1) = - 3 \cdot 0 = 0 \\ [/tex]
[tex]f(0)=(2 \cdot 0-1)(0+1) = - 1\cdot1 = - 1 \\ [/tex]
[tex]f(x) = (2\cdot1-1)(1+1) = 1\cdot2 = 2 \\ [/tex]
=>
[tex]f(-1)+f(0)+f(1) = 0 + ( - 1) + 2 = - 1 + 2 = 1 \\ [/tex]
b)
[tex]f(x)=0 = > (2x-1)(x+1) = 0 \\ 2x - 1 = 0 = > x = \frac{1}{2} \\ x + 1 = 0 = > x = - 1[/tex]
c)
[tex]f(x) \leqslant 0 = > (2x-1)(x+1) \leqslant 0[/tex]
folosim rezultatul precedent:
[tex]x \in \left[ - 1 ; \frac{1}{2} \right] \\ [/tex]
