Explicație pas cu pas:
[tex]f(x) = (1 - x)(3x + 1)[/tex]
a)
[tex]f(x) = (1 - x)(3x + 1) = 3x + 1 - 3 {x}^{2} - x \\ = 2x + 1 - 3 {x}^{2} = - 3 {x}^{2} + 2x + 1[/tex]
b)
[tex]f( - \frac{1}{2} ) = - 3 {( - \frac{1}{2})}^{2} + 2( - \frac{1}{2}) + 1 = - \frac{3}{4} - 1 + 1 = - \frac{3}{4} \\ [/tex]
[tex]f(0) = - 3 {(0)}^{2} + 2(0) + 1 = 0 + 0 + 1 = 1 \\ [/tex]
[tex]f(\frac{1}{2} ) = - 3 {(\frac{1}{2})}^{2} + 2(\frac{1}{2}) + 1 = - \frac{3}{4} + 1 + 1 = 2 - \frac{3}{4} = \frac{5}{4} \\ [/tex]
=>
[tex]f( - \frac{1}{2} ) + f(0) + f(\frac{1}{2} ) = - \frac{3}{4} + 1 + \frac{5}{4} =\frac{2}{4} + 1 = \frac{3}{2} \\ [/tex]
c)[tex]f(x) = 0[/tex]
[tex](x - 1)(3x + 1) = 0[/tex]
[tex]x - 1 = 0 = > x = 1[/tex]
[tex]3x + 1 = 0 = > x = - \frac{1}{3} \\ [/tex]
d)
[tex]f(x) \geqslant 0 = > (1 - x)(3x + 1) \geqslant 0 \\ < = > (x - 1)(3x + 1) \leqslant 0[/tex]
ne folosim de rezultatul de la punctul c)
funcția are valori negative între rădăcini
[tex] = > x \in \left[ - \frac{1}{3} ; 1 \right] \\ [/tex]
