Question

$a(x) = {x}^{3} - {x}^{2} + x - 1 \: e \: b(x) = - 3 + x + 2$
calcule
a)
$a \frac{1}{2} - b( - 1)$
b)
$a(0) + b(1)$

Answer 1

$a( \frac{1}{2}) = \frac{1}{2}^{3} - { \frac{1}{2} }^{2} + \frac{1}{2} - 1 \\\\ a( \frac{1}{2} ) = \frac{1}{8} - \frac{1}{4} + \frac{1}{2} - 1 \\ \\ a( \frac{1}{2} ) = - \frac{1}{8} - \frac{1}{2} = - \frac{5}{8}$
$b( - 1) = - 3 - 1 + 2 = - 2$

$a( \frac{1}{2} ) - b( - 1) \\\\ - \frac{5}{8} - ( - 2) = - \frac{5}{8} + 2 \\\\ \frac{11}{8}$


$a(0) = {0}^{3} - {0}^{2} + 0 - 1 = - 1$
$b(1) = - 3 + 1 + 2 = 0$

$a(0) + b(1) = - 1 + 0 = - 1$