Question

sendo f(x)
$- \frac{ 15}{ {x}^{2} } - \frac{14}{ {x}^{3} } + \frac{48}{ {x}^{4} } + \frac{32}{ {x}^{5} }$
calcule f(2).​

Answer 1

$F(x)=-\frac{15}{x^{2} } -\frac{14}{x^{3} } +\frac{48}{x^{4} } +\frac{32}{x^{5} }$

*vamos substituir todos os x por 2.

$F(2)=-\frac{15}{2^{2} } -\frac{14}{2^{3} } +\frac{48}{2^{4} } +\frac{32}{2^{5} }$

$F(2)=-\frac{15}{4} -\frac{14}{8} +\frac{48}{16} +\frac{32}{32}$

*vamos simplificar os ternos

$F(2)=-\frac{15}{4} -\frac{7}{4 } +3+1$

$F(2)=\frac{-15-7}{4} +4$

$F(2)=\frac{-22}{4} +4$

$F(2)= \frac{-11}{2} +4$

*MMC

$F(2)=\frac{-11+8}{2}$

$F(2)=-\frac{3}{2}$