Question

URGENTE, POR FAVOR!!! Seja a função (anexo) determine f (-2) - f (1) +2 f (2) / f(-1)

Answer 1

$f(x)=\frac{-x^2+x-5}{-x^3+2}$

$f(-2)=\frac{-(-2)^2+(-2)-5}{-(-2)^3+2}$

$f(-2)=\frac{-4-2-5}{-(-8)+2}$

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

$f(-2)=-\frac{11}{10}$

$f(-1)=\frac{-(-1)^2+(-1)-5}{-(-1)^3+2}$

$f(-1)=\frac{-1-1-5}{-(-1)+2}$

$f(-1)=\frac{-7}{1+2}$

$f(-1)=-\frac{7}{3}$

$f(1)=\frac{-1^2+1-5}{-1^3+2}$

$f(1)=\frac{-1+1-5}{-1+2}$

$f(1)=\frac{-5}{1}$

$f(1)=-5$

$f(2)=\frac{-2^2+2-5}{-2^3+2}$

$f(2)=\frac{-4+2-5}{-8+2}$

$f(2)=\frac{-7}{-6}$

$f(2)=\frac{7}{6}$

Agora podemos determinar aquilo que o exercício pede:

$\frac{f(-2)-f(1)+2f(2)}{f(-1)}=$

$[f(-2)-f(1)+2f(2)]\div f(-1)=$

$[-\frac{11}{10}-(-5)+2\cdot \frac{7}{6}]\div (-\frac{7}{3})=$

$[-\frac{11}{10}+5+\frac{14}{6}]\div (-\frac{7}{3})=$

$[-\frac{33}{30}+ \frac{150}{30}+ \frac{70}{30}]\div (-\frac{7}{3})=$

$\frac{187}{30}\div(-\frac{7}{3})=$

$\frac{187}{30}\cdot (-\frac{3}{7})=$

$-\frac{561}{210}=$

$-\frac{187}{70}$