Question

dada a função real definida por f(x) = x²+x, calcule: f(1)+3f(-1)-2f(1/3)

Answer 1

f(x)=x²+x

vamos calcular separados

f(1)=1²+1=1+1=2

f(-1)=(-1)²+(-1)=1-1=0

$f( \frac{1}{3} )=( \frac{1}{3} )^2+ \frac{1}{3} = \frac{1}{9}+ \frac{1}{3} = \frac{1+3}{9} = \frac{4}{9}$

agora vamos calcular

$f(1)+3f(-1)-2f( \frac{1}{3} )= \\ \\ 2+3(0)-2( \frac{4}{9}) = \\ \\ 2+0- \frac{8}{9} = \\ \\ 2- \frac{8}{9} = \frac{18-8}{9} = \fbox{ \frac{10}{9} }$