Question

alguém me ajuda nessa questão​

Answer 1

Resposta:

Explicação passo-a-passo:

São dados da questão:

$f(x)=\dfrac{3}{5}x-1\\\\g(x)=\dfrac{4}{3}x+a\\\\\\f(0)-g(0)=\dfrac{1}{3}$

$f(x)=\dfrac{3}{5}x-1\\\\f(0)=\dfrac{3}{5}.0-1\\\\f(0)=-1$

$g(x)=\dfrac{4}{3}x+a\\\\g(0)=\dfrac{4}{3}.0+a\\\\g(0)=a$

Então

$f(0)-g(0)=\dfrac{1}{3}\\\\-1-a=\dfrac{1}{3}\\\\-1-\dfrac{1}{3}=a\\\\a=-\dfrac{3}{3}-\dfrac{1}{3}\\\\a=-\dfrac{4}{3}$

$f(x)=\dfrac{3}{5}x-1\\\\f(3)=\dfrac{3}{5}.3-1\\\\f(3)=\dfrac{9}{5}-1\\\\f(3)=\dfrac{9}{5}-\dfrac{5}{5}\\\\f(3)=\dfrac{4}{5}$

$g(x)=\dfrac{4}{3}x+a\\\\g(x)=\dfrac{4}{3}x-\dfrac{4}{3}\\\\\\g(\dfrac{1}{5})=\dfrac{4}{3}.\dfrac{1}{5}-\dfrac{4}{3}\\\\g(\dfrac{1}{5})=\dfrac{4}{15}-\dfrac{4}{3}\\\\g(\dfrac{1}{5})=\dfrac{4-20}{15}\\\\\\g(\dfrac{1}{5})=-\dfrac{16}{15}\\\\\\Assim\\\\\\3.g(\dfrac{1}{5})=3.(-\dfrac{16}{15})\\\\\\3.g(\dfrac{1}{5})=-\dfrac{16}{5}\\\\\\Portanto\\\\\\f(3)-3.g(\dfrac{1}{5})=\dfrac{4}{5}-(-\dfrac{16}{5})\\\\f(3)-3.g(\dfrac{1}{5})=\dfrac{4}{5}+\dfrac{16}{5}$

$f(3)-3.g(\dfrac{1}{5})=\dfrac{20}{5}\\\\\\f(3)-3.g(\dfrac{1}{5})=4$