Question

Dada a função f:R R tal que f (x)=5x-10, qual o valor de f(-2)+f (5)/f (0)+2.f (1) ?​

Answer 1

f(x)=5x-10

se f(x)=f(-2) -> f(-2)=(5*-2)-10 -> f(-2)=-10-10 -> f(-2)= -20

se f(x)=f(5) -> f(5)=(5*5)-10 -> f(5)= 25-10 -> f(5)= 15

se f(x)=f(0) -> f(0)=(5*0)-10 -> f(0)=-10 -> f(0)= -10

se f(x)=f(1) -> f(1)=(5*1)-10 -> f(1)=-5-10 -> f(1)= -15

[f(-2)+f (5)] / [f (0)+2.f (1)] ->

[(-20)+(15)] / [(-10)+ 2*(-15)] ->

-5 / -40 = 0,125

Answer 2

$f(x) = 5x - 10\\\\f(-2) = 5.(-2) - 10 = - 10 - 10 = -20\\f(5) = 5.5 - 10 = 25 - 10 = 15\\f(0) = 5.0 - 10 = 0 - 10 = -10\\f(1) = 5.1 - 10 = 5 - 10 = -5\\\\f(-2) + \frac{f(5)}{f(0)} + 2.f(1) = -20 + \frac{15}{-10} + 2.(-5) = -20 - \frac{3}{2} - 10 = - 30 - 1,5 = -31,5$