Question

dada a função f(x)+ax+b e sabendo-se que f(3)=5 e f(-2)=_5 calcule F(1/2)

Answer 1

BOA NOITE!

$f(3)=3a+b=5\\\\f(-2)=-2a+b=-5\\\\\left \{ {{3a+b=5} \atop {-2a+b=-5}} \right. \\\\\left \{ {{3a+b=5} \atop {b=-5+2a}} \right. \\\\\\3a+(-5)+2a=5\\\\5a-5=5\\\\5a=10\\\\a=2\\\\b=-5+2(2)\\\\b=-1\\\\\\f(\frac{1}{2})= \frac{1}{2}(2)-1\\\\\ f(\frac{1}{2})= 0$

Answer 2

Resposta:

Explicação passo-a-passo:

coeficiente angular : -5-5 / -2-3 = -10/-5 = 2

f(3)=2.3+b=5

b=5-6

b= -1  

Com a e b é só substituir usando o f(1/2) no lugar de x !

f(1/2)= 2.1/2+ (-1)

f(1/2)= 1-1

f(1/2)= 0