Question

socoooroo!

considere as funções r(x)=2x+1 e g(x)=(X-3)2

(a)= r[g(3)]

(b)= g[r(-1)]

(c)= g[g(3)]

(d)= r[r(5]

Answer 1

$r(x) = 2x+1\\\\g(x) = (x-3)^2$

(a)= r[g(3)]calculando g(3) -> (substitui x por 3)  $g(3) = (3-3)^2=0$

r[g(3)] -> substitui o x na funçao r..pelo o valor de g(3) que é 0

$r[g(3) ] = 2*0 +1 = 1$
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(b)= g[r(-1)]
$r(-1) = 2*(-1) +1\\\\r(-1) = -2+1 = -1$

substituindo x por -1 na g(x)
$g[r(-1)] = (-1-3)^2\\\\g[r(-1)] = -4^2 = 16$


(c)= g[g(3)]
$g(3) = (3-3)^2 = 0$

então
$g[g(3)] = (0-3)~2 \\\\g[g(3)] = -3^2 = 9$

(d)= r[r(5]

$r(5) = 2*5 +1 = 11$

$r[r(5)] = 2*11 + 1 = 23$