Question

Given that f(x) = 3x + 1 and g(x) = the quantity of 4x plus 2 divided by 3, solve for g(f(0))

Answer 1

F(0)=1

g(1)=(4*1+2)/3 = 6/3=2

hope helped 

Answer 2

$f(x) = 3x + 1$
$g(x) = \frac{4x + 2}{3}$

$g[f(0)] = \frac{4[3(0) + 1] + 2}{3}$
$g[f(0)] = \frac{4[0 + 1] + 2}{3}$
$g[f(0)] = \frac{4(1) + 2}{3}$
$g[f(0)] = \frac{4 + 2}{3}$
$g[f(0)] = \frac{6}{3}$
$g[f(0)] = 2$