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(x) = 3x + 1`

`g(x) = (4x + 2)/(3)`


`g[f(0)] = (4[3(0) + 1] + 2)/(3)`

`g[f(0)] = (4[0 + 1] + 2)/(3)`

`g[f(0)] = (4(1) + 2)/(3)`

`g[f(0)] = (4 + 2)/(3)`

`g[f(0)] = (6)/(3)`

`g[f(0)] = 2`

Answer 2

f(0)=1

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

hope helped