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

2 Answers

4 votes

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
User Mille
by
7.4k points
3 votes
f(0)=1

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

hope helped
User Kostas Rousis
by
7.6k points

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