It is C because Since {{g(x)}={x^3+2}}g(x)=x3+2g, left parenthesis, x, right parenthesis, equals, x, start superscript, 3, end superscript, plus, 2, we can substitute {x^3+2}x3+2x, start superscript, 3, end superscript, plus, 2 in for {g(x)}g(x)g, left parenthesis, x, right parenthesis.\begin{aligned}{f(g(x))}&=3(g(x))-1 \\\\ &=3({x^3+2})-1 \\\\ &=3x^3+6-1\\\\ &=3x^3+5 \end{aligned}f(g(x))=3(g(x))−1=3(x3+2)−1=3x3+6−1=3x3+5
