fungsi f(x)=x+3 dan (fog)(x)=x²+6x+7 maka g(-1)=...

[tex](fog)(x) = {x}^{2} + 6x + 7 \\ f(g(x)) \: \: = {x}^{2} + 6x + 7 \\ g(x) + 3 = {x}^{2} + 6x + 7 \\ g(x) \: \: \: \: \: \: \: \: = {x}^{2} + 6x + 7 - 3 \\ g(x) \: \: \: \: \: \: \: \: = {x}^{2} + 6x + 4 \\ g( - 1) \: \: \: \: \: = { - 1}^{2} + 6( - 1) + 4 \\ g( - 1) \: \: \: \: \: = 1 - 6 + 4 \\ g( - 1) \: \: \: \: \: = - 1[/tex]