mentok kawan, tolong bantu

[tex]\displaystyle g(3)=3+3\\g(3)=6\\\\(f\circ g)(3)=f(g(3))\\(f\circ g)(3)=f(6)\\(f\circ g)(3)=\sqrt{2\cdot6-6\sqrt3}\\(f\circ g)(3)=\sqrt{12-6\sqrt3}\\(f\circ g)(3)=\sqrt{9+3-2\sqrt{9\cdot3}}\\(f\circ g)(3)=\sqrt9-\sqrt{3}\\\boxed{\boxed{(f\circ g)(3)=3-\sqrt{3}}}[/tex]

Matematika SMA
Fungsi Komposisi dan Invers

Pembahasan :
Fog(x) = f(g(x))
Fog(x) = √[2(x + 3) - (x + 3)√3]
Fog(3) = √[2(3 + 3) - (3 + 3)√3]
Fog(3) = √[2(6) - 6√3]
Fog(3) = √[12 - 6√3]
Fog(3) = √[12 - 2.3√3]
Fog(3) = √[12 - 2√27]

Ingat !
Bentuk √[a + b - 2√ab] = √a - √b

Maka...
Fog(3) = √[12 - 2√27]
Fog(3) = √[9 + 3 - 2√9.3]
Fog(3) = √9 - √3
Fog(3) = 3 - √3