For the real-valued functions $f(x)=\sqrt{3x+12}$ and $g(x)=x-1$, find the composition $f$ g and specify its domain using interval notation.
$$(f-g)(x)=$$

Step 1 ：Given the functions $f(x)=\sqrt{3x+12}$ and $g(x)=x-1$, we can find the composition $f(g(x))$ as follows:

Step 2 ：$f(g(x))=f(x-1)=\sqrt{3(x-1)+12}$

Step 3 ：Simplify the expression inside the square root:

Step 4 ：$f(g(x))=\sqrt{3x-3+12}=\sqrt{3x+9}$

Step 5 ：The domain of a function is the set of all possible input values (x-values) which will output real numbers. The square root function is only defined for values greater than or equal to zero. Therefore, we set the expression inside the square root greater than or equal to zero and solve for x:

Step 6 ：$3x+9\ge 0$

Step 7 ：Subtract 9 from both sides:

Step 8 ：$3x\ge -9$

Step 9 ：Divide by 3:

Step 10 ：$x\ge -3$

Step 11 ：Therefore, the domain of the function $f(g(x))=\sqrt{3x+9}$ is $[-3,\mathrm{\infty })$ in interval notation.

Step 12 ：So, the composition $f(g(x))=\sqrt{3x+9}$ and its domain is $\boxed{{\displaystyle [-3,\mathrm{\infty })}}$.