Given $$ f(x)=\sqrt{2 x} $$ and $$ g(x)=3 x+4 $$. After simplifying, a) $$ (f \circ g)(x)=\square $$ b) $$ (g \circ f)(x)=\square $$

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To find the compositions $$ (f \circ g)(x) $$ and $$ (g \circ f)(x) $$, we will substitute one function into the other.

### a) $$ (f \circ g)(x) $$

This means we need to compute $$ f(g(x)) $$.

1. Start with $$ g(x) = 3x + 4 $$.
2. Now substitute $$ g(x) $$ into $$ f(x) $$:
   $$
   f(g(x)) = f(3x + 4) = \sqrt{2(3x + 4)}
   $$
3. Simplify the expression:
   $$
   f(g(x)) = \sqrt{6x + 8}
   $$

Thus, 
$$
(f \circ g)(x) = \sqrt{6x + 8}
$$

### b) $$ (g \circ f)(x) $$

This means we need to compute $$ g(f(x)) $$.

1. Start with $$ f(x) = \sqrt{2x} $$.
2. Now substitute $$ f(x) $$ into $$ g(x) $$:
   $$
   g(f(x)) = g(\sqrt{2x}) = 3(\sqrt{2x}) + 4
   $$
3. Simplify the expression:
   $$
   g(f(x)) = 3\sqrt{2x} + 4
   $$

Thus, 
$$
(g \circ f)(x) = 3\sqrt{2x} + 4
$$

### Final Answers:
a) $$ (f \circ g)(x) = \sqrt{6x + 8} $$

b) $$ (g \circ f)(x) = 3\sqrt{2x} + 4 $$
a) $$ (f \circ g)(x) = \sqrt{6x + 8} $$ b) $$ (g \circ f)(x) = 3\sqrt{2x} + 4 $$

Given \( f(x)=\sqrt{2 x} \) and \( g(x)=3 x+4 \). After simplifying, a) \( (f \circ g)(x)=\square \) b) \( (g \circ f)(x)=\square \)

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