Express the function in the form $$ f \circ g $$. $$ H(x)=\sqrt{4+\sqrt{x}} $$ a. $$ f(x)=\sqrt{x}, g(x)=\sqrt{4+x} $$ b. $$ f(x)=\sqrt{4+x}, g(x)=\sqrt{x} $$ c. $$ f(x)=\sqrt{4-x}, g(x)=x^{2} $$ d. $$ f(x)=\sqrt{x}, g(x)=\sqrt{4-x} $$ e. $$ f(x)=\sqrt{x-4}, g(x)=\sqrt{x} $$

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To express the function $$ H(x) = \sqrt{4 + \sqrt{x}} $$ in the form $$ f \circ g $$, we need to identify functions $$ f $$ and $$ g $$ such that $$ H(x) = f(g(x)) $$.

Let's analyze the function $$ H(x) $$:

1. The outer function appears to be taking the square root of something, which suggests that $$ f(x) $$ could be $$ f(x) = \sqrt{x} $$.
2. The inner function must then be the expression inside the square root, which is $$ 4 + \sqrt{x} $$. This suggests that $$ g(x) $$ should be $$ g(x) = 4 + \sqrt{x} $$.

However, we need to check the provided options to see if any of them match this structure.

Let's evaluate the options:

a. $$ f(x) = \sqrt{x}, g(x) = \sqrt{4+x} $$
   - $$ f(g(x)) = f(\sqrt{4+x}) = \sqrt{\sqrt{4+x}} $$ (not equal to $$ H(x) $$)

b. $$ f(x) = \sqrt{4+x}, g(x) = \sqrt{x} $$
   - $$ f(g(x)) = f(\sqrt{x}) = \sqrt{4+\sqrt{x}} $$ (this matches $$ H(x) $$)

c. $$ f(x) = \sqrt{4-x}, g(x) = x^{2} $$
   - $$ f(g(x)) = f(x^2) = \sqrt{4-x^2} $$ (not equal to $$ H(x) $$)

d. $$ f(x) = \sqrt{x}, g(x) = \sqrt{4-x} $$
   - $$ f(g(x)) = f(\sqrt{4-x}) = \sqrt{\sqrt{4-x}} $$ (not equal to $$ H(x) $$)

e. $$ f(x) = \sqrt{x-4}, g(x) = \sqrt{x} $$
   - $$ f(g(x)) = f(\sqrt{x}) = \sqrt{\sqrt{x}-4} $$ (not equal to $$ H(x) $$)

The only option that correctly expresses $$ H(x) $$ in the form $$ f \circ g $$ is:

**b. $$ f(x) = \sqrt{4+x}, g(x) = \sqrt{x} $$**.
The correct answer is b. $$ f(x) = \sqrt{4+x}, g(x) = \sqrt{x} $$.

Express the function in the form .

a.
b.
c.
d.
e.

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