To express [tex]\( H(x) = \sqrt{4x + 3} \)[/tex] as a composition of two functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] such that [tex]\((f \circ g)(x) = H(x)\)[/tex], we need to identify appropriate [tex]\( f \)[/tex] and [tex]\( g \)[/tex] for which this relationship holds.
Let's break this down step-by-step:
1. Identify an intermediate function, [tex]\( g(x) \)[/tex], that simplifies the argument of the square root in [tex]\( H(x) \)[/tex].
2. Define [tex]\( f(x) \)[/tex] in such a way that it operates on the output of [tex]\( g(x) \)[/tex] to yield the final [tex]\( H(x) \)[/tex].
### Step 1: Define [tex]\( g(x) \)[/tex]
We notice that inside the square root in [tex]\( H(x) = \sqrt{4x + 3} \)[/tex], we have the expression [tex]\( 4x + 3 \)[/tex]. We can set [tex]\( g(x) \)[/tex] to simplify this part.
So, let [tex]\( g(x) = 4x + 3 \)[/tex].
### Step 2: Define [tex]\( f(x) \)[/tex]
Now, our goal is to find [tex]\( f(x) \)[/tex] such that [tex]\( f(g(x)) = H(x) \)[/tex]. Therefore, [tex]\( f \)[/tex] should be a function that, when applied to [tex]\( g(x) \)[/tex], gives us [tex]\( H(x) \)[/tex].
Since [tex]\( H(x) = \sqrt{4x + 3} \)[/tex] and we have [tex]\( g(x) = 4x + 3 \)[/tex], applying the square root to [tex]\( g(x) \)[/tex] should give us [tex]\( H(x) \)[/tex].
Therefore, let [tex]\( f(x) = \sqrt{x} \)[/tex].
### Verification
To confirm, we will compose these functions and see if it results in [tex]\( H(x) \)[/tex]:
[tex]\[ (f \circ g)(x) = f(g(x)) \][/tex]
[tex]\[ = f(4x + 3) \][/tex]
[tex]\[ = \sqrt{4x + 3} \][/tex]
[tex]\[ = H(x) \][/tex]
Hence, the functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] that satisfy [tex]\((f \circ g)(x) = H(x)\)[/tex] are:
[tex]\[ f(x) = \sqrt{x} \][/tex]
[tex]\[ g(x) = 4x + 3 \][/tex]
Therefore, the correct functions are:
[tex]\[ f(x) = \sqrt{x} \][/tex]
[tex]\[ g(x) = 4x + 3 \][/tex]
