Questions: Find functions f and g so that f ∘ g=H. H(x)=sqrt(x^2+16) Choose the correct pair of functions. A. f(x)=sqrt(x)-16, g(x)=x^2 C. f(x)=x^2, g(x)=sqrt(x)-16 B. f(x)=sqrt(x), g(x)=x^2+16 D. f(x)=x^2+16, g(x)=sqrt(x)

Solution Steps

To find the correct pair of functions \( f \) and \( g \) such that \( f \circ g = H \), we need to check each pair of functions by composing them and seeing if the result matches \( H(x) = \sqrt{x^2 + 16} \).

We are given the function \( H(x) = \sqrt{x^2 + 16} \) and need to find functions \( f \) and \( g \) such that \( f \circ g = H \). We will evaluate the following pairs of functions:

• Pair A: \( f(x) = \sqrt{x} - 16 \), \( g(x) = x^2 \)
• Pair B: \( f(x) = \sqrt{x} \), \( g(x) = x^2 + 16 \)
• Pair C: \( f(x) = x^2 \), \( g(x) = \sqrt{x} - 16 \)
• Pair D: \( f(x) = x^2 + 16 \), \( g(x) = \sqrt{x} \)

We will check each pair to see if \( f(g(x)) = H(x) \).

• For Pair A: \[ f(g(x)) = f(x^2) = \sqrt{x^2} - 16 = x - 16 \quad \text{(not equal to } H(x)\text{)} \]

• For Pair B: \[ f(g(x)) = f(x^2 + 16) = \sqrt{x^2 + 16} \quad \text{(equal to } H(x)\text{)} \]

• For Pair C: \[ f(g(x)) = f(\sqrt{x} - 16) = (\sqrt{x} - 16)^2 \quad \text{(not equal to } H(x)\text{)} \]

• For Pair D: \[ f(g(x)) = f(\sqrt{x}) = (\sqrt{x})^2 + 16 = x + 16 \quad \text{(not equal to } H(x)\text{)} \]

From the evaluations, only Pair B satisfies the condition \( f \circ g = H \). Therefore, the correct functions are:

• \( f(x) = \sqrt{x} \)
• \( g(x) = x^2 + 16 \)

Final Answer