Problem 64 Find \(f(x)\) and \(g(x)\) such ... [FREE SOLUTION] (2024)

Short Answer

inner function

The inner function, often denoted as \( g(x) \), is the function which is fed into another function. In our given problem, the goal is to identify which part of \( h(x) \) can serve as this inner function. Here, we chose \( g(x) = 7x + 2 \). This is a simpler part of \( h(x) \), making it easier to use in the composition. The inner function is called 'inner' because it is the first operation that acts on our input \( x \). By breaking down complex functions into an inner and outer function, we make them easier to analyze and work with.

outer function

The outer function \( f(x) \) is the function that takes the result of the inner function as its input. In our example, once we have \( g(x) = 7x + 2 \), we rewrite the original function in terms of this \( g(x) \). That means, \( h(x) = \frac{1}{\sqrt{g(x)}} \). Hence, the outer function becomes \( f(x) = \frac{1}{\sqrt{x}} \). The outer function essentially processes the result of \( g(x) \). It's called 'outer' because it is applied to the output of the inner function.

function composition

Function composition is the process of combining two functions where the output of one function becomes the input of another. Represented as \( (f \circ g)(x) = f(g(x)) \), function composition allows us to build more complex functions from simpler ones.
In our exercise, we composed \( f(x) \) and \( g(x) \) to reconstruct the given function \( h(x) \). By identifying \( g(x) = 7x + 2 \) and \( f(x) = \frac{1}{\sqrt{x}} \), we confirmed that: \( (f \circ g)(x) = f(g(x)) = f(7x + 2) = \frac{1}{\sqrt{7x + 2}} \), which matches \( h(x) \). Understanding function composition is key in mathematics, as it simplifies complex scenarios into manageable steps.

verification of composite function

Verification ensures that the composition of the identified functions accurately reconstructs the original function. To verify, we substitute the inner function \( g(x) \) into the outer function \( f(x) \).
In our case, we substitute \( g(x) = 7x + 2 \) into \( f(x) = \frac{1}{\sqrt{x}} \), obtaining: \( f(g(x)) = \frac{1}{\sqrt{7x + 2}} \). This calculation confirms that \( (f \circ g)(x) = h(x) = \frac{1}{\sqrt{7x + 2}} \). Verification is crucial because it validates our initial choice of \( f(x) \) and \( g(x) \), ensuring they indeed form the original function upon composition.