Question

Suppose H(x) = 2x + 9. Find two functions f and g such that (f.g)(x) = H(x). Neither function can be the identity function. (There may be more than one correct answer.) f(x) = __________ g(x) = __________

Answer 1

Let
`\(f(x) = 2\) and \(g(x) = x + 4.5\).` By multiplying these functions,
`\((f \cdot g)(x)\) results in \(2x + 9\)`, satisfying the given condition for
`\(H(x)\).`Importantly, neither
`\(f\) nor \(g\)` is the identity function, ensuring compliance with the provided instructions. This approach demonstrates the versatility of function composition in representing complex expressions through simpler components.

Step-by-step explanation:

To find functions
`\(f\) and \(g\)` such that
`\((f \cdot g)(x) = H(x) = 2x + 9\),` we need to express
`\(H(x)\)` in terms of
`\(f\) and \(g\)`. Notice that
`\(H(x) = 2x + 9\)` can be factored as
`\(H(x) = 2 \cdot (x + 4.5)\).` To achieve this, set
`\(f(x) = 2\) and \(g(x) = x + 4.5\).` When these functions are multiplied, you get
`\((f \cdot g)(x) = 2 \cdot (x + 4.5) = 2x + 9\)`, which matches the original function
`\(H(x)\)`. It's important to note that neither
`\(f\) nor \(g\)` can be the identity function, ensuring compliance with the given conditions.

This solution respects the requirement that neither function should be the identity function, which could have been
`\(f(x) = x\) or \(g(x) = x\)`. Instead,
`\(f(x) = 2\)` introduces a constant multiplier, and
`\(g(x) = x + 4.5\)`incorporates a linear term, contributing to the correct expansion of
`\(H(x)\).` This mathematical approach demonstrates the flexibility and creativity required when decomposing functions into simpler components.

Understanding the composition of functions is essential in mathematics, allowing for the representation of complex expressions through simpler building blocks. The solution provided adheres to the specified conditions, offering a clear demonstration of how to decompose
`\(H(x)\) into \(f\) and \(g\)` without using the identity function.