Problem 82 - Precalculus Essentials

Question

Express the given function \(h\) as a composition of two functions \(f\) and \(g\) so that \(h(x)=(f \circ g)(x)\). $$h(x)=\frac{1}{4 x+5}$$

Step-by-Step Solution

Answer

The function \(h(x)\) can be expressed as a composition of the functions \(f(x)=\frac{1}{x}\) and \(g(x)=4x+5\), therefore \(h(x) = (f\circ g)(x) = f(g(x))\).

Step 1: Identify the inner function g(x)

The inner function \(g(x)\) should correspond to the 'inside' part of the original function. This typically involves the variable and its transformations before other operations are applied. Here, the inside part before taking the reciprocal is \(4x+5\), so we identify \(g(x)=4x+5\). This means the given expression can be rewritten as \(\frac{1}{g(x)}\).

Step 2: Identify the outer function f(x)

Now that a function of x has been isolated in the denominator (\(g(x)\)), the 'outer' operation that is remaining in the original function is taking the reciprocal of the input '1/x'. The outer function is therefore \(f(x)=\frac{1}{x}\). This is the step of 'doing' 1/x on \(g(x)\). Therefore, the composition of the functions will be \(f(g(x))=\frac{1}{g(x)}\).

Step 3: Confirm that the Composition is Correct

Substitute \(g(x)\) into \(f(x)\) to confirm that \(f(g(x))\) yields the original function \(h(x)\). In this case, substituting \(4x+5\) into \(f(x)=\frac{1}{x}\) gives \(f(g(x))=\frac{1}{4x+5}\), which is indeed the original function.

Key Concepts

Inner Function

In the context of function composition, the **inner function** is the one that is applied first. It represents the initial transformation or operation performed on the input variable.
In the given problem, the inner function is denoted by \( g(x) \). It is derived from the part of the original function that directly modifies the variable \( x \) before any further operations.

For the function \( h(x) = \frac{1}{4x+5} \), the expression \( 4x+5 \) inside the denominator is identified as the inner function.

• The variable \( x \) undergoes multiplication by 4 and addition of 5.
• We write this inner function as \( g(x) = 4x + 5 \).

It's crucial to pick out the segment of the function formula that directly corresponds to the primary operations on \( x \). This is because it allows us to separate the function's structure into simpler parts.
By correctly identifying the inner function, we can then resolve how it fits into the larger function composition.

Outer Function

The **outer function** is the function that is applied after the inner function. It operates on the result of the inner function. In our example, it is represented by \( f(x) \).
For \( h(x) = \frac{1}{4x+5} \), once the inner function \( g(x) = 4x+5 \) has been processed, we then apply the outer function \( f \) to \( g(x) \).
This involves taking the reciprocal of the inner function's result:

• The outer function transforms any input \( x \) to \( \frac{1}{x} \).
• We define this outer function as \( f(x) = \frac{1}{x} \).

By understanding how the outer function wraps around the result of \( g(x) \), it helps us understand the overall transformation process applied to the input \( x \).
This step-by-step manipulation demonstrates how function composition can simplify complex formulas into manageable calculation processes.

Reciprocal Function

The **reciprocal function** is a specific type of function where the output is the multiplicative inverse of the input.
In mathematical terms, for any non-zero number \( x \), its reciprocal is expressed as \( \frac{1}{x} \).
This concept is crucial in understanding the outer function in our function composition.

In our scenario, the outer function \( f(x) = \frac{1}{x} \) is precisely a reciprocal function. It is used to calculate the reciprocal of the value produced by the inner function \( g(x) = 4x + 5 \).

• This means the output for input \( z = g(x) \) becomes \( f(z) = \frac{1}{z} \).
• Thus, applying this conceptual understanding, \( f(g(x)) = \frac{1}{4x+5} \) straightens out to the full reciprocal transformation.

Reciprocal functions highlight an inverse relationship, one where multiplying an input by its reciprocal results in 1, except when the input is zero (since division by zero is undefined). Understanding these functions helps in the analysis and manipulation of expressions in various areas of algebra and calculus.