Problem 42 - College Algebra

Question

Express the given function h as a composition of two functions f and g so that \(h(x)=(f \circ g)(x)\) $$h(x)=\sqrt{5 x^{2}+3}$$

Step-by-Step Solution

Answer

The functions \( f \) and \( g \) that compose \( h \) are: \(g(x) = 5x^2 + 3\) and \(f(x) = \sqrt{x}\).

Step 1: Identify the Inner Function

Notice the inner function under the square root, \(5x^2+3\). This will be \(g(x)\). So we let \(g(x)= 5x^2 +3\).

Step 2: Identify the Outer Function

The outer function will be the square root function. This can be represented as: \(f(x)=\sqrt{x}\).

Step 3: Verify the Composition

Now test if the composition of the functions \(f\) and \(g\) equals the original function \(h\). To form the composition \(f(g(x))\), we substitute \(g(x)\) into \(f(x)\) to get \(f(g(x)) = \sqrt{5x^2 +3}\), which is indeed \(h(x)\).

Key Concepts

Understanding the Inner Function

The inner function is an essential component in function composition. It forms the basis that the outer function acts upon. In the given exercise, the inner function is identified within the expression under the square root. Here, it is the expression

• \(g(x) = 5x^2 + 3\)

This inner function transforms the input \(x\) by first squaring it, then multiplying by 5, and finally adding 3. It's like laying down the foundation for the whole expression. The inner function's value is what we will subject to further transformation by the outer function.

Specifically, you can think of the inner function as setting up whatever comes inside the square root. The outer function's job is to "finish" the process of calculating \(h(x)\). Understanding this task division helps clarify why the composition of functions works the way it does.

Grasping the Outer Function

Outer functions are those functions that wrap around inner functions, performing computations on their results. In this problem, the outer function takes the form of a square root, signified by:

• \(f(x) = \sqrt{x}\)

The role of the outer function is to process the output of the inner function. It applies its own rule to the result from the inner function, which in this composition is taking the square root.

By understanding its role, you can appreciate how both the inner and outer functions contribute to the final result. The outer function encapsulates the entire process of composition, ensuring the correct result \(h(x) = \sqrt{5x^2 + 3}\) in this case. Each part of a composed function serves a purpose and when used together, they create a seamless operation.

Exploring the Square Root Function

The square root function is a common mathematical function that is vital for many types of calculations, including those involving composition. It is represented as \(\sqrt{x}\) and acts as a type of outer function in this context. When we think about the square root function as part of a composition, it simplifies the complexity of the previous transformations done by the inner function.

Understanding the square root itself is straightforward:

• It takes a non-negative input and outputs its positive square root.
• Mathematically, given by \(f(x) = \sqrt{x}\).

When applied in composition, like in this exercise, the square root function helps us "polish" the result obtained from the inner workings. The final expression \(h(x) = \sqrt{5x^2 + 3}\) is both elegant and a perfect example of how the square root can be used to finalize complex compositions straightforwardly.