Problem 90 - Precalculus : Building Concepts and Connections

Question

In Exercises \(87-96,\) find two functions \(f\) and \(g\) such that \(h(x)=(f \circ g)(x)=f(g(x)) .\) Answers may vary. $$h(x)=\sqrt[5]{-x^{3}+8}$$

Step-by-Step Solution

Answer

The two functions \(f\) and \(g\) that satisfy the condition \(h(x) = (f \circ g)(x) = f(g(x))\) where \(h(x) = \sqrt[5]{-x^{3}+8}\) are \(f(x) = \sqrt[5]{x}\) and \(g(x) = -x^{3}+8\).

Step 1 - Identify the Outer Function

Observe \(h(x) = \sqrt[5]{-x^{3}+8}\). Here, it can be seen that the outer function is the fifth root of a certain value. Thus, let \(f(x) = \sqrt[5]{x}\).

Step 2 - Identify the Inner Function

The inner function is the value that is inside the outer function. This is the equation under the fifth root in \(h(x)\). Therefore, let's designate the inner function as \(g(x) = -x^{3}+8\).

Step 3 - Confirm the Function Composition

Substitute \(g(x)\) into \(f(x)\) to confirm that the composition of the functions equals \(h(x)\). So, \(f(g(x)) = f(-x^{3}+8)\) becomes \(\sqrt[5]{-x^{3}+8}\) which matches with \(h(x)\).

Key Concepts

Outer Function

In function composition, the outer function plays the role of wrapping around the inner one. Imagine it like a gift box holding a hidden treasure. In the context of the given problem, the outer function is the part that applies an operation to the result of another function (the inner function). For the equation \(h(x) = \sqrt[5]{-x^{3}+8}\), the outer function can be represented as \(f(x) = \sqrt[5]{x}\).
This fifth root function \(f(x) = \sqrt[5]{x}\) processes the value it receives, turning it into its fifth root. It acts last in the sequence of operations, making it the outermost shell in our composition.
By identifying the outer function first, we recognize the overarching operation that defines the final shape and output of our composite function \(h(x)\). This understanding is key to interpreting how complex functions interact.

Inner Function

The inner function is what gets processed first in a composition. It sets the stage for the outer function, dictating the initial modification or transformation. In our example, \(g(x) = -x^{3}+8\) acts as the inner function within the composite function \(h(x) = (f \circ g)(x)\).
Think of the inner function as the core operation that occurs before any high-level transformations handled by the outer function. In this case, \(g(x)\) takes an input \(x\) and modifies it using the formula \(-x^{3} + 8\). This operation prepares the data for the final transformation performed by the outer function, \(f(x) = \sqrt[5]{x}\).
Understanding the inner function helps you deconstruct complex equations into understandable parts, making problem-solving much more approachable.

Fifth Root

The fifth root is a mathematical function where the target of operation is divided into five equal parts, analogous to raising a number to the power of \(1/5\). In the expression \(h(x) = \sqrt[5]{-x^{3}+8}\), the use of a fifth root is integral to defining the outer function \(f(x) = \sqrt[5]{x}\).
Just like squaring a number has its counterpart in finding a square root, cubing has a partner in cube roots, and so on. The fifth root is another member of this family of root functions, offering a powerful tool for problem-solving. It allows equations to reverse the process of quintic exponentiation, where a variable is raised to the fifth power.
In precalculus, roots, including the fifth root, are crucial for simplifying and solving polynomial equations, offering deep insights into the properties and relationships across complex mathematical expressions.

Precalculus

Precalculus is a branch of mathematics that prepares students for calculus by exploring various types of functions, equations, and foundational concepts. Understanding function composition, like the one in our example, is critical in precalculus.
This area of study serves as a stepping stone, broadening foundational knowledge and introducing advanced functions before diving into calculus' challenging concepts.

• Analyzing functions and their compositions involves breaking down complex expressions into manageable parts.
• Precalculus builds familiarity with different types of functions, such as exponential, logarithmic, and trigonometric functions.
• It plays a pivotal role in developing skills necessary for mathematical modeling and data interpretation.
  By practicing problems in precalculus, like the composition of an inner and outer function, students gain the ability to tackle more complicated calculus problems in the future with confidence and competence.