Problem 3 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Let \(j(x)=\frac{2}{3 \sqrt{4 x^{2}+3 x}} .\) Suppose that \(j(x)=h(g(f(x))) .\) Write possible formulas for \(f(x), g(x)\), and \(h(x) .\) None of \(f, g\), and \(h\) should be the identity function.

Step-by-Step Solution

Answer

The functions \(f(x)\), \(g(x)\), and \(h(x)\) that satisfy \(j(x) = h(g(f(x)))\) where none of them should be the identity function are \(f(x) = 4x^{2}+3x, g(x) = \frac{\sqrt{f(x)}}{3}\) and \(h(x) = \frac{2}{g(x)}\).

Step 1: Splitting into Composite Functions

We start off by looking at the function \(j(x)\), and observe that it is a composite of three simpler functions. It divides by 2, then takes the reciprocal, and also has within it the square root of a quadratic function. Thus, we can split the function \(j(x)\) as follows: \(j(x)=h(g(f(x)))\) where \(f(x), g(x)\), and \(h(x)\) are chosen such that when compounded, they give \(j(x)\).

Step 2: Defining \(f(x)\)

Let's start with the inside. The quadratic function inside the square root. We choose \(f(x)\) to be the function that generates this polynomial. Hence, here we have \(f(x) = 4x^{2}+3x\).

Step 3: Defining \(g(x)\)

The next step is to find \(g(x)\). Looking at \(j(x)\), after computing the quadratic, we have a square root and then a division by 3. So, we can define \(g(x) = \frac{\sqrt{f(x)}}{3}\), where \(f(x)\) is our previously defined function.

Step 4: Defining \(h(x)\)

Finally, we need to find \(h(x)\). Looking at \(j(x)\), after the square root and the division by 3, we see that a division / multiplication by 2 is performed. This is our \(h(x)\). So we define \(h(x) = \frac{2}{g(x)}\), where \(g(x)\) is our previously defined function.

Key Concepts

Calculus

When we talk about calculus, we are referring to a branch of mathematics that studies how things change. It's all about rates of change (differentiation) and accumulation of quantities (integration). To make sense of complex functions like j(x), calculus often breaks them down into simpler parts—this is where the concept of composite functions comes into play. In the exercise, we essentially performed a calculus operation by decomposing j(x) into a combination of easier functions called f(x), g(x), and h(x). This kind of problem helps build a deep understanding of how more complicated equations can be tackled piece by piece—a fundamental skill in calculus.

Quadratic Function

A quadratic function is a second-degree polynomial, typically written in the form f(x) = ax^2 + bx + c. Quadratic functions create parabolas when graphed and are fundamental in algebra and calculus. The exercise provided j(x) which included a quadratic expression under a square root. Recognizing this, we identified f(x) = 4x^2 + 3x as our quadratic component. Because quadratics are so prevalent across various applications, understanding their properties and how to manipulate them, as we did by isolating f(x), is crucial.

Square Root

The square root function is another core concept in mathematics, and it is exactly what it sounds like: finding a number that, when multiplied by itself, gives the original number. In the context of our function j(x), the square root interacts with the quadratic function inside of it. In the solution, we identified the innermost function f(x) and then applied a square root next, through g(x), to peel away the layers of our composite function. Understanding how to handle square roots—especially when they're part of more complex expressions—is essential in calculus.

Reciprocal Function

Lastly, the reciprocal function is simply the inverse of a given number or function, which means we take 1 divided by the number or function. For our exercise, h(x) = 2/g(x) explains how we turned the output of g(x) into its reciprocal and then multiplied by 2. This kind of operation is commonplace when working with rates or in situations where inversion is naturally part of the model we're working with. Recognizing and manipulating reciprocal functions is a handy skill in mathematics, helping simplify complex expressions into more manageable forms.