Problem 32 - College Algebra with Corequisite Support

Question

For the following exercises, find functions \(f(x)\) and \(g(x)\) so the given function can be expressed as \(h(x)=f(g(x))\). \(h(x)=\frac{1}{\left(3 x^{2}-4\right)^{-3}}\)

Step-by-Step Solution

Answer

Let \(f(u) = u^{-3}\) and \(g(x) = 3x^2 - 4\).

Step 1: Identify the Inner Function

First, we need to observe the structure of the function \(h(x) = \frac{1}{\left(3x^2 - 4\right)^{-3}}\). Notice the expression \(3x^2 - 4\) inside the parentheses. We can define this expression as the inner function. Let \(g(x) = 3x^2 - 4\). This simplifies the argument of the function we are looking to decompose.

Step 2: Identify the Outer Function

Next, we notice the form of the function \(h(x)\) after substituting \(g(x)\) for \(3x^2 - 4\). With \(g(x) = 3x^2 - 4\), the expression becomes \(h(x) = \frac{1}{(g(x))^{-3}}\). This is the same as raising \(g(x)\) to the power of 3, since \(\left(\frac{1}{x^{-3}}\right) = x^{3}\). Therefore, define the outer function as \(f(u) = u^{-3}\), with \(u = g(x)\).

Step 3: Verify the Composition of Functions

Now that we have the functions \(f(u) = u^{-3}\) and \(g(x) = 3x^2 - 4\), verify that the composition \(f(g(x))\) gives us the original function \(h(x)\). Substitute \(g(x)\) into \(f\), giving \(f(g(x)) = f(3x^2 - 4) = (3x^2 - 4)^{-3}\). This matches \(h(x)\) as \(\frac{1}{(3x^2 - 4)^{-3}}\) simplifies to \((3x^2 - 4)^{3}\), confirming the composition is correct.

Key Concepts

Inner Function

To understand the concept of an inner function, let's take a closer look at the function composition process. When examining a complex function, the inner function acts as a stepping stone within the overall composition. It particularly represents the first transformation applied to the variable.

Consider the function we want to decompose: \(h(x) = \frac{1}{(3x^2 - 4)^{-3}}\). Here, identifying the inner function requires us to spot the expression that is first applied to the variable \(x\). This part is located inside the parentheses. In our case, the expression \(3x^2 - 4\) is isolated and assigned as the inner function \(g(x)\). This choice simplifies subsequent operations.

The inner function \(g(x) = 3x^2 - 4\) is essential for simplifying and organizing the overall function into smaller, manageable parts. By recognizing and setting aside this step, we can efficiently manage the complexities involved in more challenging functions.

Outer Function

Once the inner function is identified, it's time to define the outer function. The outer function operates on the result of the inner function. This means that after the inner transformation is applied to \(x\), the outer function provides the final step in shaping the overall function.

In the example of our function, remember that after we define \(g(x)=3x^2-4\), the remaining structure of the expression becomes \(\frac{1}{(g(x))^{-3}}\). To interpret this properly, it helps to see \(\frac{1}{x^{-3}}\) as \((x^3)\). Hence, we define the outer function as \(f(u) = u^3\), with \(u = g(x)\).

The outer function is equally crucial, as it dictates how the inner composition is altered to arrive at the final form \(h(x)\). With both functions pinned down, a clearer comprehension journey is set out, easing the often-daunting task of function composition.

Decomposition of Functions

Decomposition of functions is an ingenious method revealing the layered structure of a complex function. It allows us to break down a cumbersome function into easily interpretable components.

By identifying each part, such as the inner and outer functions, decomposition simplifies the manipulation of functions, making calculations more manageable. Among the advantages of this technique is its effectiveness in verifying the behavior and structure of composite functions.

• Breaking a function into constituent parts makes it easier to handle.
• The method helps in spotting and correcting errors in function composition.
• Decomposition aids in understanding the impact of each stage of transformation on the original function.

Function decomposition is particularly valuable in calculus and algebra, where it forms the foundational skill for tackling more complicated mathematical problems. In our example, decomposing \(h(x)=\frac{1}{(3x^2 - 4)^{-3}}\) into \(f(g(x))\) (\(f(u)=u^3\) and \(g(x)=3x^2-4\)), not only simplifies the function but reinforces our comprehension and analytical capabilities essential for more advanced mathematics.