Problem 89 - A Graphical Approach to Precalculus with Limits

Question

Consider the function \(h\) as defined. Find functions \(f\) and \(g\) such that \((f \circ g)(x)=h(x)\). (There are several possible ways to do this.) $$h(x)=\sqrt{x^{2}-1}$$

Step-by-Step Solution

Answer

Choose \(f(x) = \sqrt{x}\) and \(g(x) = x^2 - 1\).

Step 1: Understanding the Composition

We need to find functions \(f(x)\) and \(g(x)\) such that \((f \circ g)(x) = h(x) = \sqrt{x^2 - 1}\). In a composition, \((f \circ g)(x) = f(g(x))\). This means that the function \(g(x)\) will be the inner function and \(f(x)\) will be the outer function applied to the result of \(g(x)\).

Step 2: Choosing the Inner Function \(g(x)\)

Let's find a function \(g(x)\) such that it produces an expression that could logically be inside a square root. In this case, we select \(g(x) = x^2 - 1\). This simplifies our target function \(h(x)\) by isolating the operations inside the square root.

Step 3: Defining the Outer Function \(f(x)\)

Now that we have \(g(x) = x^2 - 1\), we need an outer function \(f(x)\) that, when applied to \(g(x)\), yields \(h(x)\). Thus, we choose \(f(x) = \sqrt{x}\), since \(f(g(x)) = \sqrt{g(x)} = \sqrt{x^2 - 1}\), which is exactly \(h(x)\).

Step 4: Verification of the Function Composition

Verify by substituting \(g(x)\) into \(f(x)\): \[ (f \circ g)(x) = f(g(x)) = f(x^2 - 1) = \sqrt{x^2 - 1} \]. This confirms that \(f(x) = \sqrt{x}\) and \(g(x) = x^2 - 1\) are the correct functions that meet the condition \((f \circ g)(x) = h(x)\).

Key Concepts

Inner Function

When dealing with function composition, the concept of the "inner function" is essential. In mathematical expressions, the inner function is the function that is "nested" inside another function. For example, when you have a composition

• \( (f \circ g)(x) \), it means \( f(g(x)) \).

Here, \(g(x)\) is considered the inner function. It operates on the input \(x\) first before any other function is applied. To define an inner function, think about what needs to happen first in the order of operations.

In our exercise, we chose \(g(x) = x^2 - 1\) as the inner function. This decision was based on isolating the operations we want to perform before applying the square root.

• This step transforms the input \(x\) into something more manageable (\(x^2 - 1\)),
• setting the stage for further processing by the next function.

By understanding the role of the inner function, you can simplify complex compositions and tackle them one step at a time.

Outer Function

The "outer function" takes on the important task of refining or transforming the result of an inner function. In essence, it is the last step in the process for defining composed functions. When you see

• \( (f \circ g)(x) \), the outer function is \(f(x)\).

The outer function must be chosen strategically to attain the desired final expression from an inner function's output.

In our problem, we defined the outer function as \(f(x) = \sqrt{x}\). The responsibility of \(f(x)\) is to take the result of \(g(x) = x^2 - 1\) and then apply the square root to it:

• This is the essential last step which will provide us with \(h(x) = \sqrt{x^2 - 1}\),
• matching the originally targeted function.

Clearly understanding the purpose and application of the outer function in function composition helps master complex operations step-by-step.

Square Root Function

The square root function is frequently used in mathematics to denote the principal square root of a number. Throwing out some key characteristics, the square root function, denoted as \( \sqrt{x} \), has the following properties:

• It is only defined for non-negative values of \(x\) (i.e., \(x \geq 0\)).
• The result of \( \sqrt{x} \) is always non-negative, providing the positive root.

In our composition exercise, the square root function plays a critical role. It acts as the outer function \(f(x) = \sqrt{x}\) and ensures that the output of the inner function developed a legitimate input for it to process.
The square root simplifies the expression \(x^2 - 1\) by transforming it into \( \sqrt{x^2 - 1} \), creating a straightforward picture of the composition's endpoint.

Therefore, grasping the usage and constraints of the square root function, assures correct application within any function composition exercise.