Problem 79 - Precalculus Essentials

Question

Express the given function \(h\) as a composition of two functions \(f\) and \(g\) so that \(h(x)=(f \circ g)(x)\). $$h(x)=|2 x-5|$$

Step-by-Step Solution

Answer

The given function \(h(x)=|2 x-5|\) can be modeled as the composition of two functions \(f\) and \(g\) where \(g(x) = 2x - 5\) and \(f(x) = |x|\). Therefore, \(h(x)=(f \circ g)(x)=|2 x-5|\).

Step 1: Identify the Inner Function

The first step is to identify the 'inner' function \(g(x)\). This is the function that is first applied to the input \(x\). Notice that inside the absolute value, we have the linear function \(2x - 5\). So, we can take \(g(x) = 2x - 5\).

Step 2: Identify the Outer Function

The second step is to identify the 'outer' function \(f(x)\). This function is applied to the result of the 'inner' function. In this case, the outer operation is the absolute value operation. So, we can take \(f(x) = |x|\).

Step 3: Express the Composition

The final step is to express the composition of two functions, \(f\) and \(g\). This can be written as \(h(x)=(f \circ g)(x)\). In our case, the composition would be expressed as \(|2x-5| = |(2x - 5)|\). So, \(h(x)\) can be expressed as the composition of functions \(f\) and \(g\), \(h(x) = (f \circ g)(x)\).

Key Concepts

Absolute Value Function

When we deal with an absolute value function, we're interested in the 'distance' a number is from zero, regardless of its direction on the number line. The absolute value of a number is always non-negative. This function can be expressed as \(|x|\thinspace\), where \(x\) can be any real number.

For example, the absolute value of both \(3\) and \(-3\) is \(3\), because both points are three units away from zero. In the context of our exercise, the absolute value function takes the place of the 'outer' function. This means that we first apply another operation (in the inner function) before taking the absolute value of the result.

The role of the absolute value in function composition is integral, as it ensures that no matter what the inside function computes, the overall output will be non-negative. See, in our case, the absolute value alters the output of the linear function \(2x - 5\thinspace\), so that it always delivers a positive result or zero.

Inner and Outer Functions

Understanding the concepts of inner and outer functions is key for mastering function composition. The inner function is the one applied first, directly to the given input. The outer function is then applied to the result of the inner function.

This is much like the process of baking a cake where the initial mix of ingredients (inner function) is then baked (outer function) to produce the final cake (composite function). In mathematical representation, if \(f(x)\thinspace\) is the outer function and \(g(x)\thinspace\) is the inner function, then their composite function is described by \((f \thinspace \thinspace \circ \thinspace \thinspace g)(x) = f(g(x))\thinspace\).

In the exercise provided, \(g(x) = 2x - 5\thinspace\) represents the first step applied to \(x\thinspace\), and \(f(x) = |x|\thinspace\) represents the subsequent, or outer, step that leads us to the final expression of \(h(x)\thinspace\).

Piecewise-Defined Functions

Another intriguing concept is that of piecewise-defined functions. These are functions that have different definitions or rules for different intervals of their domain. Essentially, they can be seen as a combination of multiple 'pieces' that come together to form a single, more complex function.

For example, a piecewise function could be defined to equal \(x^2\thinspace\) for \(x\) values less than zero, and \(2x\thinspace\) for \(x\) values greater than or equal to zero. This allows for functions that adapt their behavior according to the specific value of \(x\thinspace\).

In relation to absolute value, one can think of it as a simple two-piece function that splits at zero: for all positive values of \(x\), \(|x|\) is just \(x\), and for all negative values of \(x\), it's \(-x\). This piecewise nature of absolute value makes it especially interesting when combined with other functions, as we need to take into account how each 'piece' interacts with the inner function. This wasn't explicitly required in the given exercise, but is useful information for understanding more complex compositions that involve absolute values.