Problem 48 - Algebra and Trigonometry

Question

Find \(f \circ g \circ h\) $$ f(x)=\sqrt{x}, \quad g(x)=\frac{x}{x-1}, \quad h(x)=\sqrt[3]{x} $$

Step-by-Step Solution

Answer

\( f(g(h(x))) = \sqrt{\frac{\sqrt[3]{x}}{\sqrt[3]{x} - 1}} \).

Step 1: Understanding composite functions

The expression \( f \circ g \circ h \) represents the composition of three functions: \( f(x) \), \( g(x) \), and \( h(x) \). We will evaluate these functions from the innermost function outward, meaning we first find \( h(x) \), follow with \( g(h(x)) \), and finally \( f(g(h(x))) \).

Step 2: Calculate the innermost function

Start by finding \( h(x) \). Since \( h(x) = \sqrt[3]{x} \), the innermost function is \( h(x) \). There are no operations to be performed here as \( h(x) \) is already expressed in terms of \( x \).

Step 3: Compose with the middle function

Now apply \( g(x) \) to \( h(x) \). Recall that \( g(x) = \frac{x}{x-1} \) and \( h(x) = \sqrt[3]{x} \). Substitute \( h(x) \) into \( g(x) \) to get \( g(h(x)) = \frac{\sqrt[3]{x}}{\sqrt[3]{x} - 1} \).

Step 4: Compose with the outer function

Finally, apply \( f(x) \) to the result from \( g(h(x)) \). Given \( f(x) = \sqrt{x} \), substitute \( g(h(x)) \) into \( f(x) \) to obtain:\[f(g(h(x))) = \sqrt{\frac{\sqrt[3]{x}}{\sqrt[3]{x} - 1}}.\]

Step 5: Conclusion

Thus, the composite function \( f \circ g \circ h \) is given by:\[ f(g(h(x))) = \sqrt{\frac{\sqrt[3]{x}}{\sqrt[3]{x} - 1}}.\]

Key Concepts

Function Composition

Function composition is much like following a set of instructions in a multi-step process. Imagine you have three functions, each with its own rule as how input is to be transformed. In our example, this could be likened to combining ingredients in a recipe, where each function changes the mixture until we reach the final dish. Function composition involves applying one function to the result of another, in a sequential manner. It is represented using the "composition operator" symbol: \( \circ \).

For instance, with functions \( f(x) \), \( g(x) \), and \( h(x) \), the notation \( f \circ g \circ h \) means that we first apply \( h \) to the input \( x \), then apply \( g \) to the output of \( h(x) \), and finally apply \( f \) to the result from \( g(h(x)) \). By unpacking the process layer by layer, much like peeling the layers of an onion, the task of function composition becomes clearer.

• Innermost function: \( h(x) \)
• Next, apply: \( g(h(x)) \)
• Final step: \( f(g(h(x))) \)

Domains of Functions

The domain of a function is the set of all possible inputs (usually represented as \( x \)) for which the function is defined. Understanding the domain is critical, especially when dealing with radical or fraction functions. Each function must adhere to its domain rules to ensure the operations are valid and defined. In the case with our composite \( f \circ g \circ h \), each step of the composition process must respect the domain of the involved functions.
- For \( h(x) = \sqrt[3]{x} \), the cube root function is defined for all real numbers, so it has no domain restrictions.- For \( g(x) = \frac{x}{x-1} \), the division operator prohibits the denominator from being zero, meaning \( x eq 1 \), which defines the domain restriction.- For \( f(x) = \sqrt{x} \), it requires that the input must be non-negative, meaning \( x \geq 0 \).

When combined into the composite function, we should consider all these constraints together. Each function's domain influences the final composition, limiting the range of values \( x \) in order for \( f(g(h(x))) \) to be defined. By this, ensuring we are meeting all required conditions for validity and correctness of the overall formulation.

Radical Functions

Radical functions are a specific type of function that include square roots, cube roots, or higher-order roots. Understanding these functions is important because they often have restrictions on their domain due to the nature of roots.

The component functions in our exercise include radicals which affect how we treat them within our operations and compositions.

• For \( f(x) = \sqrt{x} \) – this is a square root function, meaning it is only defined for non-negative numbers \( x \), so \( x \geq 0 \). This is because square roots of negative numbers are not defined in the set of real numbers.
• For \( h(x) = \sqrt[3]{x} \) – unlike similar functions with even roots, cube roots are defined for all real numbers \( x \). Hence, \( h(x) \) does not impose additional domain restrictions aside from those of the other functions involved.

Radical functions play a key role in altering the domain considerations for the overall composite function, thus influencing behaviors of functions within the composite form. Understanding how the square and cube roots restrict or allow values ensures correct implementation and interpretation of such composite functions.