Problem 52 - Applied Calculus

Question

For each pair of functions \(f(x)\) and \(g(x)\), find a. \(f(g(x))\) b. \(g(f(x))\) and c. \(f(f(x))\) $$ f(x)=\sqrt{x} ; g(x)=x^{3}-1 $$

Step-by-Step Solution

Answer

a. \( \sqrt{x^3 - 1} \); b. \( x^{3/2} - 1 \); c. \( x^{1/4} \).

Step 1: Understand the Problem

We are given two functions: \( f(x) = \sqrt{x} \) and \( g(x) = x^3 - 1 \). The problem asks us to find three compositions: \( f(g(x)) \), \( g(f(x)) \), and \( f(f(x)) \). This means we need to substitute one function into another and simplify if possible.

Step 2: Find f(g(x))

To find \( f(g(x)) \), substitute \( g(x) \) into \( f(x) \). So, \( f(g(x)) = f(x^3 - 1) \). Since \( f(x) = \sqrt{x} \), we have \( f(g(x)) = \sqrt{x^3 - 1} \). This is the composition of \( f \) and \( g \).

Step 3: Find g(f(x))

For \( g(f(x)) \), substitute \( f(x) \) into \( g(x) \). Thus, \( g(f(x)) = g(\sqrt{x}) \). Since \( g(x) = x^3 - 1 \), we replace \( x \) with \( \sqrt{x} \) to get \( g(f(x)) = (\sqrt{x})^3 - 1 = x^{3/2} - 1 \).

Step 4: Find f(f(x))

To find \( f(f(x)) \), substitute \( f(x) \) into itself. So, \( f(f(x)) = f(\sqrt{x}) \). Since \( f(x) = \sqrt{x} \), we have \( f(f(x)) = \sqrt{\sqrt{x}} = x^{1/4} \). This represents the composition of \( f \) with itself.

Key Concepts

Square Root Function

The square root function is a type of function described by the form \( f(x) = \sqrt{x} \). This function plays an essential role in mathematics, representing the value that, when squared, gives the original input, \( x \). Due to its properties, the square root function is only defined for non-negative numbers.- The primary focus is its operation on numbers—it essentially reduces a square to its side length.
- For example, \( \sqrt{4} = 2 \) because \( 2^2 = 4 \).
Generally, the graph of this function forms a curve starting at the origin and extending to the right as \( x \) increases.
The square root function undergoes further operations when used in function compositions, which makes it a versatile tool in various algebraic manipulations.

Cubic Function

The cubic function is expressed mathematically as \( g(x) = x^3 - 1 \). This function entails raising a variable \( x \) to the third power and often involves adding or subtracting a constant.- A critical aspect of cubic functions is their unique ability to display inflection points and change in curvature, unlike linear or quadratic functions.
- These functions are defined for all real numbers, so you can input any real number to get a valid output.
When looking at its graph, a cubic function typically shows a significant incline or decline, depicting the increase (or decrease) of \( x^3 \) behavior across different intervals.
Additionally, employing cubic functions in compositions lets us observe how power manipulation transforms expressions, given that cubing plays a crucial role in expanding algebraic terms and solving equations.

Composite Functions

When we talk about composite functions, we refer to a new function created by applying one function to the results of another. This concept is crucial because it represents performing multiple operations on an input sequentially.
- Composite functions are typically denoted as \( (f \circ g)(x) = f(g(x)) \). This notation means you first apply \( g(x) \) and then use the result as the input for \( f(x) \).
- It allows for more complex expressions to be simplified or transformed into manageable calculations.
Understanding how to break down and solve composite functions requires identifying which function to apply first and then evaluating each subsequent function step-by-step.
In our example, the expressions \( f(g(x)) = \sqrt{x^3 - 1} \), \( g(f(x)) = x^{3/2} - 1 \), and \( f(f(x)) = x^{1/4} \) are all results of pairing different functions together adeptly, showcasing this concept in action.