Problem 9 - Thomas Calculus

Question

Let \(f(x)=x-3, g(x)=\sqrt{x}, \quad h(x)=x^{3},\) and \(j(x)=2 x .\) Express each of the functions in Exercises 9 and 10 as a composite involving one or more of \(f, g, h,\) and \(j .\) $$ \begin{array}{ll}{\text { a. } y=\sqrt{x}-3} & {\text { b. } y=2 \sqrt{x}} \\\ {\text { c. } y=x^{1 / 4}} & {\text { d. } y=4 x} \\ {\text { e. } y=\sqrt{(x-3)^{3}}} & {\text { f. } y=(2 x-6)^{3}}\end{array} $$

Step-by-Step Solution

Answer

Compositions are: a. \(f(g(x))\), b. \(j(g(x))\), c. \(g(g(x))\), d. \(j(j(x))\), e. \(g(h(f(x)))\), f. \(h(f(j(x)))\).

Step 1: Understand Function Composition

Function composition involves creating a function by passing the output of one function as the input to another. Given functions are: \( f(x) = x - 3 \), \( g(x) = \sqrt{x} \), \( h(x) = x^3 \), and \( j(x) = 2x \). We need to express each target function as a composition of these.

Step 2: Composite for a. \( y = \sqrt{x} - 3 \)

Decompose \( y = \sqrt{x} - 3 \) as the composition \( f(g(x)) \). The function \( g(x) = \sqrt{x} \) is first applied, then \( f(x) = x - 3 \) is applied to the result.

Step 3: Composite for b. \( y = 2\sqrt{x} \)

Decompose \( y = 2\sqrt{x} \) as the composition \( j(g(x)) \). The function \( g(x) = \sqrt{x} \) is first applied, then \( j(x) = 2x \) is applied to the result.

Step 4: Composite for c. \( y = x^{1/4} \)

Since \( y = x^{1/4} = (x^{1/2})^{1/2} \), decompose this as \( g(g(x)) \). The function \( g(x) = \sqrt{x} \) is applied twice.

Step 5: Composite for d. \( y = 4x \)

Rewrite \( y = 4x = 2(2x) \) using \( j(x) = 2x \). Decompose as the composition \( j(j(x)) \). The function \( j(x) = 2x \) is applied twice.

Step 6: Composite for e. \( y = \sqrt{(x-3)^3} \)

This is composed as \( g(h(f(x))) \). First, apply \( f(x) = x - 3 \), then \( h(x) = x^3 \), and finally \( g(x) = \sqrt{x} \).

Step 7: Composite for f. \( y = (2x-6)^3 \)

Express \( y = (2x-6)^3 \) as the composition \( h(f(j(x))) \). First, use \( j(x) = 2x \), then \( f(x) = x - 3 \), and finally \( h(x) = x^3 \).

Key Concepts

Mathematical Functions

In mathematics, a function is essentially a relationship between a set of inputs and a set of permissible outputs where each input is related to exactly one output. Functions are like machines: you insert a value (input), something happens inside the function (the rule within), and you get an output. They are denoted by letters such as \( f(x), g(x), \) and so forth, where \( x \) represents the input variable.

Each function has a specific rule that tells you what to do with the input you provide. For instance:

• The function \( f(x) = x - 3 \) subtracts 3 from \( x \).
• The function \( g(x) = \sqrt{x} \) calculates the square root of \( x \).
• The function \( h(x) = x^3 \) raises \( x \) to the power of 3.
• The function \( j(x) = 2x \) multiplies \( x \) by 2.

Understanding functions and their specific operations helps you to manipulate and combine them effectively in various mathematical problems. This foundational knowledge equips you for exploring more advanced concepts like function composition.

Composite Functions

Composite functions are created when you apply one function to the results of another. It's like chaining together multiple operations to form a new function. This process helps in breaking down complex mathematical scenarios into simpler parts. Let's see how this works through some examples from the exercise.

Suppose you have a function \( f(g(x)) \): here, you first apply \( g(x) \), and the output of \( g(x) \) becomes the input for \( f(x) \). One example is \( y = \sqrt{x} - 3 \), which can be expressed as \( f(g(x)) \) if \( g(x) = \sqrt{x} \) and \( f(x) = x - 3 \).

Another example is \( j(g(x)) \) for the function \( y = 2\sqrt{x} \), where you first use \( g(x) = \sqrt{x} \) and then \( j(x) = 2x \). This shows how combining functions allows you to express more complicated relationships using simpler, already-defined functions.

• Understanding how to decompose and recompose functions is vital in mathematics and calculus.
• This technique is widely used in solving equations, optimizing systems, and modeling real-world phenomena.

Calculus Concepts

In calculus, the concept of function composition plays a critical role in understanding and solving complex problems. It allows mathematicians to explore relationships and changes between functions, and is particularly useful when working with derivatives and integrals.

When dealing with derivative calculation, composite functions introduce us to the chain rule, which is essential for differentiating composed functions. For example, if you have a composite function like \( g(h(x)) \), then according to the chain rule, the derivative is \( g'(h(x)) \cdot h'(x) \). This allows you to differentiate composite functions efficiently.

• The chain rule helps in understanding how changes in one variable affect others, especially in physics and engineering problems.
• In integrals, decomposing a complex function into composite functions can simplify the integration process, making it easier to find the total area under a curve.

Understanding calculus concepts such as function composition helps in unlocking deeper insights into how functions behave and interact. This forms a foundation for tackling advanced topics and solving intricate mathematical tasks.