Problem 81 - College Algebra

Question

For the following exercises, use \(f(x)=x^{3}+1\) and \(g(x)=\sqrt[3]{x-1}\). Find \((f \circ g)(2)\) and \((g \circ f)(2)\).

Step-by-Step Solution

Answer

(f ∘ g)(2) = 2 and (g ∘ f)(2) = 2.

Step 1: Understand Composition of Functions

The composition of two functions involves plugging one function into the other. To find \(f \circ g\)(x), we replace every instance of \(x\) in \(f(x)\) with \(g(x)\). Similarly, for \(g \circ f\)(x), replace \(x\) in \(g(x)\) with \(f(x)\).

Step 2: Substitute g(x) into f(x) for (f ∘ g)(x)

Given \(f(x) = x^3 + 1\) and \(g(x) = \sqrt[3]{x-1}\), the composition \(f \circ g\)(x) = f(g(x)). Substituting for \(g(x)\), we have \((f \circ g)(x) = (\sqrt[3]{x-1})^3 + 1\)."

Step 3: Simplify (f ∘ g)(x)

Simplify the expression \(\sqrt[3]{x-1}^3+1\) to \(x-1+1\), which results in \(x\). Thus, \(f \circ g)(x) = x\).

Step 4: Evaluate (f ∘ g)(2)

Since \(f \circ g(x) = x\), we substitute \(x = 2\) and find \(f \circ g(2) = 2\).

Step 5: Substitute f(x) into g(x) for (g ∘ f)(x)

Now, consider \(g \circ f)(x) = g(f(x))\). Replace \(f(x)\) with \(x^3 + 1\), so \((g \circ f)(x) = \sqrt[3]{(x^3 + 1)-1}\)."

Step 6: Simplify (g ∘ f)(x)

Simplify \(\sqrt[3]{x^3 + 1 - 1}\) to \(\sqrt[3]{x^3}\), which equals \(x\). Therefore, \((g \circ f)(x) = x\).

Step 7: Evaluate (g ∘ f)(2)

Substitute \(x = 2\) into \(g \circ f(x) = x\). Thus, \(g \circ f(2) = 2\).

Key Concepts

Function Notation

Function notation is a helpful way to represent and handle functions in mathematics.
It is expressed as a relation between inputs and outputs. For example, the notation \(f(x)\) indicates that \(f\) is a function with \(x\) as its input.
If we assign a value to \(x\), the function will return a specific value or result based on the defined rule for \(f\).
In our exercise, we have two functions, \(f(x) = x^3 + 1\) and \(g(x) = \sqrt[3]{x-1}\). This notation helps us easily manage these functions and understand how to evaluate them.

• keep track of the function names
• understand the variable being manipulated
• plug values into these functions efficiently

By using function notation, we can precisely execute operations like composition of functions, transformations, and evaluations without ambiguity.

Cube Roots

Cube roots are the inverse operation of cubing a number. They help us find which number, when multiplied by itself three times, gives the original number.
Symbolically, the cube root of a number \(x\) is represented by \(\sqrt[3]{x}\).
This concept is essential in our exercise, particularly in the function \(g(x) = \sqrt[3]{x-1}\), which involves finding the cube root of \(x-1\).
If \(x = 8\), then \(\sqrt[3]{8} = 2\), since \(2\times2\times2 = 8\).
Understanding cube roots allows us to work with expressions that involve cubed terms efficiently.Some key points to remember about cube roots:

• They can be positive or negative because \((-2) \times (-2) \times (-2) = -8\)
• Cube roots reverse cubing, returning the original base number
• Use cube root properties to simplify expressions and solve equations

This understanding facilitates working with functions like \(g(x)\) and simplifies the computations involved in function compositions.

Function Evaluation

Function evaluation involves finding the output of a function for a given input.
This process is straightforward once we're familiar with the function notation and rules.
When evaluating a function, substitute the input value into the function's equation.
In our example, for \((f \circ g)(2)\), we first evaluate \(g(2)\), and then take that result to plug into \(f\).
Similarly, for \((g \circ f)(2)\), evaluate \(f(2)\) first, and substitute it into \(g\).Here’s how it is done step by step:

• Identify the correct order of function compositions
• Evaluate the innermost function first, then move to the outer function
• Simplify any contained expressions during substitution

This results in evaluating \((f \circ g)(2) = 2\) and \((g \circ f)(2) = 2\), because both function compositions simplify to \(x\), making the evaluation straightforward and efficient.