Problem 87 - College Algebra with Corequisite Support

Question

For the following exercises, find the composition when \(f(x)=x^{2}+2\) for all \(x \geq 0\) and \(g(x)=\sqrt{x-2}\). \((f \circ g)(6) ; \quad(g \circ f)(6)\)

Step-by-Step Solution

Answer

Both \((f \circ g)(6)\) and \((g \circ f)(6)\) equal 6.

Step 1: Understand Function Composition

Function composition involves applying one function to the results of another. The notation \((f \circ g)(x)\) means we first apply \(g(x)\) and then apply \(f\) to the result. Conversely, \((g \circ f)(x)\) means we apply \(f(x)\) first, and then \(g\) to that result.

Step 2: Calculate \(g(6)\) for \((f \circ g)(6)\)

First, find \(g(6)\) since it will be used in the expression \((f \circ g)(6)\). The function \(g(x) = \sqrt{x-2}\) gives us:\[ g(6) = \sqrt{6-2} = \sqrt{4} = 2 \]

Step 3: Calculate \(f(g(6))\) for \((f \circ g)(6)\)

Now that we have \(g(6) = 2\), substitute it into \(f(x)\):\[ f(g(6)) = f(2) \]Since \(f(x) = x^2 + 2\), we have:\[ f(2) = 2^2 + 2 = 4 + 2 = 6 \]

Step 4: Calculate \(f(6)\) for \((g \circ f)(6)\)

Calculate \(f(6)\) as the first part of \((g \circ f)(6)\). The function \(f(x) = x^2 + 2\) gives us:\[ f(6) = 6^2 + 2 = 36 + 2 = 38 \]

Step 5: Calculate \(g(f(6))\) for \((g \circ f)(6)\)

Substitute \(f(6) = 38\) into \(g(x)\):\[ g(f(6)) = g(38) \]Since \(g(x) = \sqrt{x-2}\), we have:\[ g(38) = \sqrt{38 - 2} = \sqrt{36} = 6 \]

Step 6: Conclude the Results

For the composition \((f \circ g)(6)\), the result is 6. For the composition \((g \circ f)(6)\), the result is also 6.

Key Concepts

Composition of Functions

Function composition is an essential concept in mathematics. It's like creating a new function by applying one function to the results of another. If you see the notation \((f \circ g)(x)\), it means you first apply the function \(g\) to the value \(x\), and then apply function \(f\) to the result of \(g(x)\). It’s like following a sequence of steps, almost like a "function pipeline."

• The order of functions is important: \((f \circ g)(x)\) isn't the same as \((g \circ f)(x)\).
• When you change the order, you usually get a different result because the functions interact in various ways.

For example, in our exercise, when calculating \((f \circ g)(6)\), you first evaluate \(g(6)\), and then use this result in \(f\). In contrast, for \((g \circ f)(6)\), you compute \(f(6)\) first and then apply \(g\) to this result. Understanding the sequence helps in performing the correct calculations.

Domain and Range

The domain of a function is the set of all possible inputs (or \(x\)-values) for which the function is defined. Meanwhile, the range is the set of all possible outputs (or \(y\)-values). Composition of functions impacts both domain and range since the output of the first function becomes the input of the second.

• Each function in the composition needs to be defined for particular values or conditions.
• The domain of \((f \circ g)(x)\) is determined by the input values \(x\) that are valid for \(g\) and where \(f\) can act on the output of \(g(x)\).
• Similarly, for \((g \circ f)(x)\), \(x\) must be in the domain of \(f\), and \(g\) must be defined for \(f(x)\).

In this exercise, understanding domain restrictions like \(x \geq 2\) for \(g(x) = \sqrt{x-2}\) ensures that you only use valid inputs for calculations of \((f \circ g)(x)\) or \((g \circ f)(x)\). Pay attention to any such criteria in exercises involving composition.

Algebraic Functions

Algebraic functions are mathematical expressions that include variables and constants, often involving operations like addition, subtraction, multiplication, and division. They form the foundational blocks in calculus and higher mathematics.

• The function \(f(x) = x^2 + 2\) is an example of a polynomial function, which is a type of algebraic function.
• Polynomial functions map any real number to another, forming smooth curves.
• In contrast, the function \(g(x) = \sqrt{x-2}\) includes a radical, and radical functions have specific conditions like non-negative radicands.

In our composition exercises, handling algebraic functions like these effectively involves substituting and simplifying. For example, when evaluating \(f(g(x))\), you substitute the expression from \(g(x)\) into \(f(x)\), making sure to follow arithmetic rules carefully. Always watch for domain restrictions posed by radicals or denominators to avoid undefined expressions.