Problem 23 - Calculus with Concepts in Calculus

Question

Find the domain and rule of \(g \circ f\) and \(f \circ g\). \(f(x)=\sqrt{x}\) and \(g(x)=x^{2}-5 x+6\)

Step-by-Step Solution

Answer

The domain of \(g \circ f\) is \(x \geq 0\), and \( f \circ g \) is defined on \((- fty, 2] \cup [3, \\infty)\).

Step 1: Understand the Composition

The composition of two functions, such as \( g \circ f \) and \( f \circ g \), involves applying one function to the results of another. Specifically, \( g \circ f(x) = g(f(x)) \) means substituting \( f(x) \) into \( g(x) \). Similarly, \( f \circ g(x) = f(g(x)) \). Let's find the expressions for these compositions.

Step 2: Calculate g(f(x))

First, substitute \( f(x) = \sqrt{x} \) into \( g(x) \). Hence, \( g(f(x)) = g(\sqrt{x}) = (\sqrt{x})^2 - 5(\sqrt{x}) + 6 = x - 5\sqrt{x} + 6 \).

Step 3: Find Domain of g(f(x))

The domain of \( g(f(x)) \) is determined by the domain of \( f(x) \), which is constrained by the square root. The expression under the square root, \( x \), must be non-negative, so the domain of \( g(f(x)) \) is \( x \geq 0 \).

Step 4: Calculate f(g(x))

Now, substitute \( g(x) = x^2 - 5x + 6 \) into \( f(x) \). Hence, \( f(g(x)) = f(x^2 - 5x + 6) = \sqrt{x^2 - 5x + 6} \).

Step 5: Find Domain of f(g(x))

The domain of \( f(g(x)) \) requires that the expression within the square root is non-negative. Solve \( x^2 - 5x + 6 \geq 0 \). Factor the quadratic to \((x-2)(x-3) \geq 0\), which gives intervals \((-\infty, 2] \cup [3, \infty)\), making this the domain of \( f(g(x)) \).

Key Concepts

Domain of a Function

The domain of a function refers to all the possible input values for which the function is defined. For each function, there are specific mathematical rules to determine its domain. For example, the domain for the function \( f(x) = \sqrt{x} \) includes all values where the expression under the square root is non-negative, which means \( x \geq 0 \).
In function composition, the domain plays a crucial role because you need to ensure that the results of one function are valid inputs for another. When finding the domain of a composed function like \( g \circ f \), keep in mind:

• The domain of \( f(x) \) must satisfy the function's own rule.
• After applying \( f(x) \), the result should still be a valid input for \( g(x) \).

When finding \( f(g(x)) \), aside from \( g(x) \)'s natural domain, ensure the range of \( g(x) \) fits into \( f(x)'s \) domain. This can make domain determination for composed functions complex, especially with restrictions such as radicals or denominators. Understanding these constraints is key to accurately determining the domain of any function or function composition.

Quadratic Functions

Quadratic functions take the form \( g(x) = ax^2 + bx + c \) and graph as parabolas, which can open either upward or downward depending on the sign of \( a \). These functions are defined for all real numbers, meaning their domain is \( (-\infty, \infty) \).
However, when working with quadratics within other contexts, such as under square roots or in rational functions, certain restrictions apply. For example, in our exercise, \( g(x) = x^2 - 5x + 6 \) factors to \((x-2)(x-3)\).
The roots of this quadratic are the points where the function crosses the x-axis. When evaluating or composing functions, these roots play a role in determining intervals over which inequalities hold true.
Using the roots, when solving \( x^2 - 5x + 6 \geq 0 \) for \( f(g(x)) \) helps us understand where values are non-negative.

• This leads to the solution involving intervals \((-\infty, 2] \cup [3, \infty)\).

Understanding how to factor and solve quadratic inequalities is essential when working with mixed function types in compositions.

Radical Functions

Radical functions contain roots, such as square roots, and their domains often involve ensuring that expressions under the root are non-negative.
Take \( f(x) = \sqrt{x} \) as an example; it is only defined for \( x \geq 0 \) because you cannot take the square root of a negative number in real numbers.
When these functions are part of compositions, the entire expression under the radical must evaluate to a non-negative number. Thus, when \( f(g(x)) = \sqrt{x^2-5x+6} \), you need to solve the inequality \( x^2-5x+6 \geq 0 \):

• This gives us the domain where the quadratic expression yields non-negative values, namely \((-\infty, 2] \cup [3, \infty)\).

Handling radicals requires careful attention to ensure expressions are valid throughout their domains. This is critical for ensuring composition results are mathematically sound and fit the conditions under radical signs.