Problem 57 - Algebra and Trigonometry

Question

If \(f(x)=x^{2}-6\) and \(g(x)=\sqrt{x+2},\) find \(g(f(x))\) and state its domain.

Step-by-Step Solution

Answer

\(g(f(x)) = \sqrt{x^2 - 4}\) with the domain \((-\infty, -2] \cup [2, \infty)\).

Step 1 - Understand Composition of Functions

The problem requires finding the composition of two functions, which means substituting one function into another. For this exercise, substitute the function f(x) into g(x), that is find g(f(x)).

Step 2 - Substitute f(x) into g(x)

The given functions are: \[ f(x) = x^2 - 6 \] and \[ g(x) = \sqrt{x + 2} \] To find \( g(f(x)) \), substitute f(x) into g: \[ g(f(x)) = g(x^2 - 6) \] Since g(x) is defined as \( \sqrt{x + 2} \), substitute \(x^2 - 6\) for \(x\): \[ g(x^2 - 6) = \sqrt{(x^2 - 6) + 2} \] This simplifies to: \[ g(f(x)) = \sqrt{x^2 - 4} \]

Step 3 - Find the Domain of g(f(x))

The domain of g(f(x)) must satisfy the condition of the square root: the expression inside the square root must be non-negative. Set up the inequality: \[ x^2 - 4 \geq 0 \] Solve for x: \[ x^2 \geq 4 \] Take the square root of both sides: \[ |x| \geq 2 \] This means: \[ x \leq -2 \quad \text{or} \quad x \geq 2 \] Hence, the domain of g(f(x)) is: \[ (-\infty, -2] \cup [2, \infty) \]

Key Concepts

Domain of Functions

The domain of a function is essentially all the possible input values (x-values) that the function can accept. Each function has its own domain that restricts the values of x that can be used to produce a valid output. For instance, the domain of the function must be defined so there are no mathematical errors like division by zero or taking the square root of a negative number.

In the exercise, we start by determining the domain of the composition function, which combines two other functions. To find the domain of the composite function, you need to consider where both individual functions are defined.

For the composite function g(f(x)), you need to ensure that:

• The output of the inner function f(x) is within the domain of the outer function g.
• The result of g(f(x)) meets any other requirements that g(x) has, such as being within the domain that g(x) allows.

Square Roots

Square roots are a mathematical operation where we look for a number that, when multiplied by itself, gives us the original number. For example, the square root of 9 is 3, since 3 multiplied by 3 equals 9.

When dealing with the domain of functions involving square roots, remember that the value inside the square root (the radicand) must be greater than or equal to zero. This is because you cannot get a real number as the square root of a negative number.

In our exercise, the function g(x) is defined as \(\text{g(x)} = \sqrt{x + 2}\). Therefore, the expression inside the square root must be non-negative. When we substituted the expression from f(x) into g(x), we obtained:

• g(f(x)) = \sqrt{x^2 - 4}
• To ensure this square root is valid, x^2 - 4 must be greater than or equal to zero.

This results in the inequality \[ x^2-4 \geq 0 \], which has critical points at x = -2 and x = 2. Solving this, we get the intervals \(( -\infty, -2 ]\) and \([ 2, \infty )\) for which the function is defined.

Function Substitution

Function substitution is a process where one function is substituted into another. This is a crucial step in composing functions. Let's say you have two functions f(x) and g(x), and you want to find the composition g(f(x)).

Here's how it works in practice:

• You start with an inner function f(x) and an outer function g(x).
• Substitute the expression that defines f(x) into every instance of x in g(x).

In our exercise, f(x) is \( x^2 - 6 \), and we need to substitute this expression into g(x), defined as \(\text{g(x) = \sqrt{x+2}} \). This means we replace x in g(x) with x^2 - 6:

• g(f(x)) = \sqrt{ (x^2 - 6) + 2 }
• This simplifies to g(f(x)) = \sqrt{ x^2 - 4 }

Function substitution ensures that we end up with a new single expression that represents the composition of both original functions. Learning to substitute functions correctly is very important for solving complex mathematical problems that involve multiple functions operating together.