Question

Given \(f(x)=\frac{2}{x-5}\) and \(g(x)=\sqrt{x+2},\) find the domain of \((f \circ g)(x)\)

Short answer

The domain of \( (f \circ g)(x) \) is \( [-2, 23) \cup (23, \infty) \).

Step-by-step solution

Identify the Domain of g(x)

The function \( g(x) = \sqrt{x+2} \) has a domain where the expression under the square root is non-negative. Therefore, we need to solve: \[ x + 2 \geq 0 \]Solving this inequality, we get:\[ x \geq -2 \]So, the domain of \( g(x) \) is \( x \geq -2 \).

Determine g(x) Output for f(x)

To determine the domain of \( (f \circ g)(x) \), we need to consider the domain of \( f(x) \) with respect to the outputs of \( g(x) \). Since \( g(x) = \sqrt{x+2} \), the outputs will be non-negative values. So, \( t = g(x) \geq 0 \) and we need to consider \( f(t) = \frac{2}{t-5} \).

Identify the Domain of f(g(x))

The function \( f(x) = \frac{2}{x-5} \) is defined for all values of \( x \) except \( x = 5 \), because the denominator becomes zero. Therefore, for \( f(g(x)) \) to be defined, \( g(x) eq 5 \). Since \( g(x) = \sqrt{x+2} \), we must have:\[ \sqrt{x+2} eq 5 \]Solving for \( x \), we get:\[ x + 2 eq 25 \]\[ x eq 23 \]

Combine Domain Conditions

The domain of \( (f \circ g)(x) \) is values of \( x \) that satisfy both domain restrictions: \( x \geq -2 \) (from the domain of \( g(x) \)) and \( x eq 23 \) (from \( f(g(x)) \)). Hence, the domain of \( (f \circ g)(x) \) is:\( [-2, 23) \cup (23, \infty) \).

Key concepts

Domain of Functions

The domain of a function is the set of all possible input values that allow the function to work without any issues. Think of it like the ingredients list for a recipe; you can only make the dish if you have all the required ingredients. For example, for the function \( g(x) = \sqrt{x+2} \), the expression under the square root, \( x+2 \), must always be non-negative. This means we need to solve the inequality \( x + 2 \geq 0 \), leading to the domain \( x \geq -2 \).

For the function \( f(x) = \frac{2}{x-5} \), we need to avoid dividing by zero because division by zero is undefined in mathematics. Therefore, \( x eq 5 \). Understanding these domain restrictions is key to finding the domain of composite functions.

Composite Functions

Composite functions combine two functions, where the output of one function becomes the input for another. Symbolically, we represent this as \((f \circ g)(x)\), which reads as “\(f\) of \(g\) of \(x\)”.

In our example, we need to find the domain of \( (f \circ g)(x) \) where \( f(x) = \frac{2}{x-5} \) and \( g(x) = \sqrt{x+2} \). First, evaluate the function \( g(x) \). The output from \( g(x) \) will be the input for \( f(x) \).

Since \( g(x) = \sqrt{x+2} \), the output values of \( g(x) \) are non-negative (because the square root of a real number is always non-negative). That means we need to evaluate \( f(t) = \frac{2}{t-5} \) for non-negative \( t \) values.

Inequalities

Handling inequalities is crucial for determining the domains of functions. An inequality tells us which values make an expression true. For example, when we say \( x + 2 \geq 0 \), we're finding all \( x \)-values that keep the expression under the square root non-negative.

Solving \( x + 2 \geq 0 \) gives \( x \geq -2 \). This inequality means our function \( g(x) = \sqrt{x+2} \) is defined for all \( x \)-values greater than or equal to \( -2 \).

Similarly, for the composite function, we must ensure that the argument of the function \( f \), which is \( g(x) \), is never equal to 5 because \( f \) is undefined at 5. This leads us to solve \( \sqrt{x+2} eq 5 \). Solving gives us \( x \eq 23 \). Combining these inequalities, the domain of the composite function \( (f \circ g)(x) \) is \( x \geq -2 \) excluding \( x=23 \).

Square Roots

Square roots often cause domain restrictions because you can't take the square root of a negative number and get a real number. The expression under the square root must be zero or positive.

Consider the function \( g(x) = \sqrt{x+2} \). To ensure that \( g(x) \) is real and defined for all \( x \), the term \( x+2 \) must satisfy the inequality \( x+2 \geq 0 \).

This requirement tells us that \( x \) must be greater than or equal to \( -2 \). By understanding these restrictions, we can accurately determine the domain for functions that involve square roots and ensure the functions are correctly defined.