91Ó°ÊÓ

The domain of a function is essentially the set of all possible inputs for which the function is defined. To find the domain, we need to consider the type of function we are dealing with. For instance, with a square root function like \( f(x) = \sqrt{x+1} \), the expression inside the square root must be non-negative. Therefore, we solve the inequality \( x + 1 \geq 0 \) to get \( x \geq -1 \).
This means that the domain of \( f(x) \) is all real numbers greater than or equal to \(-1\). Similarly, for the function \( g(x) = x + 2 \), there is no restriction, so the domain is all real numbers.

• For \( (f + g)(x) = \sqrt{x+1} + x + 2 \), the domain is determined by \( f(x) \), so \( x \geq -1 \).
• For \( (f - g)(x) = \sqrt{x+1} - x - 2 \), the condition \( x \geq -1 \) still applies.
• For \( (f \cdot g)(x) = (x+2)\sqrt{x+1} \), again, \( x \geq -1 \).
• For \( (f / g)(x) = \frac{\sqrt{x+1}}{x+2} \), in addition to \( x \geq -1 \), \( x eq -2 \) since the denominator cannot be zero.

Understanding how to find the sum and difference of functions is crucial. When combining functions, simply add or subtract their corresponding outputs. Specifically, for two functions \( f(x) \) and \( g(x) \), we use:

• The sum: \((f+g)(x) = f(x) + g(x)\)
• The difference: \((f-g)(x) = f(x) - g(x)\)

In our example, adding \( f(x) = \sqrt{x+1} \) and \( g(x) = x+2 \) results in:
\[(f+g)(x) = \sqrt{x+1} + x + 2\]
Similarly, subtracting the two gives us:
\[(f-g)(x) = \sqrt{x+1} - x - 2\]
The domain for both expressions is constrained by the condition \( x \geq -1 \), derived from the requirement that the square root term is defined.

When you multiply or divide functions, you create new relationships between inputs and outputs. To compute the product and quotient, use the following rules for two functions \( f(x) \) and \( g(x) \):

• The product: \((f\cdot g)(x) = f(x) \cdot g(x)\)
• The quotient: \((f/g)(x) = \frac{f(x)}{g(x)}\), given \( g(x) eq 0 \).

In this problem, the product of \( f \) and \( g \) leads to:
\[(f \cdot g)(x) = (x+2)\sqrt{x+1}\]
The domain here is \( x \geq -1 \), since the square root part dictates positivity. For the quotient, dividing them:
\[(f / g)(x) = \frac{\sqrt{x+1}}{x+2}\]
is valid only if \( x+2 eq 0 \), thus \( x eq -2 \), which modifies the domain to include \( x \geq -1 \) and exclude \( x = -2 \). This produces a domain of all real numbers from \(-1\) (inclusive), except for \(-2\).

Square root functions, like \( f(x) = \sqrt{x} \), hold unique properties that affect their domains and graphs. Here, \( f(x) = \sqrt{x+1} \). The key rule is that the expression inside the root must be zero or positive, ensuring the function has real values.
We solve:

• Inside part non-negative: \( x+1 \geq 0 \)
• Leading to solution: \( x \geq -1 \)

This restriction means any function operation involving \( \sqrt{x+1} \) should respect this domain. The result should not evaluate the square root of a negative number, which is why understanding root function domains is crucial for function operations with square roots. Each operation in this exercise followed the same inclusion \( x \geq -1 \) except the quotient where \( x eq -2 \) must be added, all derived from ensuring the square root and potential division by zero issues are managed properly.