Question

The domain of \(f\) is the set \(A=\\{-2,-1,0,1,2\\}\) Write the function as a set of ordered pairs. \(f(x)=\sqrt{x+2}\)

Short answer

The function as a set of ordered pairs is \{(-2, 0), (-1, 1), (0, \sqrt{2}), (1, \sqrt{3}), (2,2)\}.

Step-by-step solution

Calculation for \( x=-2 \) in \( f(x) \)

Find the value of the function when \( x \) equals to -2. Plug in \( x = -2 \) into \( f(x)=\sqrt{x+2} \), which gives \( f(-2)=\sqrt{-2+2} = 0 \). The ordered pair for this is \( (-2, 0) \).

Calculation for \( x=-1 \) in \( f(x) \)

Find the value of the function when \( x \) equals to -1. Plug in \( x = -1 \) into \( f(x)=\sqrt{x+2} \), which gives \( f(-1)=\sqrt{-1+2} = 1 \). The ordered pair for this is \( (-1,1) \).

Calculation for \( x=0 \) in \( f(x) \)

Find the value of the function when \( x \) equals to 0. Plug in \( x = 0 \) into \( f(x)=\sqrt{x+2} \), which gives \( f(0)=\sqrt{0+2} = \sqrt{2} \). The ordered pair for this is \( (0, \sqrt{2}) \).

Calculation for \( x=1 \) in \( f(x) \)

Find the value of the function when \( x \) equals to 1. Plug in \( x = 1 \) into \( f(x)=\sqrt{x+2} \), which gives \( f(1)=\sqrt{1+2} = \sqrt{3} \). The ordered pair for this is \( (1, \sqrt{3}) \).

Calculation for \( x=2 \) in \( f(x) \)

Find the value of the function when \( x \) equals to 2. Plug in \( x = 2 \) into \( f(x)=\sqrt{x+2} \), which gives \( f(2)=\sqrt{2+2} = 2 \). The ordered pair for this is \( (2,2) \).

Key concepts

Ordered Pairs

In mathematics, the concept of ordered pairs is fundamental to understanding many different types of relations, including functions. An ordered pair \( (x, y) \) consists of two elements where the order is significant; the first element is the input or domain value, and the second element is the output or range value.

For example, in the function \( f(x)=\sqrt{x+2} \) given the domain set \( A=\{-2,-1,0,1,2\} \) as in the exercise, we calculate the corresponding range values, resulting in specific ordered pairs \( (x, f(x)) \) for each \( x \) in the domain. These pairs essentially form a one-to-one relationship between members of the domain and the function's results. So, the function written as a set of ordered pairs based on the prescribed domain becomes \( \{(-2, 0), (-1, 1), (0, \sqrt{2}), (1, \sqrt{3}), (2, 2)\} \) which clearly delineates the links between each input and its corresponding output.

Function Evaluation

Function evaluation is a process where we find the output of a function given a particular input. To 'evaluate' the function \( f(x) \) at a specific value of \( x \) means to plug that value into the formula of the function and perform the algebraic operations to simplify and find the result.

This step-by-step process is crucial when determining the set of ordered pairs as seen in the solution steps. For instance, to evaluate the square root function \( f(x) = \sqrt{x+2} \) when \( x = -1 \) we replaced the \( x \) with \( -1 \) to get \( f(-1) = \sqrt{-1 + 2} = 1 \) and identified the ordered pair associated with \( x = -1 \) as \( (-1, 1) \) in the function's graph. Evaluating functions accurately is essential since it underpins a crucial aspect of analyzing functions and understanding their behavior.

Square Root Functions

Square root functions are a type of radical function where the principal square root operation is applied to the input value. The standard form of a square root function is \( f(x) = \sqrt{x} \) or, more generally, \( f(x) = \sqrt{ax + b} \) where \( a \) and \( b \) are constants.

In the context of the provided exercise, we are dealing with the function \( f(x) = \sqrt{x + 2} \) which means for each input \( x \) we calculate the square root of the sum of \( x \) and 2. The domain of these functions is all values of \( x \) for which the expression under the square root is non-negative, as square roots of negative numbers are not real. For the function \( f(x) = \sqrt{x + 2} \) the domain starts at \( x = -2 \) because that's the point where the expression inside the square root becomes zero (the least value for which the square root is defined in real numbers). This is why in the problem's solution, the domain is correctly selected to consist of only those values that keep the radicand (the number under the square root) non-negative.