Problem 88
Question
Identify the domain and then graph each function. \(f(x)=\sqrt{x+1} ;\) use the following table. $$ \begin{array}{|c|c|} \hline x & {f(x)} \\ \hline-1 & {} \\ \hline 0 & {} \\ \hline 3 & {} \\ \hline 8 & {} \\ \hline \end{array} $$
Step-by-Step Solution
Verified Answer
Domain: \(x \geq -1\); points to plot: (-1,0), (0,1), (3,2), (8,3).
1Step 1: Identify the Domain
To find the domain of the function \(f(x) = \sqrt{x+1}\), we need to ensure that the expression inside the square root, \(x+1\), is non-negative. This means \(x+1 \geq 0\). Solving this inequality gives us \(x \geq -1\). Thus, the domain of the function is all real numbers \(x\) such that \(x \geq -1\).
2Step 2: Evaluate the Function at Specified Points
To fill the table, we will calculate \(f(x)\) for each given \(x\) value:1. For \(x = -1\): \[ f(-1) = \sqrt{-1+1} = \sqrt{0} = 0 \] 2. For \(x = 0\): \[ f(0) = \sqrt{0+1} = \sqrt{1} = 1 \]3. For \(x = 3\): \[ f(3) = \sqrt{3+1} = \sqrt{4} = 2 \]4. For \(x = 8\): \[ f(8) = \sqrt{8+1} = \sqrt{9} = 3 \]
3Step 3: Complete the Table
Using the calculations from Step 2, complete the table:\[\begin{array}{|c|c|}\hline x & f(x) \\hline-1 & 0 \\hline0 & 1 \\hline3 & 2 \\hline8 & 3 \\hline\end{array}\]
4Step 4: Graph the Function
To graph the function, plot the points calculated in Step 3 on a coordinate system where:- \( x = -1, f(x) = 0 \)- \( x = 0, f(x) = 1 \)- \( x = 3, f(x) = 2 \)- \( x = 8, f(x) = 3 \)Connect the dots smoothly, keeping in mind that the graph starts at \(x = -1\) and continues to the right, as the domain is \(x \geq -1\).
Key Concepts
Square Root FunctionGraphing FunctionsInequalitiesEvaluating Functions
Square Root Function
A square root function is a type of algebraic function that involves the square root of a variable. It generally takes the form \( f(x) = \sqrt{x} \), but it can also involve shifted or transformed expressions like \( f(x) = \sqrt{x + 1} \). These functions are interesting because they are defined only when the value inside the square root is non-negative.
The precise reason is that the square root of a negative number is not defined in the real number system, which means the expression under the square root must be greater than or equal to zero.
This makes square root functions less straightforward than linear functions, often resulting in a limited domain. For example, for \( f(x) = \sqrt{x + 1} \), you must solve \( x + 1 \geq 0 \) to find that the domain includes all real numbers \( x \) such that \( x \geq -1 \).
The range, however, is determined by the values that the square root can take on, starting from zero and increasing to positive values.
The precise reason is that the square root of a negative number is not defined in the real number system, which means the expression under the square root must be greater than or equal to zero.
This makes square root functions less straightforward than linear functions, often resulting in a limited domain. For example, for \( f(x) = \sqrt{x + 1} \), you must solve \( x + 1 \geq 0 \) to find that the domain includes all real numbers \( x \) such that \( x \geq -1 \).
The range, however, is determined by the values that the square root can take on, starting from zero and increasing to positive values.
Graphing Functions
Graphing a function offers a visual insight into the behavior and properties of that function. When graphing \( f(x) = \sqrt{x+1} \), we start by identifying crucial characteristics such as the domain and range.
To graph this function, you'll need to calculate several points by inputting \( x \)-values into the function and plotting the corresponding \( f(x) \)-values on a coordinate graph. These points are plotted as ordered pairs, like \( (-1, 0) \), \( (0, 1) \), \( (3, 2) \), and \( (8, 3) \).
This continuous curve should increase slowly and steadily. The curve reflects the relationship between the \( x \) and \( f(x) \)-values. Do note that the graph starts at \( x = -1 \) and extends indefinitely to the right because \( x \geq -1 \).
To graph this function, you'll need to calculate several points by inputting \( x \)-values into the function and plotting the corresponding \( f(x) \)-values on a coordinate graph. These points are plotted as ordered pairs, like \( (-1, 0) \), \( (0, 1) \), \( (3, 2) \), and \( (8, 3) \).
- Start at the point where \( x = -1 \), which is the smallest \( x \)-value allowed by the domain.
- Continue to plot points for increasing \( x \)-values, and then connect them smoothly with a curve.
This continuous curve should increase slowly and steadily. The curve reflects the relationship between the \( x \) and \( f(x) \)-values. Do note that the graph starts at \( x = -1 \) and extends indefinitely to the right because \( x \geq -1 \).
Inequalities
Inequalities are mathematical expressions involving the symbols \( > \), \( < \), \( \geq \), and \( \leq \). They are used to express ranges of values that a variable can take.
In the context of functions, inequalities are crucial for determining the domain, particularly for square root functions. Since a square root requires the argument to be non-negative, we often deal with inequalities like \( x + 1 \geq 0 \).
To solve \( x + 1 \geq 0 \), subtract 1 from both sides to isolate \( x \), yielding \( x \geq -1 \). This inequality shows that only values \( x \geq -1 \) are suitable as inputs for the function \( f(x) = \sqrt{x+1} \).
Understanding such inequalities is fundamental in mathematics, as they help signal which real-number inputs are valid for certain operations or transformations.
In the context of functions, inequalities are crucial for determining the domain, particularly for square root functions. Since a square root requires the argument to be non-negative, we often deal with inequalities like \( x + 1 \geq 0 \).
To solve \( x + 1 \geq 0 \), subtract 1 from both sides to isolate \( x \), yielding \( x \geq -1 \). This inequality shows that only values \( x \geq -1 \) are suitable as inputs for the function \( f(x) = \sqrt{x+1} \).
Understanding such inequalities is fundamental in mathematics, as they help signal which real-number inputs are valid for certain operations or transformations.
Evaluating Functions
Evaluating a function involves calculating the output \( f(x) \) for specific input values \( x \). For the function \( f(x) = \sqrt{x+1} \), evaluation is done by substituting chosen \( x \)-values into the function and solving.
Consider the value \( x = 3 \). Substitute 3 into the function: \( f(3) = \sqrt{3+1} = \sqrt{4} = 2 \). This tells us that, when \( x \) is 3, \( f(x) \) results in 2.
Creating a table of such computations helps visualize relationships between different \( x \)-values and their outcomes. For instance:
Consider the value \( x = 3 \). Substitute 3 into the function: \( f(3) = \sqrt{3+1} = \sqrt{4} = 2 \). This tells us that, when \( x \) is 3, \( f(x) \) results in 2.
Creating a table of such computations helps visualize relationships between different \( x \)-values and their outcomes. For instance:
- For \( x = -1 \), \( f(-1) = \sqrt{0} = 0 \)
- For \( x = 0 \), \( f(0) = \sqrt{1} = 1 \)
- For \( x = 8 \), \( f(8) = \sqrt{9} = 3 \)
Other exercises in this chapter
Problem 87
a. Add: \(\sqrt{3}+\sqrt{3}\) b. Multiply: \(\sqrt{3} \cdot \sqrt{3}\) c. Describe the differences in parts (a) and (b).
View solution Problem 87
Find the midpoint of the line segment whose endpoints are given. See Example 7 $$ (6,-8),(2,4) $$
View solution Problem 88
Use rational expressions to write as a single radical expression. $$ \frac{\sqrt[4]{a}}{\sqrt[5]{a}} $$
View solution Problem 88
Simplify. \(\frac{\frac{1}{y}+\frac{4}{5}}{-\frac{3}{20}}\)
View solution