Chapter 1: Problem 17
For the following exercises, find the domain, range, and all zeros/intercepts, if any, of the functions. $$ f(x)=-1+\sqrt{x+2} $$
Short Answer
Expert verified
Domain: \([-2, \infty)\); Range: \([-1, \infty)\); Zero: \(x = -1\).
Step by step solution
01
Identify the Domain
The function is \( f(x) = -1 + \sqrt{x+2} \), which involves a square root. The expression under the square root, \( x+2 \), must be greater than or equal to zero for the function to be defined. Set up the inequality: \( x+2 \geq 0 \). Solving this inequality gives \( x \geq -2 \). Therefore, the domain of the function is all \( x \) values that are greater than or equal to \(-2\), or \([-2, \infty)\).
02
Determine the Range
The range of \( f(x) = -1 + \sqrt{x+2} \) is determined by the values \( -1 + \sqrt{x+2} \) can take. The minimum of \( \sqrt{x+2} \) is zero (when \( x = -2 \)), making \( f(x) = -1 \). As \( x \) increases, \( \sqrt{x+2} \) becomes larger, so \( f(x) \) gets larger without bound. Therefore, the range is \([-1, \infty)\).
03
Find the Zeros/Intercepts
To find the zeros, set \( f(x) = 0 \): \(-1 + \sqrt{x+2} = 0\). Solve for \( x \): \[ \sqrt{x+2} = 1 \] Square both sides: \[ x+2 = 1^2 = 1 \] \[ x = 1 - 2 = -1 \] So, the zero of the function is at \( x = -1 \). The y-intercept occurs when \( x = 0 \): \( f(0) = -1 + \sqrt{0+2} = -1 + \sqrt{2} \), which is not a zero but an intercept on the y-axis.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Domain of a Function
The domain of a function tells us which values we can plug into the function that will not cause any mathematical errors. For our function, \( f(x) = -1 + \sqrt{x+2} \), we must be cautious about the square root. The expression inside the square root, \( x+2 \), must be greater than or equal to zero to avoid taking the square root of a negative number, which is undefined in the real number system.
We set up the inequality: \( x+2 \geq 0 \). Solving this inequality involves subtracting 2 from both sides, resulting in \( x \geq -2 \).
This means that the domain of this function includes all real numbers from \( -2 \) to positive infinity. We write this domain in interval notation as \([-2, \infty)\). Here, \(-2\) is included in the domain because the expression is zero at \( x = -2 \), and the square root of zero is zero.
We set up the inequality: \( x+2 \geq 0 \). Solving this inequality involves subtracting 2 from both sides, resulting in \( x \geq -2 \).
This means that the domain of this function includes all real numbers from \( -2 \) to positive infinity. We write this domain in interval notation as \([-2, \infty)\). Here, \(-2\) is included in the domain because the expression is zero at \( x = -2 \), and the square root of zero is zero.
Range of a Function
The range of a function is about the possible output values that the function can produce. For \( f(x) = -1 + \sqrt{x+2} \), we start by considering the smallest and largest possible values that can result from the expression \( \sqrt{x+2} \).
At \( x = -2 \), the expression equals zero, so \( f(-2) = -1 + 0 = -1 \). This value represents the minimum output, or the smallest the function can be.
At \( x = -2 \), the expression equals zero, so \( f(-2) = -1 + 0 = -1 \). This value represents the minimum output, or the smallest the function can be.
- As \( x \) increases, the value of \( \sqrt{x+2} \) becomes larger, potentially reaching very large numbers. Since there is no upper bound for \( x \), the outputs of the function can increase without limit.
Zeros of a Function
Zeros of a function are the input values where the function outputs zero. These are also referred to as the "roots" or "x-intercepts."
To find the zeros of \( f(x) = -1 + \sqrt{x+2} \), we set the function equal to zero: \(-1 + \sqrt{x+2} = 0\).
To find the zeros of \( f(x) = -1 + \sqrt{x+2} \), we set the function equal to zero: \(-1 + \sqrt{x+2} = 0\).
- Adding 1 to both sides gives us: \( \sqrt{x+2} = 1 \).
- Squaring both sides removes the square root: \( x+2 = 1^2 = 1 \).
- Finally, subtracting 2 from both sides results in \( x = -1 \).
Square Root Functions
Square root functions have unique properties because they involve the square root of an expression. These functions have a distinctive curved graph and specific domain restrictions due to the square root.
For example, in the function \( f(x) = -1 + \sqrt{x+2} \), the square root affects both the domain and the shape of the graph.
For example, in the function \( f(x) = -1 + \sqrt{x+2} \), the square root affects both the domain and the shape of the graph.
- The domain is limited to x-values that make the quantity inside the square root non-negative. Thus, any changes inside the radical affect the domain directly.
- The graph of the function starts at a certain point (here, \( x=-2 \)) and continues to rise that impacts the function's minimum or starting value (here, \(-1\)).