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Chapter 2: Problem 23

Exer \(21-32:\) Find the domain of \(f\) $$f(x)=\sqrt{16-x^{2}}$$

Short Answer

Expert verified
The domain of \(f(x) = \sqrt{16 - x^2}\) is \([-4, 4]\).

Step by step solution

01

Understand the Domain of a Square Root Function

The domain of a function is the set of input values (x-values) for which the function is defined. For a square root function like \(f(x) = \sqrt{16 - x^2}\), the expression inside the square root must be greater than or equal to zero. This is because the square root of a negative number is not defined in the set of real numbers.
02

Set Up an Inequality for the Radicand

The radicand in this function is \(16 - x^2\). We need this to be greater than or equal to zero: \(16 - x^2 \geq 0\). Solving this inequality will provide the domain of the function.
03

Rearrange the Inequality

Start by rearranging the inequality: \(16 \geq x^2\). This implies that \(x^2 \leq 16\).
04

Solve for x

Solve the inequality \(x^2 \leq 16\) by taking the square root of both sides. Remember to consider both the positive and negative roots since square roots will give \(|x| \leq 4\), which means \(-4 \leq x \leq 4\).
05

Write the Domain in Interval Notation

The domain of the function is all \(x\) values from \(-4\) to \(4\), inclusive. In interval notation, this is written as \([-4, 4]\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Square Root Function
The square root function is a key concept in mathematics, often used in various equations and functions. It is important to understand that the square root of a number or an expression refers to a value that, when multiplied by itself, gives the original number or expression. In our exercise, the function is given by \(f(x) = \sqrt{16 - x^2}\).
  • To ensure the function is real and defined, the expression inside the square root, known as the radicand, must be non-negative.
  • This is because the square root of a negative number is not a real number.
Therefore, for the function \(f(x)\) to be defined, the condition \(16 - x^2 \geq 0\) must be satisfied. Understanding why the radicand must be non-negative helps us determine the domain of the function, or all possible \(x\)-values for which \(f(x)\) is defined.
Inequality Solving
Solving inequalities is a critical skill when working with functions, especially those involving square roots. In this case, the inequality \(16 - x^2 \geq 0\) must be addressed to find the domain of the function.
  • First, rearrange the inequality as \(16 \geq x^2\). This helps in simplifying the analysis, leading to solutions where \(x^2 \leq 16\).
  • To solve \(x^2 \leq 16\), take the square root of both sides. It鈥檚 crucial to consider both the positive and negative roots, resulting in \(|x| \leq 4\).
By decomposing this absolute value inequality, we find two separate inequalities: \(-4 \leq x \leq 4\). Solving these shows all possible \(x\)-values meeting the initial condition. Mastery of inequality solving allows us to precisely define where functions are valid and operational.
Interval Notation
Interval notation is a concise way of representing a range of numbers and is crucial for expressing domains of functions. Once we've solved the inequality and determined the possible \(x\)-values, we can neatly represent the domain using interval notation.
  • Given the range \(-4 \leq x \leq 4\), we include both endpoints because these are solutions where the expression inside the square root remains zero and valid.
  • In interval notation, this domain is expressed as \([-4, 4]\).
The square brackets indicate that the endpoints are included in the domain, reflecting the inclusive nature of our original inequality \(-4 \leq x \leq 4\). Interval notation is a succinct way to communicate domain information clearly and effectively across mathematical contexts.

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Most popular questions from this chapter

Find (a) ( \(f \circ g)(x)\) and the domain of \(f \circ g\) and (b) \((g \circ f)(x)\) and the domain of \(g \circ f\) $$f(x)=\frac{x+2}{x-1}, \quad \quad \quad \quad \quad \quad g(x)=\frac{x-5}{x+4}$$

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