Understanding the Domain of a Function

The question asks for the domain of the function \(f(x)=\sqrt{2-x}+\sqrt{2+x}\). The domain of a function is the set of all possible input values (x-values) for which the function is defined and produces a real number output.

For a function involving square roots, the expression under the square root symbol must be non-negative (greater than or equal to zero) because the square root of a negative number is not a real number. The given function is a sum of two square root terms. For the function \(f(x)\) to be defined, both terms must be defined independently.

Determining the Domain for Each Term

Let's consider the first term: \(\sqrt{2-x}\). For this term to be defined, the expression inside the square root, \(2-x\), must be greater than or equal to zero.

• Set up the inequality: \(2-x \ge 0\)
• Subtract 2 from both sides: \(-x \ge -2\)
• Multiply both sides by -1 and reverse the inequality sign: \(x \le 2\)

So, the first term \(\sqrt{2-x}\) is defined for all \(x\) such that \(x \le 2\). In interval notation, this is \((-\infty, 2]\).

Now, let's consider the second term: \(\sqrt{2+x}\). For this term to be defined, the expression inside the square root, \(2+x\), must be greater than or equal to zero.

• Set up the inequality: \(2+x \ge 0\)
• Subtract 2 from both sides: \(x \ge -2\)

So, the second term \(\sqrt{2+x}\) is defined for all \(x\) such that \(x \ge -2\). In interval notation, this is \([-2, \infty)\).

Finding the Intersection of Domains

The function \(f(x)=\sqrt{2-x}+\sqrt{2+x}\) is defined only when both terms are defined. Therefore, the domain of \(f(x)\) is the intersection of the domains of the two individual terms.

We need to find the values of \(x\) that satisfy both \(x \le 2\) AND \(x \ge -2\).

• From the first term: \(x \le 2\)
• From the second term: \(x \ge -2\)

Combining these two conditions, we get \(-2 \le x \le 2\).

In interval notation, this range is represented as \([-2, 2]\). This means the function \(f(x)\) is defined for all real numbers \(x\) from -2 to 2, including -2 and 2.

Comparing with Options

Let's look at the given options and compare them with our calculated domain \([-2, 2]\):

The domain of the function \(f(x)=\sqrt{2-x}+\sqrt{2+x}\) is \([-2, 2]\).

Revision Table: Key Concepts for Domain

Additional Information: Domain of Functions

Understanding the domain is crucial because it tells us for which input values a function is meaningful and produces a real output. Different types of functions have different restrictions on their domains.

• Algebraic Functions: These involve algebraic operations (addition, subtraction, multiplication, division, roots). For these, we typically need to avoid division by zero and taking the square root (or any even root) of a negative number.
• Transcendental Functions: These include trigonometric, exponential, and logarithmic functions. They have their own specific domain restrictions (e.g., \(\log(x)\) is defined only for \(x > 0\)).

When a function is a combination of different types of functions, like our example \(f(x)\) which is a sum of two square root functions, the domain is found by determining the restrictions for each part and finding the values that satisfy all restrictions simultaneously (the intersection of the individual domains).

For \(f(x)=\sqrt{2-x}+\sqrt{2+x}\), we needed \(2-x \ge 0\) AND \(2+x \ge 0\). Solving these inequalities gave us \(x \le 2\) and \(x \ge -2\). The intersection of these conditions is \(-2 \le x \le 2\), which corresponds to the interval \([-2, 2]\).