Problem 11 - Precalculus Mathematics for Calculus

Question

Find the domain of the function. $$f(x)=\sqrt{x}+\sqrt{1-x}$$

Step-by-Step Solution

Answer

The domain of the function is \([0, 1]\).

Step 1: Understand the Restriction on the Square Root

For the square root function \(\sqrt{x}\), the expression inside the root must be non-negative. This means \(x \geq 0\).

Step 2: Apply the First Condition

The first condition \(x \geq 0\) implies that the domain of \(\sqrt{x}\) is all real values of \(x\) that are zero or positive, i.e., \([0, \infty)\).

Step 3: Understand the Restriction on the Other Square Root

Similarly, for \(\sqrt{1-x}\), the expression inside the root must also be non-negative. Therefore, \(1-x \geq 0\) which leads to \(x \leq 1\).

Step 4: Apply the Second Condition

The second condition \(x \leq 1\) implies that the domain of \(\sqrt{1-x}\) is all real values of \(x\) that are less than or equal to one, i.e., \((-\infty, 1]\).

Step 5: Determine the Intersection of Domains

The domain of the function \(f(x) = \sqrt{x} + \sqrt{1-x}\) is the set of values that satisfy both conditions simultaneously: \(x \geq 0\) and \(x \leq 1\). Therefore, the intersection of \([0, \infty)\) and \((-\infty, 1]\) is \([0, 1]\).

Step 6: State the Domain

The domain of the function is the interval \([0, 1]\), so \(x\) can take any value from 0 to 1 inclusive.

Key Concepts

Square Root Function

The square root function is fundamental when dealing with expressions like \( \sqrt{x} \) and \( \sqrt{1-x} \). For functions involving square roots, one crucial aspect is ensuring that the numbers under the square root are non-negative. This is because the square root of a negative number is not defined in the set of real numbers—such expressions are typically undefined or handled in the complex number domain. Let's break this down a bit more:

• The square root function \( \sqrt{x} \) implies that the expression \( x \) must be greater than or equal to zero (\( x \geq 0 \)).
• Similarly, \( \sqrt{1-x} \) requires that \( 1-x \) is non-negative, which translates to \( x \leq 1 \).

These conditions ensure that for any function involving square roots, like \( f(x) = \sqrt{x} + \sqrt{1-x} \), the expressions inside each square root are non-negative, helping define the domain where the function is valid.

Intersection of Intervals

Intersection of intervals is a technique used to find common values that satisfy multiple inequalities or constraints at the same time. Imagine you want to find the x-values that make both \( \sqrt{x} \) and \( \sqrt{1-x} \) valid. First, we identify the individual intervals:

• The domain for \( \sqrt{x} \) is \([0, \infty)\), meaning x can be zero or any positive number.
• The domain for \( \sqrt{1-x} \) is \(( -\infty, 1] \), meaning x can be any negative number up to and including one.

To determine where both conditions hold true, we find where these intervals overlap. The overlap or intersection of \([0, \infty)\) and \(( -\infty, 1] \) is the interval \([0, 1]\). This intersection represents the x-values where both \( \sqrt{x} \) and \( \sqrt{1-x} \) are defined, giving us the domain of the overall function \( f(x) = \sqrt{x} + \sqrt{1-x} \).

Non-Negative Constraints

Non-negative constraints are conditions that ensure any expression under a square root remains valid for real number calculations. When dealing with square roots, for example, the expression inside must be greater than or equal to zero. This principle is applied to find the domain of functions involving square roots:

• For \( \sqrt{x} \), the non-negative constraint is \( x \geq 0 \).
• For \( \sqrt{1-x} \), the constraint becomes \( 1-x \geq 0 \), simplifying to \( x \leq 1 \).

These non-negative constraints are crucial in determining where a function is valid and can provide real number outputs. By applying these constraints, we ensure the overall function, like \( f(x) = \sqrt{x} + \sqrt{1-x} \), behaves correctly over its defined domain. The resulting domain, \([0, 1]\), is therefore the set of x-values where both constraints are satisfied simultaneously.