What is under the root must be ≥ 0

\(\sqrt{\text{expression}}\) → require: \(\text{expression} \geq 0\)

💡 Why?

Because no real number has a negative square!

For example: \(\sqrt{-4}\) does not exist (in the real numbers)

Step 1: What is under the root? → x

Step 2: Require: \(x \geq 0\)

Domain: \(x \geq 0\) or \([0, \infty)\)

Example 1: \(f(x) = \sqrt{x - 3}\)

Condition: \(x - 3 \geq 0\)

Solution: \(x \geq 3\)

Domain: \([3, \infty)\)

Example 2: \(f(x) = \sqrt{5 - x}\)

Condition: \(5 - x \geq 0\)

Solution: \(5 \geq x\) or \(x \leq 5\)

Domain: \((-\infty, 5]\)

Example 3: \(f(x) = \sqrt{2x + 6}\)

Condition: \(2x + 6 \geq 0\)

Solution: \(2x \geq -6\) → \(x \geq -3\)

Domain: \([-3, \infty)\)

Example 4: \(f(x) = \sqrt{-3x + 12}\)

Condition: \(-3x + 12 \geq 0\)

Solution: \(12 \geq 3x\) → \(x \leq 4\)

Domain: \((-\infty, 4]\)

💡 Method: Solve a quadratic inequality!

1. Find the roots of the quadratic expression

2. Sketch the parabola

3. Mark where the parabola is above the x-axis (≥ 0)

Example 5: \(f(x) = \sqrt{x^2 - 9}\)

Condition: \(x^2 - 9 \geq 0\)

Roots: \(x^2 = 9\) → \(x = 3\) or \(x = -3\)

Parabola with a > 0 (smile): positive "outside"

Domain: \(x \leq -3\) or \(x \geq 3\)

In interval notation: \((-\infty, -3] \cup [3, \infty)\)

Example 6: \(f(x) = \sqrt{-x^2 + 4}\)

Condition: \(-x^2 + 4 \geq 0\)

Roots: \(-x^2 + 4 = 0\) → \(x^2 = 4\) → \(x = \pm 2\)

Parabola with a < 0 (frown): positive "inside"

Domain: \(-2 \leq x \leq 2\) or \([-2, 2]\)

Example 7: \(f(x) = \sqrt{x^2 - 4x + 3}\)

Condition: \(x^2 - 4x + 3 \geq 0\)

Factoring: \((x-1)(x-3) \geq 0\)

Roots: \(x = 1\) or \(x = 3\)

Parabola with a > 0 → positive "outside"

Domain: \(x \leq 1\) or \(x \geq 3\)

Example 8: \(f(x) = \sqrt{x^3}\)

Condition: \(x^3 \geq 0\)

When is x³ positive? When x is positive! (odd power preserves sign)

Domain: \(x \geq 0\)

Example 9: \(f(x) = \sqrt{x^4 - 16}\)

Condition: \(x^4 - 16 \geq 0\)

Solution: \(x^4 \geq 16\) → \(|x|^4 \geq 16\) → \(|x| \geq 2\)

Domain: \(x \leq -2\) or \(x \geq 2\)

Example 10: \(f(x) = \sqrt{(x-1)(x+2)(x-4)}\)

Condition: \((x-1)(x+2)(x-4) \geq 0\)

Roots: \(x = -2, 1, 4\)

Check signs on each interval (or use a sign table):

Domain: \([-2, 1] \cup [4, \infty)\)

1. Even root (√, ⁴√, …) → requires expression ≥ 0

2. Odd root (∛, ⁵√, …) → no restriction! defined for all x

3. If there is a root plus something else:

For example \(f(x) = \sqrt{x-2} + 3x\)

The root is the "bottleneck" — it determines the domain!