Find the Domain of √(ax + b) – Square Root of a Linear Function

For a function \( f(x) = \sqrt{ax + b} \) to have real values, the radicand (expression under the square root) must satisfy:

\( ax + b \ge 0 \)

The domain is the set of all \( x \) that satisfy this linear inequality. The solution depends on the sign of coefficient \( a \).

📌 The graph below shows the function \( f(x) = \sqrt{ax + b} \) (blue) and its radicand (green line). The function exists only where the green line is on or above the x-axis.

✧ Domain of √(ax + b) ✧

Enter coefficients a and b of the linear expression

Linear Expression: \( ax + b \)

\( f(x) = \sqrt{ax + b} \)   →   \( f(x) = \sqrt{2x - 4} \)

⚠️ If a = 0, the function becomes \( f(x) = \sqrt{b} \), a constant. Its domain is all real numbers if b ≥ 0, empty otherwise.
📐 Domain of \( f(x) = \sqrt{ax + b} \)
\( [2, \infty) \)

📖 Step-by-Step Solution

STEP 1: Set the radicand greater than or equal to zero
\( 2x - 4 \ge 0 \)
STEP 2: Solve the linear inequality \( ax + b \ge 0 \)
STEP 3: Write the domain in interval notation
\( [2, \infty) \)
💡 Graphical Interpretation
The blue curve is \( f(x) = \sqrt{ax + b} \), the green line is \( y = ax + b \). The function exists only where the green line is on or above the x-axis.
📊 Interactive Graph
f(x) = √(ax + b) y = ax + b (radicand) Domain boundary point