Problem 26 - College Algebra Essentials

Question

Find the domain of each function. $$h(x)=\sqrt{x-3}+\sqrt{x+4}$$

Step-by-Step Solution

Answer

The domain of the function \(h(x) = \sqrt{x-3} + \sqrt{x+4}\) is all x-values such that \(x \geq 3\).

Step 1: Identify the Conditions for Each Square Root

Here, the function h(x) is composed of two square root functions: \(\sqrt{x-3}\) and \(\sqrt{x+4}\). We need to ensure each square root function can accept values that make it non-negative. Therefore, we have two conditions: \(x-3 \geq 0\) and \(x+4 \geq 0\).

Step 2: Solve the Conditions

Solve each inequality to find the minimum value that x can take for each square root. For \(x-3 \geq 0\), solving for x gives \(x \geq 3\). For \(x+4 \geq 0\), solving for x gives \(x \geq -4\).

Step 3: Evaluate Constraints for Domain

We have two constraints for x from Step 2: \(x \geq 3\) and \(x \geq -4\). The domain of the function must satisfy BOTH constraints. Therefore, when comparing these two constraints, the larger lower limit is the real lower limit of the domain. Hence the function is only useful for \(x \geq 3\). This means we include all numbers greater or equal to 3 to be the domain of the function.