Problem 25 - Precalculus Essentials

Question

Find the domain of each function. $$h(x)=\sqrt{x-2}+\sqrt{x+3}$$

Step-by-Step Solution

Answer

The domain of the function \(h(x)\) is \(x \geq 2\).

Step 1: Set up Inequalities

We first set up inequalities for each term under the square root, stating that they must be greater than or equal to 0: \[(x-2) \geq 0\] and \[(x+3) \geq 0\].

Step 2: Solve the Inequalities

Next, we solve both inequalities for \(x\). The inequality \(x-2 \geq 0\) becomes \(x \geq 2\) when calculated, and the inequality \(x+3 \geq 0\) becomes \(x \geq -3\) when simplified.

Step 3: Find the Common Domain

Now, we have to find the domain that satisfies both inequalities. The domain \(x \geq 2\) means that \(x\) can be any number greater than or equal to 2. The domain \(x \geq -3\) means that \(x\) can be any number greater than or equal to -3. A number that satisfies both conditions is a number that is greater than or equal to 2. Therefore, the domain of the function \(h(x)\) is \(x \geq 2\).