Problem 26 - Precalculus Essentials

Question

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

Step-by-Step Solution

Answer

The domain of \(h(x) = \sqrt{x-3}+\sqrt{x+4}\) is \(x \ge 3\)

Step 1: Understand the condition for domain of square root functions

For a function with square roots, as the one at hand, a primary consideration to determine their domains is that the expressions inside the square roots must be nonnegative. This is because in the set of real numbers, which is where we're working, there are no real square roots of negative numbers.

Step 2: Set up inequality for first square root

Start with the requirement for the first square root: \(x-3 \ge 0\). This means that \(x\) (the possible values in the domain) must be greater than or equal to 3.

Step 3: Set up inequality for second square root

Similarly, for the requirement of the second square root: \(x+4 \ge 0\). This means that \(x\) must be greater than or equal to -4.

Step 4: Identify the common values

Since the function \(h(x)\) includes BOTH the square roots and should be defined for all the values, we must consider the common values satisfying both inequalities. Clearly, \(x\) must be greater than or equal to 3, as that includes the range from -4 upwards.