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

Short Answer

Expert verified
The domain of the function \(h(x)=\sqrt{x-2}+\sqrt{x+3}\) is \(x \geq 2\).

Step by step solution

01

Setting each Root Greater or Equal to Zero

To find the domain, let's set the expression under each square root greater or equal to zero. So we start by setting \(x-2 \geq 0\) and \(x+3 \geq 0\).
02

Solving Inequalities

Solving these inequalities results in two solutions for x: \(x \geq 2\) from \(x-2 \geq 0\) and \(x \geq -3\) from \(x+3 \geq 0\).
03

Intersection of Solution

Since both conditions have to be satisfied (i.e., both square roots need to be defined), the final domain is the intersection of both solution sets, which means taking the greatest lower limit. Our domain is then \(x \geq 2\).

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Most popular questions from this chapter

In Exercises \(105-108,\) you will be developing functions that model given conditions. A company that manufactures bicycles has a fixed cost of \(\$ 100,000 .\) It costs \(\$ 100\) to produce each bicycle. The total cost for the company is the sum of its fixed cost and variable costs. Write the total cost, \(C,\) as a function of the number of bicycles produced, \(x .\) Then find and interpret \(C(90)\)

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