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Chapter 1: Problem 25

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

Short Answer

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

Step by step solution

01

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 \[(x+3) \geq 0\].
02

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.
03

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 \(x \geq 2\).

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Most popular questions from this chapter

Write the standard form of the equation of the circle with the given center and radius. $$(x+1)^{2}+y^{2}=25$$

A tangent line to a circle is a line that intersects the circle at exactly one point. The tangent line is perpendicular to the radius of the circle at this point of contact. Write an equation in point-slope form for the line tangent to the circle whose equation is \(x^{2}+y^{2}=25\) at the point (3,-4).

Solve by the quadratic formula: \(5 x^{2}-6 x-8=0\) (Section P.7, Example 10)

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