why do you not need the infinity symbol? are solutions inbetween negative root 6 and root 6?

Great questions! Let's clarify both points.

### Why don't we need the infinity symbol?
The domain of a function refers to the set of all possible input values (in this case, \(x\)) for which the function is defined. When dealing with square roots, the function is only defined when the expression inside the square root is non-negative.

For the function \( h(x) = \sqrt{6 - x^2} \), we determined that \( x^2 \leq 6 \), which leads to \( -\sqrt{6} \leq x \leq \sqrt{6} \). This means that \( x \) can take any value between \(-\sqrt{6}\) and \(\sqrt{6}\) inclusive, but it cannot be smaller than \(-\sqrt{6}\) or greater than \(\sqrt{6}\). 

In this context, the infinity symbol is not needed because the domain is a closed interval: \( [-\sqrt{6}, \sqrt{6}] \). This interval contains all real numbers between \(-\sqrt{6}\) and \(\sqrt{6}\), and nothing beyond these boundaries.

### Are the solutions between \(-\sqrt{6}\) and \(\sqrt{6}\)?
Yes, the solutions are all the values of \( x \) that lie between \(-\sqrt{6}\) and \(\sqrt{6}\), inclusive. This means the function \( h(x) = \sqrt{6 - x^2} \) is defined for any real number \( x \) in the interval \([- \sqrt{6}, \sqrt{6}]\). Every point within this interval is a valid input for \( x \) because it satisfies the inequality \( 6 - x^2 \geq 0 \), ensuring that the expression under the square root is non-negative.

To summarize:
- The interval \([- \sqrt{6}, \sqrt{6}]\) includes all the values between \(-\sqrt{6}\) and \(\sqrt{6}\), including the endpoints.
- The function is defined for all \( x \) within this interval, so no infinity symbols are needed.

Would you like more details or have any other questions?

Here are 5 related questions:
1. How do you find the range of the function \( h(x) = \sqrt{6 - x^2} \)?
2. What happens to the domain if \( h(x) = \sqrt{x^2 - 6} \) instead?
3. How would you express the domain if the function were \( h(x) = \frac{1}{\sqrt{6 - x^2}} \)?
4. Can \( h(x) = \sqrt{6 - x^2} \) ever equal 0? If so, at what \( x \)-values?
5. How does the graph of \( h(x) = \sqrt{6 - x^2} \) look within the domain?

**Tip:** When expressing domains, square brackets \([ ]\) indicate that endpoints are included, while parentheses \(( )\) indicate they are not.

Explore why the infinity symbol is not needed in the domain of h(x) = sqrt(6 - x^2) and whether solutions lie between -sqrt(6) and sqrt(6). This detailed explanation covers square roots, function domains, and inequalities.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation