find the domain and range of 1 over root 25squred minus xsquared

Let's consider the function:

\[
f(x) = \frac{1}{\sqrt{25^2 - x^2}} = \frac{1}{\sqrt{625 - x^2}}
\]

### Domain:
The domain consists of all \( x \) values for which the function is defined. Since the square root is in the denominator, we need to ensure that:

1. \( 625 - x^2 \geq 0 \) (the expression inside the square root must be non-negative).
2. \( \sqrt{625 - x^2} \neq 0 \) (the denominator must be non-zero).

Let's solve these conditions step by step.

1. \( 625 - x^2 \geq 0 \):

\[
625 \geq x^2
\]
\[
x^2 \leq 625
\]
Taking square roots of both sides:

\[
-25 \leq x \leq 25
\]

2. \( \sqrt{625 - x^2} \neq 0 \):

\[
625 - x^2 \neq 0
\]
\[
x^2 \neq 625
\]
\[
x \neq 25 \quad \text{and} \quad x \neq -25
\]

Thus, the domain is:

\[
x \in (-25, 25)
\]

### Range:
The range consists of all possible values of \( f(x) \).

- The expression inside the square root, \( \sqrt{625 - x^2} \), achieves its maximum value when \( x = 0 \), giving \( \sqrt{625} = 25 \).
- As \( x \) approaches \( \pm 25 \), the expression inside the square root approaches zero.

Thus, the values of \( f(x) \) range from:

\[
f(x) = \frac{1}{\sqrt{625 - x^2}}
\]

The minimum value occurs when \( \sqrt{625 - x^2} \) is largest (i.e., when \( x = 0 \)), giving \( f(0) = \frac{1}{25} \). As \( x \) approaches \( \pm 25 \), \( f(x) \) becomes very large, approaching infinity.

Therefore, the range of the function is:

\[
f(x) \in (0, \frac{1}{25}]
\]

### Summary:
- **Domain**: \( (-25, 25) \)
- **Range**: \( (0, \frac{1}{25}] \)

Would you like more details or have any questions about this?

Here are five related questions you can explore:
1. What is the domain of \( \frac{1}{\sqrt{a^2 - x^2}} \) for any constant \( a \)?
2. How would the domain change if the function were \( \frac{1}{25 - x^2} \)?
3. What happens to the range if we change the square root to a cube root in the function?
4. How does the graph of \( f(x) = \frac{1}{\sqrt{625 - x^2}} \) behave near the boundary points \( x = \pm 25 \)?
5. Can we extend this function to complex numbers for values of \( x \) outside the domain?

**Tip**: When finding the domain of a square root, remember to ensure the expression inside it is non-negative, as square roots of negative numbers are undefined in the real number system.

Learn how to find the domain and range of the function f(x) = 1 over √(25^2 - x^2). This problem explores key mathematical concepts such as square roots, inequalities, and function behavior. Suitable for students in Grades 9-12.

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