To find the domain and range of the function $f(x) = \frac{1}{\sqrt{25 - x^2}}$, we need to ensure that the expression inside the square root is non-negative, and that the denominator is non-zero.

1. Domain:

The function is defined when:

• The expression inside the square root, $25 - x^2$, is non-negative, meaning $25 - x^2 \geq 0$.
• The denominator must not be zero, so $\sqrt{25 - x^2} \neq 0$.

Let's solve these conditions step by step.

Step 1: Solve $25 - x^2 \geq 0$:

$25 \geq x^2 \implies -5 \leq x \leq 5$ This gives the interval $x \in [-5, 5]$, which ensures the square root is non-negative.

Step 2: Ensure the denominator is non-zero:

For the denominator to be non-zero, $\sqrt{25 - x^2} \neq 0$, which means $25 - x^2 \neq 0$. This simplifies to: $x^2 \neq 25 \implies x \neq \pm 5$

Thus, the domain is all $x$ such that $-5 < x < 5$.

Final Domain:

$x \in (-5, 5)$

2. Range:

To find the range of $f(x) = \frac{1}{\sqrt{25 - x^2}}$, we need to consider the behavior of the function within its domain.

• As $x$ approaches $-5$ or $5$, the term $\sqrt{25 - x^2}$ approaches zero, meaning $f(x)$ approaches infinity.
• As $x$ approaches 0, $\sqrt{25 - 0^2} = \sqrt{25} = 5$, so $f(0) = \frac{1}{5}$.

Thus, the function can take any value from $\frac{1}{5}$ to $+\infty$.

Final Range:

$f(x) \in \left( \frac{1}{5}, +\infty \right)$

Summary:

• Domain: $(-5, 5)$
• Range: $\left( \frac{1}{5}, +\infty \right)$

Would you like more details or have any questions? Here are some related questions:

1. How do we handle square roots in other types of functions?
2. What happens when $x$ is exactly 5 or -5?
3. How do we find the range of a rational function in general?
4. What transformations would shift the domain and range of this function?
5. How would the domain and range change if the square root expression were $25 - x^4$?

Tip: Always check both the non-negativity of square roots and the non-zero condition for denominators when dealing with such functions.