Let's go through each function and find their domain and range step by step.

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(a) $f(x) = \frac{\sqrt{5x - 10}}{x^2 - 36}$

Domain:

• For the square root expression $\sqrt{5x - 10}$, we need the expression inside the square root to be non-negative (i.e., $5x - 10 \geq 0$). Solving for $x$:

$5x - 10 \geq 0$ $5x \geq 10$ $x \geq 2$

• For the denominator $x^2 - 36$, we need it to not equal zero, since division by zero is undefined. Solving for when $x^2 - 36 = 0$:

$x^2 - 36 = 0$ $x^2 = 36$ $x = \pm 6$

Thus, $x = 6$ and $x = -6$ are excluded from the domain.

Domain:

The domain is $x \geq 2$, excluding $x = 6$. So, the domain is: $\boxed{[2, 6) \cup (6, \infty)}$

Range:

• The numerator $\sqrt{5x - 10}$ is always non-negative for $x \geq 2$.
• The denominator $x^2 - 36$ can take any value except zero.
• As $x \to 2^+$, $\sqrt{5x - 10} \to 0$, and the denominator approaches $-36$. Therefore, $f(x)$ approaches $0/(-36) = 0$.
• As $x \to \infty$, the numerator grows, and the denominator also grows large, so $f(x)$ approaches 0.

Therefore, $f(x)$ can take any real value, except zero.

Range:

The range is all real numbers except 0: $\boxed{(-\infty, 0) \cup (0, \infty)}$

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(b) $f(x) = x^2$

Domain:

The function $f(x) = x^2$ is a polynomial, and polynomials are defined for all real numbers. So, the domain is: $\boxed{(-\infty, \infty)}$

Range:

Since $x^2$ is always non-negative (i.e., $x^2 \geq 0$ for all $x$), the range is: $\boxed{[0, \infty)}$

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(c) $f(x) = \frac{\sqrt{x + 1}}{x - 4}$

Domain:

• For the square root expression $\sqrt{x + 1}$, we need $x + 1 \geq 0$, which gives: $x \geq -1$

• For the denominator $x - 4$, we need $x \neq 4$ (since division by zero is undefined).

Domain:

The domain is $x \geq -1$, excluding $x = 4$. So, the domain is: $\boxed{[-1, 4) \cup (4, \infty)}$

Range:

• As $x \to -1^+$, the numerator $\sqrt{x + 1} \to 0$, and the denominator approaches $-5$, so $f(x) \to 0$.
• As $x \to 4^-$, the numerator remains finite, and the denominator approaches 0 from the negative side, causing $f(x) \to -\infty$.
• As $x \to \infty$, the numerator grows, and the denominator also grows, so $f(x)$ approaches 0.

Therefore, $f(x)$ can take any value except 0, with $f(x)$ having a negative range near $x = 4$.

Range:

The range is all real numbers except 0: $\boxed{(-\infty, 0) \cup (0, \infty)}$

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Summary:

• (a) Domain: $[2, 6) \cup (6, \infty)$, Range: $(-\infty, 0) \cup (0, \infty)$
• (b) Domain: $(-\infty, \infty)$, Range: $[0, \infty)$
• (c) Domain: $[-1, 4) \cup (4, \infty)$, Range: $(-\infty, 0) \cup (0, \infty)$

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Would you like more details on any of the steps? Here are some related questions for further exploration:

1. How does the square root function affect the domain of a function?
2. What happens to the range of a rational function as the denominator approaches zero?
3. How do you handle functions that have both a square root and a denominator?
4. What are some techniques for finding the domain and range of rational functions?
5. How does the power of $x$ in a polynomial affect the range of the function?

Tip: When dealing with rational functions, always check for points where the denominator is zero, as these will not be in the domain.