Let's analyze the domain of each function step by step:

(a) $f(x) = \frac{1}{\sqrt{x} + |x|}$

Finding the domain:

• The square root function $\sqrt{x}$ requires $x \geq 0$ since the square root of a negative number is undefined in the real number system.
• The absolute value function $|x|$ is defined for all real numbers $x$.

Now we analyze the denominator $\sqrt{x} + |x|$:

• For $x \geq 0$, $\sqrt{x} = x$ and $|x| = x$, so $\sqrt{x} + |x| = 2x$, which is 0 when $x = 0$.
• At $x = 0$, the denominator becomes zero, which makes the function undefined.

Therefore, the function is defined for $x > 0$. The domain of $f(x)$ is: $\boxed{(0, \infty)}$

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(b) $f(x) = \frac{1}{\sqrt{x^2 - x - 6}}$

Finding the domain:

• The square root function $\sqrt{x^2 - x - 6}$ requires that the expression inside the square root, $x^2 - x - 6$, must be non-negative: $x^2 - x - 6 \geq 0$ We solve this inequality by factoring: $x^2 - x - 6 = (x - 3)(x + 2) \geq 0$

Now, we solve $(x - 3)(x + 2) \geq 0$ using a sign chart:

• The roots are $x = 3$ and $x = -2$.
• The intervals to test are $(-\infty, -2)$, $(-2, 3)$, and $(3, \infty)$.

Testing each interval:

• For $x < -2$, both $x - 3$ and $x + 2$ are negative, so $(x - 3)(x + 2) > 0$.
• For $-2 < x < 3$, one factor is negative and the other is positive, so $(x - 3)(x + 2) < 0$.
• For $x > 3$, both factors are positive, so $(x - 3)(x + 2) > 0$.

Thus, $x^2 - x - 6 \geq 0$ holds for $x \in (-\infty, -2] \cup [3, \infty)$.

Also, since the denominator cannot be zero, we exclude points where $x^2 - x - 6 = 0$, i.e., $x = -2$ and $x = 3$.

Therefore, the domain of $f(x)$ is: $\boxed{(-\infty, -2) \cup (3, \infty)}$

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Summary of domains:

• (a) $f(x) = \frac{1}{\sqrt{x} + |x|}$ has the domain $(0, \infty)$.
• (b) $f(x) = \frac{1}{\sqrt{x^2 - x - 6}}$ has the domain $(-\infty, -2) \cup (3, \infty)$.

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

1. How would the domain change if $f(x) = \sqrt{x} + |x|$ instead of being in the denominator?
2. What happens if we remove the absolute value in the first function?
3. Can the domain of a function ever include complex numbers?
4. How do you handle quadratic inequalities like in part (b) without factoring?
5. How does the domain change if we shift or scale the quadratic expression?

Tip: Always check for points where the denominator becomes zero or where the function inside a square root is negative, as these restrict the domain.