The image shows two problems related to finding the domain of functions involving square roots and fractions. Both problems appear to have incorrect answers marked with red crosses.

Let's analyze each case:

Problem (c)

Given function: $\sqrt{3 - 4x} \cdot (8x - 4)$

• The expression inside the square root must be non-negative: $3 - 4x \geq 0$ Solving for $x$: $x \leq \frac{3}{4}$

• Also, we must consider the product $(8x - 4)$. However, this does not affect the domain since no restrictions are placed by it alone on $x$, as the square root is not applied to the entire product.

• Therefore, the domain is: $(-\infty, \frac{3}{4}]$

The domain provided is $\left(-\infty, \frac{1}{2}\right] \cup \left[\frac{3}{4}, \infty\right)$, which is incorrect because $\frac{3}{4}$ is the upper bound, not part of a union. The correct domain should be $(-\infty, \frac{3}{4}]$.

Problem (d)

Given function: $\frac{1}{\sqrt{3 - 4x} \cdot (8x - 4)}$

• Here, the expression inside the square root must also be non-negative: $3 - 4x > 0 \quad \text{(since the denominator cannot be zero)}$ Solving for $x$: $x < \frac{3}{4}$

• Additionally, the entire denominator $\sqrt{3 - 4x} \cdot (8x - 4)$ must not be zero. Therefore, we must exclude the value of $x$ that makes $8x - 4 = 0$: $8x - 4 = 0 \implies x = \frac{1}{2}$

• The domain excludes both $x = \frac{1}{2}$ and any $x \geq \frac{3}{4}$. Thus, the domain is: $(-\infty, \frac{1}{2}) \cup \left(\frac{1}{2}, \frac{3}{4}\right)$

The given domain in the image is the union of intervals $\left(-\infty, \frac{1}{2}\right] \cup \left[\frac{3}{4}, \infty\right)$, which is incorrect. The correct domain excludes $\frac{1}{2}$ and is within $(-\infty, \frac{3}{4})$, excluding $\frac{1}{2}$.

Summary

Both answers provided are incorrect due to misunderstandings in interpreting the restrictions imposed by the square root and the entire denominator in problem (d).

Would you like more details on the steps or have any questions?

Here are five related questions to further explore:

1. What happens to the domain if the square root is applied to the entire expression in problem (c)?
2. How do we handle the domain when dealing with inequalities involving square roots?
3. Why does the presence of a square root in a denominator lead to more restrictive domain conditions?
4. Can the domain include points where the denominator equals zero if the numerator is also zero?
5. How would the domain change if the expression inside the square root had a different sign?

Tip: Always carefully consider whether an expression can lead to undefined situations like division by zero or taking the square root of a negative number when determining domains.