Let's analyze the functions given in the image and find their domains of definition.

(i) $g(x) = \frac{x^3}{x^2 - 3x + 2}$

This is a rational function. The domain is all real values of $x$ except where the denominator becomes zero. To find the values that make the denominator zero, we need to solve the equation:

$x^2 - 3x + 2 = 0$

Factoring the quadratic:

$x^2 - 3x + 2 = (x - 1)(x - 2) = 0$

So, $x = 1$ and $x = 2$ make the denominator zero. Therefore, the domain of $g(x)$ is:

$x \in (-\infty, 1) \cup (1, 2) \cup (2, \infty)$

(ii) $F(t) = \sqrt{2 - \sqrt{t}}$

For the function $F(t)$, we must ensure that the expressions inside both square roots are non-negative.

1. The inner square root, $\sqrt{t}$, requires that $t \geq 0$.
2. The expression inside the outer square root, $2 - \sqrt{t}$, must also be non-negative: $2 - \sqrt{t} \geq 0$ Solving this inequality: $\sqrt{t} \leq 2$ Squaring both sides: $t \leq 4$

Combining both conditions, we get $0 \leq t \leq 4$.

Thus, the domain of $F(t)$ is: $t \in [0, 4]$

Would you like any further clarification or more details?

Related Questions:

1. How do you solve more complicated quadratic equations when finding domains of rational functions?
2. What are the implications of having complex numbers in the domain of a square root function?
3. How would the domain of $g(x)$ change if the denominator were a cubic polynomial instead of quadratic?
4. Can you explain how transformations affect the domain of a square root function?
5. How do you deal with domains for functions involving higher-degree radicals (e.g., cube roots)?

Tip:

When dealing with rational functions, always check the denominator for values that could cause division by zero, as those will not be in the domain.