The exercise in the image asks to find the domain and range of the following functions:

1. $f(x) = 1 + x^2$
2. $f(x) = 1 - \sqrt{x}$
3. $F(x) = \sqrt{5x + 10}$
4. $g(x) = \sqrt{x^2 - 3x}$
5. $f(t) = \frac{4}{3 - t}$
6. $G(t) = \frac{2}{t^2 - 16}$

Let's solve each one for domain and range:

1. $f(x) = 1 + x^2$

   • Domain: $x \in (-\infty, \infty)$, since the quadratic function is defined for all real numbers.
   • Range: $y \geq 1$, as $x^2$ is always non-negative, and the minimum value is $1$ when $x = 0$.

2. $f(x) = 1 - \sqrt{x}$

   • Domain: $x \geq 0$, because the square root function is defined for non-negative values.
   • Range: $y \leq 1$, since $\sqrt{x} \geq 0$, the maximum value occurs when $x = 0$, and the function decreases as $x$ increases.

3. $F(x) = \sqrt{5x + 10}$

   • Domain: $5x + 10 \geq 0$ ⟶ $x \geq -2$.
   • Range: $y \geq 0$, since the square root function always returns non-negative values.

4. $g(x) = \sqrt{x^2 - 3x}$

   • Domain: $x^2 - 3x \geq 0$ ⟶ $x(x - 3) \geq 0$, which gives $x \leq 0$ or $x \geq 3$ by solving the inequality.
   • Range: $y \geq 0$, since it's a square root function.

5. $f(t) = \frac{4}{3 - t}$

   • Domain: $3 - t \neq 0$ ⟶ $t \neq 3$.
   • Range: All real numbers except $y = 0$, because the function never equals zero.

6. $G(t) = \frac{2}{t^2 - 16}$

   • Domain: $t^2 - 16 \neq 0$ ⟶ $t \neq \pm 4$.
   • Range: All real numbers except $y = 0$, as the denominator never makes the function zero.

Would you like further details or explanations?

Related Questions:

1. How do you find the domain of a square root function?
2. What is the range of a quadratic function?
3. How does the domain of a rational function differ from that of a square root function?
4. Can the range of a rational function include zero? Why or why not?
5. How do you solve inequalities to find the domain of a function?

Tip: Always check for values that make the denominator zero or cause square roots to produce negative values when determining the domain.