Here’s a detailed solution to the questions provided in the image:

1. Find the range of $f(x) = x^2$ for the domain $-2 \leq x \leq 3$:

The function $f(x) = x^2$ is a quadratic function that opens upwards. To find the range:

1. Evaluate $f(x)$ at the endpoints of the domain:

   • When $x = -2$: $f(-2) = (-2)^2 = 4$
   • When $x = 3$: $f(3) = 3^2 = 9$

2. Check if there are other values of $x$ in the domain where $f(x)$ reaches a minimum or maximum. Since $f(x) = x^2$, the minimum occurs at $x = 0$ (the vertex within the domain):

   • When $x = 0$: $f(0) = 0^2 = 0$

Range of $f(x)$: $[0, 9]$.

2. Given $h(x) = x^2 - 5$, find the range of $h(x)$ for real $x$:

The function $h(x) = x^2 - 5$ is also a quadratic function that opens upwards, shifted downward by 5 units. Its vertex occurs at $x = 0$, and at the vertex:

• $h(0) = 0^2 - 5 = -5$

Since $x^2 \geq 0$ for all real $x$, $h(x)$ will always satisfy:

• $h(x) \geq -5$.

Range of $h(x)$: $[ -5, \infty )$.

3. Given $g(x) = (x - 1)^2$, find the range of $g(x)$:

The function $g(x) = (x - 1)^2$ is a quadratic function that opens upwards, with its vertex at $x = 1$. At the vertex:

• $g(1) = (1 - 1)^2 = 0$

Since $(x - 1)^2 \geq 0$ for all real $x$, the function is non-negative.

Range of $g(x)$: $[0, \infty)$.

Do you want more detailed steps or further explanations? Here are some related questions to expand your understanding:

1. How do transformations like shifting and stretching affect the range of a quadratic function?
2. Can you derive the vertex formula for a quadratic function using calculus?
3. What happens to the range of $h(x)$ if a constant $c$ is added (e.g., $h(x) + c$)?
4. How does restricting the domain impact the range of a function like $g(x) = (x - 1)^2$?
5. Can you generalize the range of a quadratic function $ax^2 + bx + c$ for all possible values of $a$?

Tip: For any quadratic function, always locate the vertex first to determine the minimum or maximum value and use the domain to further constrain the range.