Let's go through the problem step by step:

The function given is: $f(x) = \frac{1}{2}x^3 - 3x^2 + 5x + 2$

(a) Find the average rate of change between the points (1, 4.5) and (3.62, 4.5).

The formula for the average rate of change of a function $f(x)$ between two points $x_1$ and $x_2$ is:

$\text{Average Rate of Change} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$

Here, $x_1 = 1$, $x_2 = 3.62$, and $f(x_1) = f(1) = 4.5$, $f(x_2) = f(3.62) = 4.5$.

So, the average rate of change:

$\text{Average Rate of Change} = \frac{4.5 - 4.5}{3.62 - 1} = \frac{0}{2.62} = 0$

Thus, the average rate of change is 0.

(b) Does the Mean Value Theorem (MVT) apply to $f$ over the interval [1, 3.62]?

The Mean Value Theorem (MVT) states that if a function is continuous and differentiable on the closed interval $[a, b]$, then there exists at least one point $c$ in the open interval $(a, b)$ such that:

$f'(c) = \frac{f(b) - f(a)}{b - a}$

For the MVT to apply:

1. Continuity: The given function $f(x) = \frac{1}{2}x^3 - 3x^2 + 5x + 2$ is a polynomial, which is continuous everywhere.
2. Differentiability: Polynomials are also differentiable everywhere.

Thus, $f(x)$ is both continuous and differentiable over the interval $[1, 3.62]$, so the Mean Value Theorem applies.

(c) If so, find the value(s) guaranteed by the Mean Value Theorem.

To find the value(s) of $c$ guaranteed by the Mean Value Theorem, we need to find $c$ such that:

[ f'(c) = \frac{f(3.62) - f(1)}{3.62