Let's break down and solve the problem step by step:

a. Using the Intermediate Value Theorem

The Intermediate Value Theorem (IVT) states that if a function $f(x)$ is continuous on a closed interval $[a, b]$, and if $f(a)$ and $f(b)$ have opposite signs, then there exists at least one $c$ in the open interval $(a, b)$ such that $f(c) = 0$.

The given equation is: $f(x) = x^3 - 5x^2 + 2x + 2$ We need to show that this equation has a solution on the interval $[-1, 5]$.

1. Check continuity of the function:
   The function $f(x) = x^3 - 5x^2 + 2x + 2$ is a polynomial. Since all polynomials are continuous everywhere, the function is continuous on the given interval $[-1, 5]$.

2. Evaluate $f(x)$ at the endpoints:

   • At $x = -1$: $f(-1) = (-1)^3 - 5(-1)^2 + 2(-1) + 2 = -1 - 5 - 2 + 2 = -6$
   • At $x = 5$: $f(5) = 5^3 - 5(5^2) + 2(5) + 2 = 125 - 125 + 10 + 2 = 12$

   Since $f(-1) = -6$ and $f(5) = 12$, the function changes signs between $-1$ and $5$. By the Intermediate Value Theorem, there must be at least one solution to $f(x) = 0$ in the interval $(-1, 5)$.

b. Using a graphing utility to find solutions

Using a graphing tool or calculator, you would input the function $f(x) = x^3 - 5x^2 + 2x + 2$ and find the x-values where the graph crosses the x-axis, which represents the solutions to $f(x) = 0$ in the interval $[-1, 5]$.

c. Illustrating with an appropriate graph

A graph of $f(x) = x^3 - 5x^2 + 2x + 2$ on the interval $[-1, 5]$ would show a curve that crosses the x-axis at least once between $-1$ and $5$, confirming that there is a solution in that interval. The exact solution can be approximated graphically using the graphing tool.

Continuity Question:

The second part of the problem asks for which values of $x$ the function $f(x) = x^3 - 5x^2 + 2x$ is continuous.

Since $f(x)$ is a polynomial, it is continuous for all real numbers. Therefore, the correct answer is:

B. It is continuous for all x.

Let me know if you'd like more detailed steps or explanations!

Here are some related questions to explore further:

1. How can we use the Intermediate Value Theorem in other types of problems?
2. What types of functions are not continuous on certain intervals?
3. How can we approximate solutions to polynomial equations graphically?
4. What is the significance of differentiability in relation to continuity?
5. How does the degree of a polynomial affect its number of roots?

Tip: The Intermediate Value Theorem is especially useful for finding roots in cases where you cannot solve the equation algebraically.