It looks like the image is related to a problem involving the Intermediate Value Theorem (IVT) and the analysis of a cubic function $f(x) = -x^3 + 6x^2 - 7x$. Let's break down the components:

1. Using the Intermediate Value Theorem:

   • The Intermediate Value Theorem states that if a function $f(x)$ is continuous on a closed interval $[a, b]$ and $N$ is any value between $f(a)$ and $f(b)$, then there exists a $c \in (a, b)$ such that $f(c) = N$.
   • The goal is to demonstrate that the equation has a solution in the given interval $[-1, 1.5]$ by applying this theorem. First, ensure that the function is continuous on the interval (which it is, as it is a polynomial).

2. Graphing utility to solve the equation:

   • After applying the IVT, using a graphing utility would help visualize the roots of the equation on the interval $[-1, 1.5]$.

3. Illustrate with a graph:

   • You would graph $f(x) = -x^3 + 6x^2 - 7x$ on this interval to see where the function crosses the x-axis, confirming the solutions.

From the image, the following values are given:

• Function at the left endpoint $x = -1$: $f(-1) = 14$.
• Function at the right endpoint $x = 1.5$: $f(1.5) = 0$ (based on the box shown in the image).

The next steps involve:

• Using the fact that $f(-1) = 14$ and $f(1.5) = 0$, the Intermediate Value Theorem confirms that there is a root between $x = -1$ and $x = 1.5$ because the function changes from positive to zero.

Do you want a more detailed solution or an explanation of the graph? Here are some follow-up questions to consider:

1. How does the Intermediate Value Theorem help find the roots of polynomial functions?
2. Can you describe why continuity is crucial in applying the IVT?
3. How would you solve the equation $-x^3 + 6x^2 - 7x = 3$ algebraically?
4. Why is $f(x) = -x^3 + 6x^2 - 7x$ continuous for all $x$?
5. What can be concluded from the IVT when $f(-1) = 14$ and $f(1.5) = 0$?

Tip: Always verify the continuity of a function before applying the Intermediate Value Theorem to ensure its conditions hold.