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Let's break down the problem from the image step by step:

### a. Intermediate Value Theorem (IVT) Application:
The problem asks you to apply the **Intermediate Value Theorem** (IVT) to show that the equation \( x^3 + x + 3 = 0 \) has a solution in the interval \( (-2, 1) \).

The **Intermediate Value Theorem** states:
- If \( f(x) \) is continuous on a closed interval \( [a, b] \),
- And \( N \) is a value strictly between \( f(a) \) and \( f(b) \),
- Then there is at least one \( c \in (a, b) \) such that \( f(c) = N \).

Here, the function is \( f(x) = x^3 + x + 3 \), and we are given the interval \( (-2, 1) \).

#### Step 1: Check Continuity
The function \( f(x) = x^3 + x + 3 \) is a polynomial, and polynomials are continuous everywhere, so it's continuous on \( (-2, 1) \). This satisfies the first condition of the IVT.

#### Step 2: Evaluate the Function at the Endpoints
Evaluate \( f(x) \) at \( x = -2 \) and \( x = 1 \):
- \( f(-2) = (-2)^3 + (-2) + 3 = -8 - 2 + 3 = -7 \)
- \( f(1) = (1)^3 + 1 + 3 = 1 + 1 + 3 = 5 \)

Now we have:
- \( f(-2) = -7 \)
- \( f(1) = 5 \)

#### Step 3: Apply IVT
Since \( f(-2) = -7 \) and \( f(1) = 5 \), and 0 lies between -7 and 5, by the Intermediate Value Theorem, there exists at least one \( c \in (-2, 1) \) such that \( f(c) = 0 \). Therefore, the equation \( x^3 + x + 3 = 0 \) has a solution in this interval.

### b. Graphing Utility and Solutions
To find the exact solution(s), you can use a graphing calculator or utility to graph the function \( f(x) = x^3 + x + 3 \) and identify where the graph crosses the x-axis (i.e., where \( f(x) = 0 \)) in the interval \( (-2, 1) \).

### c. Graph Illustration
Once the function is graphed, you should illustrate the curve of \( x^3 + x + 3 \) on the interval \( (-2, 1) \), ensuring that it passes through the x-axis, confirming the existence of the solution.

#### Summary:
- \( f(x) \) is continuous on \( (-2, 1) \).
- \( f(-2) = -7 \), \( f(1) = 5 \), and 0 lies between -7 and 5.
- Therefore, by the Intermediate Value Theorem, a solution exists in the interval \( (-2, 1) \).

Would you like further details on this, or help with solving any part? 

Here are five related questions to expand your understanding:
1. How does the Intermediate Value Theorem work for non-polynomial functions?
2. Can there be more than one solution in the interval \( (-2, 1) \)?
3. What other methods can be used to solve \( x^3 + x + 3 = 0 \) besides the IVT?
4. How would the graph of \( x^3 + x + 3 \) look outside the interval \( (-2, 1) \)?
5. How do you verify the exact root found using graphing tools?

**Tip:** Always remember that polynomials are continuous everywhere, so the IVT can easily apply to any interval for polynomial functions!

Use the Intermediate Value Theorem to show that the following equation has a solution on the given interval: x^3 + x + 3 = 0; Interval: (-2,1). Evaluate the function at the endpoints and illustrate the solution with a graph.

Explore how the Intermediate Value Theorem applies to the polynomial x^3 + x + 3 within the interval (-2,1), including continuity checks, evaluations at endpoints, and graphical illustrations confirming the existence of a solution.

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