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The problem provided in the image asks to determine how many positive values of \( b \) make the limit of the given function \( f(x) \) equal to 2. The function is:

\[
f(x) = 0.1x^4 - 0.5x^3 - 3.3x^2 + 7.7x - 1.99
\]

We are tasked with finding the number of positive values of \( b \) such that:

\[
\lim_{x \to b} f(x) = 2
\]

This means we need to solve the equation:

\[
f(b) = 2
\]

Substituting the function into this equation, we get:

\[
0.1b^4 - 0.5b^3 - 3.3b^2 + 7.7b - 1.99 = 2
\]

Simplifying:

\[
0.1b^4 - 0.5b^3 - 3.3b^2 + 7.7b - 1.99 - 2 = 0
\]

\[
0.1b^4 - 0.5b^3 - 3.3b^2 + 7.7b - 3.99 = 0
\]

Now, we need to find the positive values of \( b \) that satisfy this equation. I will solve this equation numerically to determine the number of positive solutions.The polynomial equation \( 0.1b^4 - 0.5b^3 - 3.3b^2 + 7.7b - 3.99 = 0 \) has three positive real roots:

1. \( b \approx 0.95 \)
2. \( b \approx 1.05 \)
3. \( b \approx 8.00 \)

Thus, there are **three positive values** of \( b \) for which \( \lim_{x \to b} f(x) = 2 \). The correct answer is **C. Three**.

Would you like more details or have any questions?

Here are five related questions for further exploration:
1. How do we find the roots of a general polynomial equation?
2. What is the significance of using limits in calculus?
3. How does the degree of a polynomial influence the number of possible solutions?
4. What methods exist for solving higher-degree polynomials by hand?
5. Can a polynomial function have both real and imaginary roots?

**Tip:** When solving polynomial equations numerically, always check the nature of the roots (real vs. imaginary) and their signs to determine relevant solutions for the problem.

The function f(x) = 0.1x^4 - 0.5x^3 - 3.3x^2 + 7.7x - 1.99. For how many positive values of b does lim(x→b) f(x) = 2?

Learn how to solve the polynomial equation f(x) = 0.1x^4 - 0.5x^3 - 3.3x^2 + 7.7x - 1.99 for positive values of b where the limit lim(x→b) f(x) = 2. Explore the concepts of limits and polynomial roots. Suitable for students in Grades 11-12.

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