Evaluate $$ f(x)=x^{3}-5 x^{2}+3 x+6 $$ at 1 and 3 to determine if the Intermediate Value Theorem guarantees that a zero exists between the two values. a) $$ f(1)=5 $$ b) $$ f(3)= $$

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Sure, let's evaluate the function $$ f(x) = x^3 - 5x^2 + 3x + 6 $$ at $$ x = 3 $$ to proceed with the Intermediate Value Theorem (IVT) analysis.

### b) Evaluating $$ f(3) $$:

$$
f(3) = 3^3 - 5 \cdot 3^2 + 3 \cdot 3 + 6
$$
$$
f(3) = 27 - 45 + 9 + 6
$$
$$
f(3) = (27 + 9 + 6) - 45
$$
$$
f(3) = 42 - 45
$$
$$
f(3) = -3
$$

### Conclusion Using the Intermediate Value Theorem:

- **At $$ x = 1 $$:** $$ f(1) = 5 $$ (positive)
- **At $$ x = 3 $$:** $$ f(3) = -3 $$ (negative)

Since $$ f(x) $$ is a continuous function (being a polynomial) and it changes sign between $$ x = 1 $$ and $$ x = 3 $$, the Intermediate Value Theorem guarantees that there exists at least one value $$ c $$ in the interval $$ (1, 3) $$ such that $$ f(c) = 0 $$.

**Answer Summary:**

b) $$ f(3) = -3 $$
b) $$ f(3) = -3 $$ The Intermediate Value Theorem guarantees that there is at least one zero between 1 and 3 since $$ f(1) = 5 $$ and $$ f(3) = -3 $$ have opposite signs.

Evaluate at 1 and 3 to determine if the Intermediate Value Theorem
guarantees that a zero exists between the two values.
a)
b)

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