Key Concepts and Formulas

• Local Extrema and First Derivative: A polynomial function $p(x)$ has local extrema (maximum or minimum) at points $x=c$ where its first derivative $p'(c) = 0$. These points are called critical points.
• Degree of Derivative: If $p(x)$ is a polynomial of degree $n$, then its derivative $p'(x)$ is a polynomial of degree $n-1$.
• Indefinite Integration (Antidifferentiation): The indefinite integral of a function $f(x)$, denoted by $\int f(x) \, dx$, gives a family of functions whose derivative is $f(x)$. We must remember to add the constant of integration, denoted by $C$.

Step-by-Step Solution

Step 1: Define the polynomial and its derivative

Since $p(x)$ is a polynomial of degree 3, we can express its derivative $p'(x)$ as a polynomial of degree 2. Also, we know that $p'(1) = 0$ and $p'(2) = 0$ because $p(x)$ has local extrema at $x=1$ and $x=2$. Therefore, $x=1$ and $x=2$ are roots of $p'(x)$. We can express $p'(x)$ as: $p'(x) = k(x-1)(x-2)$ where $k$ is a constant. The goal is to find $p(0)$.

Step 2: Integrate to find p(x)

Integrate $p'(x)$ to obtain $p(x)$: $p(x) = \int p'(x) \, dx = \int k(x-1)(x-2) \, dx = k \int (x^2 - 3x + 2) \, dx$ $p(x) = k \left( \frac{x^3}{3} - \frac{3x^2}{2} + 2x \right) + C$ where $C$ is the constant of integration.

Step 3: Use the given information to find k and C

We are given that $p(1) = 8$ and $p(2) = 4$. Substitute these values into the expression for $p(x)$:

For $x = 1$, $p(1) = 8$: $8 = k \left( \frac{1}{3} - \frac{3}{2} + 2 \right) + C = k \left( \frac{2 - 9 + 12}{6} \right) + C = \frac{5k}{6} + C$ So, $\frac{5k}{6} + C = 8 \hspace{1cm} (1)$

For $x = 2$, $p(2) = 4$: $4 = k \left( \frac{8}{3} - \frac{12}{2} + 4 \right) + C = k \left( \frac{8}{3} - 6 + 4 \right) + C = k \left( \frac{8 - 18 + 12}{3} \right) + C = \frac{2k}{3} + C$ So, $\frac{2k}{3} + C = 4 \hspace{1cm} (2)$

Now we have a system of two equations with two unknowns, $k$ and $C$. Subtract equation (2) from equation (1): $\left( \frac{5k}{6} + C \right) - \left( \frac{2k}{3} + C \right) = 8 - 4$ $\frac{5k}{6} - \frac{4k}{6} = 4$ $\frac{k}{6} = 4$ $k = 24$

Substitute $k = 24$ into equation (2): $\frac{2(24)}{3} + C = 4$ $16 + C = 4$ $C = -12$

Step 4: Determine p(x)

Now we have the values of $k$ and $C$, so we can write the complete expression for $p(x)$: $p(x) = 24 \left( \frac{x^3}{3} - \frac{3x^2}{2} + 2x \right) - 12$ $p(x) = 8x^3 - 36x^2 + 48x - 12$

Step 5: Calculate p(0)

Finally, we need to find $p(0)$: $p(0) = 8(0)^3 - 36(0)^2 + 48(0) - 12$ $p(0) = -12$

Common Mistakes & Tips

• Remember the constant of integration, $C$, when integrating. Forgetting it will lead to an incorrect answer.
• Carefully substitute the values of $x$ and $p(x)$ to form the equations needed to solve for the unknown constants.
• Double-check your algebra and arithmetic throughout the process.

Summary

We started by finding the derivative of the polynomial in terms of an unknown constant $k$. Integrating the derivative gave us the general form of the polynomial, including another unknown constant $C$. Using the given information about the local extrema, we created a system of two equations to solve for $k$ and $C$. With these constants, we could determine the specific polynomial and evaluate it at $x=0$, which resulted in $p(0) = -12$.

Final Answer

The final answer is $\boxed{-12}$, which corresponds to option (C).