Derivative of Minimum Degree Polynomial with Given Extrema

To find the polynomial $p(x)$ of minimum degree, we note that $p(x)$ has local extrema at $x = 1$ and $x = 3$. Therefore, $x = 1$ and $x = 3$ must be critical points, which means $p'(1) = 0$ and $p'(3) = 0$.

The simplest derivative $p'(x)$ satisfying these conditions is a quadratic polynomial, making $p(x)$ a cubic polynomial (degree $3$).

Let the derivative of $p(x)$ be represented as: $p'(x) = a(x - 1)(x - 3) = a(x^2 - 4x + 3)$ where $a$ is a non-zero real constant.

Integrating $p'(x)$ with respect to $x$, we get: $p(x) = a \left( \frac{x^3}{3} - 2x^2 + 3x \right) + C$ where $C$ is the constant of integration.

We are given $p(3) = 2$ and $p(1) = 6$:

1. Using $p(3) = 2$: $p(3) = a \left( \frac{3^3}{3} - 2(3)^2 + 3(3) \right) + C = 2$ $a(9 - 18 + 9) + C = 2 \implies C = 2$

2. Using $p(1) = 6$: $p(1) = a \left( \frac{1^3}{3} - 2(1)^2 + 3(1) \right) + 2 = 6$ $a \left( \frac{4}{3} \right) + 2 = 6 \implies a \left( \frac{4}{3} \right) = 4 \implies a = 3$

Since $a = 3 > 0$, we verify the extrema using the second derivative test: $p''(x) = 3(2x - 4)$

• At $x = 1$, $p''(1) = -6 < 0$, which confirms a local maximum at $x = 1$.
• At $x = 3$, $p''(3) = 6 > 0$, which confirms a local minimum at $x = 3$.

Thus, the derivative of the polynomial is: $p'(x) = 3(x - 1)(x - 3)$

Now, evaluating $p'(0)$: $p'(0) = 3(0 - 1)(0 - 3) = 3(-1)(-3) = 9$