Let $p(x)$ be a real polynomial of least degree which has a local maximum at $x = 1$ and a local minimum at $x = 3$. If $p(1) = 6$ and $p(3) = 2$, then $p'(0)$ is

Let p(x) be a real polynomial of least° which has a local maximum at x = 1 and a local minimum at x = 3. If p(1) = 6 and p(3) = 2, then p'(0) is

$\text{local max at } x=1, \quad \text{local min at } x=3$

$p(1)=6, \quad p(3)=2$

$\text{To find: } p'(0)$

$\text{local max/min} \Rightarrow p'(1)=0, \quad p'(3)=0$

$\text{least degree} \Rightarrow p'(x) = k(x-1)(x-3)$

$p'(x) = k(x^2 - 4x + 3)$

$p(x) = k\left(\dfrac{x^3}{3} - 2x^2 + 3x\right) + C$

$p(3) = k(9 - 18 + 9) + C = C = 2 \Rightarrow C = 2$

$p(1) = k\left(\dfrac13 - 2 + 3\right) + 2 = k\cdot\dfrac43 + 2 = 6 \Rightarrow k = 3$

$p'(x) = 3(x-1)(x-3)$

$p'(0) = 3(-1)(-3) = 9$

$\boxed{9}$