# Math Help - polynomial with max $min 1. ## polynomial with max$ min

Let $p(x)$ be a real polynomials of Least Degree Which has Local Maximum at $x=1$ and Local Minimum

at $x=3$.If $p(1)=6$ and $p(3)=2$. Then $p^{'}(0)=$

2. ## Re: polynomial with max $min Originally Posted by jacks Let $p(x)$ be a real polynomials of Least Degree Which has Local Maximum at $x=1$ and Local Minimum at $x=3$.If $p(1)=6$ and $p(3)=2$. Then $p^{'}(0)=$ $p(x)=x^3-6x^2+9x+2 ~\text { so }~ p'(x)=3x^2-12x+9$ Hence : $p'(0)=9$ 3. ## Re: polynomial with max$ min

Hello, jacks!

$\text{Let }p(x)\text{ be a polynomial of least degree which has local max at }x=1$
$\text{and local minimum at }x=3.\;\text{ If }p(1)=6\text{ and }p(3)=2\text{, then }p'(0)= \_\_$

The desired polynomial is a cubic: . $p(x) \:=\:ax^3 + bx^2 + cx + d$
Its derivative is: . $p'(x) \:=\:3ax^2 + 2bx + c$

We are given:

. . $\begin{array}{ccccc}p(1) = 6\!:& a + b + c + d &=& 6 \\ \\[-4mm] p(3) = 2\!:& 27a + 9b + 3c + d &=&2 \\ \\[-4mm] p'(1) = 0\!: & 3a + 2b + c &=&0 \\ \\[-4mm] p'(3) = 0\!: & 27a + 6b + c &=&0 \end{array}$

Solve the system of equations: . $a = 1,\;b = -6,\;c = 9,\;d = 2$

The cubic is: . $p(x) \:=\:x^3 - 6x^2 + 9x + 2$
. . Its derivative is: . $p'(x) \:=\:3x^2 - 12x + 9$

Therefore: . $p'(0) \:=\:3(0^2) - 12(0) + 9 \:=\:9$