# Thread: polynomial with max $min 1. ## polynomial with max$ min

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

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

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

Hello, jacks!

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

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

We are given:

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

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

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

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