# Finding $p'(0)$ for the polynomial of least degree which has a local maximum at $x=1$ and a local minimum at $x=3$

The question goes as follows:

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...

What I did first, naturally, was consider that $p(x)$ is a cubic. But, all conditions cannot be satisfied simultaneously if $p(x)$ is a cubic.

Next, I considered $p(x)$ to be a $4$ degree polynomial, solved and got $p'(0)=15$. But the answer provided was $p'(0)=9$.

• Why do you think a cubic doesn't work? – Daniel Fischer Aug 16 '14 at 16:50
• @DanielFischer I made a mistake.. I considered $f'(x)$ to be $(x-1)(x-3)$, whereas I should have considered it to be $a(x-1)(x-3)$... – user1001001 Aug 16 '14 at 16:57
• maxima is the plural form of maximum. You want a singular form here (I edited). – user147263 Aug 16 '14 at 18:34

Basically you need to setup system of equations and solve the coefficients of $p(x)$

Since $p(x)$ has two turning points, the minimum possible degree is $3$

Say $p(x) = ax^3+bx^2+cx+d$

$\implies p'(x) = 3ax^2+2bx+c$

from the problem we have :

$p(1) = 6$

$p(3) = 2$

$p'(1) = 0$

$p'(3)=0$

four equations and four unknowns - can be easily solved

$p'(1)=0, p'(3)=0, p(1)=6, p(3)=2$.

We must have $\partial p' \ge 2$, so try $p'(x) = c(x-1)(x-3)$. This gives $p(x) = p(1)+c\int_1^x p'(t)dt = 6+{c \over 3} (x-4)(x-1)^2$. Setting $p(3) = 2$ gives $c=3$ and so we have $p(x) = 6+(x-4)(x-1)^2$.

Computing $p'(0)$ gives 9.

As an alternative to finding the coefficients of $p$ explicitly, start with the "standard" zig-zagging cubic $$f(x) = 2x^3 - 3x^2$$ which has a maximum at $(0,0)$ and a minimum at $(1,-1)$.

Now just scale and translate to get $p(x) = 6 + 4f(\frac{x-1}2)$. Then you don't even need to simplify to find $p'(0)$, just the chain rule.

Hint:

$p(x)$ is in fact a cubic.

The system of equations you have to solve is:

$$\begin{eqnarray}p'(1) = 0&\Rightarrow&3a& + &2b& + &c& & &=& 0\\p(1) = 6&\Rightarrow&a&+&b&+&c&+&d &=& 6\\p'(3) = 0&\Rightarrow&27a&+&6b&+&c& & &=&0\\ p'(3) = 2&\Rightarrow&27a&+&9b&+&3c&+&d&=&2 \end{eqnarray}$$

Note that $p'(0) = c$

(You should end up with $p(x) = x^3-6x^2+9x+2$)