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

Method: Since the smallest polynomial whose derivative gives 2 root would be a cubic equation. So I assumed my function to be

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

Solving the condition above gives me $b=-6a$ and $c=11a$. I am stuck after this. Can anyone tell me how to proceed from here?

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    $\begingroup$ Well, you also know that $P(1)=6,P(3)=2$. $\endgroup$ – lulu Sep 23 '18 at 14:02
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    $\begingroup$ Note: I don't agree that $c=11a$. You have $P'(x)=3ax^2+2bx+c=\lambda(x-3)(x-1)=\lambda x^2-4\lambda x+3\lambda\implies a=3\lambda,\,2b=-4\lambda=-12a,\,c=3\lambda=9a$. No? $\endgroup$ – lulu Sep 23 '18 at 14:06
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Hint: Consider the System $$a+b+c+d=6$$ $$27a+9b+3c+d=2$$ $$3a+2b+c=0$$ $$27a+6b+c=0$$

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  • $\begingroup$ Graubner,So,P'(0)=9. Am i right? $\endgroup$ – Dhamnekar Winod Sep 23 '18 at 14:45
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HINT

You attempt was right to assume a function of the type $P(x)=ax^3+bx^2+cx+d$ as the solution. Rewriting the conditions as $P(1)=6, P(3)=2, P'(1)=0$ and $P'(3)=0$ leads to a system of four equations with four variables $a,b,c$ and $d$.

Can you do it on your own from hereon?

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If you use lulu's idea, then you should get the answer quicker because you will be solving a simultaneous system in two variables, instead of four variables. First, if $P(x)$ is of lowest degree, then we expect $P(x)$ to be a cubic polynomial. Since $P'(x)$ is a quadratic polynomial with two roots $x=1$ and $x=3$, we have $$P'(x)=3a(x-1)(x-3)=3ax^2-12ax+9a$$ for some constant $a$ (the factor $3$ is there simply to make the integration easier). Thus, $$P(x)=ax^3-6ax^2+9ax+c$$ for some constant $c$. Since $P(1)=6$ and $P(3)=2$, we get $$4a+c=6\text{ and }c=2\,,$$ so $a=1$ and $c=2$. That is, $$P(x)=x^3-6x^2+9x+2\,,$$ with $$P'(x)=3(x-1)(x-3)=3x^2-12x+9\text{ and }P''(x)=6(x-2)\,.$$ This makes $P''(1)=-6<0$ and $P''(3)=6>0$, so $x=1$ and $x=3$ are the local maximum and the local minimum, respectively. Hence, $P(x)$ is indeed the desired polynomial.

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