A polynomial $P(x)$ of degree $5$ with lead coefficient one,increases in the $(-\infty,1)$ and $(3,\infty)$ and decreases in the interval $(1,3)$

Question

A polynomial function $P(x)$ of degree $5$ with leading coefficient one,increases in the interval $(-\infty,1)$ and $(3,\infty)$ and decreases in the interval $(1,3)$. Given that $P(0)=4$ and $P'(2)=0$, find the value of $P'(6)$.

I noticed that $1,2,3$ are the roots of the polynomial $P'(x)$.$P'(x)$ is a fourth degree polynomial. So

Let $$P'(x)=(x-1)(x-2)(x-3)(ax+b)$$ Now I expanded $P'(x)$ and integrated it to get $P(x)$ and use the given condition $P(0)=4$, but I am not able to calculate $a,b$.

My previous (deleted) comment is well-summarized and corrected by the answer of @HagenVonEitzen. — awllower, Apr 10 '16 at 14:51

score 7 · Accepted Answer · answered Apr 10 '16 at 14:49

7

As $P(x)=x^5+\ldots$, we have $P'(x)=5x^4+\ldots$. This gives us $a=5$. Moreover, the facts that $P$ is decreasing on $(1,3)$ and $P'(2)=0$ imply that $x=2$ must in fact be a multiple root of $P'$. We conclude that $2a+b=0$, so $b=-10$.

answered Apr 10 '16 at 14:49

Hagen von Eitzen

374,180

1

Note that the factoid about $P(0)$ was totally irrelevant to determining $P'$ – Hagen von Eitzen Apr 10 '16 at 14:52
I got it thank you.@Hagen von Eitzen – learner_avid Apr 10 '16 at 15:21

score -2 · Answer 2 · answered Dec 15 '20 at 03:24

-2

The polynomial of degree 5, P(x) has a leading coefficient 1, has roots of multiplicity 2 at x=5 and x=0, and a root of multiplicity 1 at x=-2 find a possible formula for P(x).

answered Dec 15 '20 at 03:24

Nish J.

1

A polynomial $P(x)$ of degree $5$ with lead coefficient one,increases in the $(-\infty,1)$ and $(3,\infty)$ and decreases in the interval $(1,3)$

2 Answers2