If $a + \frac{b}{2} + \frac{c}{3} + \frac{d}{4} = 0$ then prove the function $$P(x)=a+bx+cx^2+dx^3$$ has a root somewhere between $0$ and $1$.

If it has a root between and $0$ and $1$ then I could show that it changes signs between $0$ and $1$. But some cubics have roots where they don't change signs.

$$P(1)=a+b+c+d\\P(0)=a$$ So $$P(1)=a+b+c+d=\\=a+b+c+d-(a + \frac{b}{2} + \frac{c}{3} + \frac{d}{4})\\=\frac{b}{2} + \frac{2c}{3} + \frac{3d}{4}\\P(0)=a=a-(a + \frac{b}{2} + \frac{c}{3} + \frac{d}{4})\\= -\frac{b}{2} -\frac{c}{3} - \frac{d}{4}$$ I feel like I'm close.

Although I'm sure there is a solution using the formula for cubic roots, I don't think I am supposed to use it, because the next part of the question (which I will hopefully be able to do myself) asks to generalize the result to any polynomial.

  • $\begingroup$ Maybe the two expressions have opposite signs still? $\endgroup$ – Joao Noch Nov 7 '17 at 0:03
  • $\begingroup$ Problem source is James Stewart Calculus 6E Ch5 (Integrals) $\endgroup$ – Joao Noch Nov 7 '17 at 0:04
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    $\begingroup$ Because you need at least one coeff to be non zero you can get a root between 1/2 and 1. Because 1>1/3>1/4 and 1>1/4>1/8 then P(1)=a+b+c+d>a1+b2+c3+d4=0>a1+b2+c4+d8=P(1/2)P(1)=a+b+c+d>a1+b2+c3+d4=0>a1+b2+c4+d8=P(1/2) wait: i supposed they were positive, i am left with the rest still $\endgroup$ – user499752 Nov 7 '17 at 0:14

.Consider the polynomial defined by : $Q(x) = ax + \frac{bx^2}{2} + \frac{cx^3}{3} + \frac{dx^4}{4}$. Note that $Q(0) = 0$, since $Q(x)$ has constant term $0$. Furthermore, $Q(1) = a + \frac b2 + \frac c3 + \frac d4 = 0$. Therefore, we have located two roots of $Q$, $0$ and $1$.

Therefore, by Rolle's theorem, the derivative of $Q$ has a root between $0$ and $1$. But you can see that $Q'(x) = P(x)$, so that $P(x)$ has a root between $0$ and $1$.

  • $\begingroup$ I'm liking it. That explains why it's in a calculus book (integral section) as well. $\endgroup$ – Joao Noch Nov 7 '17 at 0:14
  • $\begingroup$ Thank you. This answer hit me when I saw the regularity of the expression $a + \frac b2 + \frac c3 + \frac d4 = 0$. $\endgroup$ – Teresa Lisbon Nov 7 '17 at 0:16

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