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Well,
if I assume all $a_i$'s to be positive

then $f(0)$ would be $a_0= - \{ a_1/2 + a_2/3..... \} $ , a negative value
and $f(1)= a_1+a_2...... $, a positive value so it does have a root.

Is it good enough?

  • 1
    How can $a_0,...,a_n$ be positive? Look at the hypothesis. – 2017-02-10
  • 0
    Ouchhh.........! – 2017-02-10
  • 3
    Hint: Look at the integral of $f$ on $[0,1]$, – 2017-02-10
  • 0
    Got it. Thanks. – 2017-02-10

3 Answers 3

2

This is a straight-forward application of Rolle's Theorem.

Consider $f(x)=a_0x+\frac{a_1x^2}{2}+\frac{a_2x^3}{3}+\frac{a_3x^4}{4}+ \ldots +\frac{a_nx^{n+1}}{n+1}$

So, according to the theorem, in the interval $[0,1]$

  1. Since $f(x)$ is a polynomial, it is continuous over the interval.
  2. Since $f(x)$ is a polynomial, it is differentiable over the interval.
  3. $f(0)=f(1)=0$

So, by the theorem, $f'(c)=0$ for some $c \in (0,1)$.

That is, $f'(x)=a_0+a_1x+a_2x^2+a_3x^3+ \ldots +a_nx^n$ has a root in $(0,1)$.

0

consider the polynomial you have . just integrate it and then you see that the integrated polynomial has a root at $0$ and a root at $1$ .The root at $1$ is due to the condition given.Now just use the Rolle's theorem aur use the fact that between any two roots of a polynomial there is one root of it's derivative.

0

Let $f(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n$. If we look at the integral of $f(x)$ in the interval $(0,1)$,

$$ g(x) = \int_0^1 f(x) dx = a_0 + \frac{a_1}2 + \cdots + \frac{a_n}{n+1}. $$

We're given $g(x) = 0$ which means that the area under the curve $f(x)$ in the interval $(0,1)$ is zero. This can happen if $f(x) = 0 \ \forall x\in (0,1)$ or the area above and below the $x$-axis are equal. The first case is not possible and the second implies that $f(x)$ has a roots in $(0,1)$.