# Intermediate Value Theorem, an application problem [duplicate]

The question is: if $$\frac{a_0}{1}+\frac{a_1}{2}+\dots+\frac{a_n}{n+1}=0$$ then, $a_0+a_1x+\dots+a_nx^n=0$ for some $x$ in the interval $[0,1]$.

My approach is to let $f(x)=a_0+a_1x+\dots+a_nx^n$ and the find $x_0$ and $x_1$ such that, for example, $f(x_0)\leq0$ and $f(x_1)\geq0$ and then use IVT to make the conclusion.

However, I do not know how to use the given condition, especially I am not sure about if $n$ is odd or even and if the leading coefficient, $a_n$ is positive or negative.

Any help will be appreciated. Thanks so much.

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## marked as duplicate by David Mitra, amWhy calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 27 '14 at 15:05

Hint: The given equation says that $\displaystyle\int_0^1 f(x)\,dx = 0$.

Is it possible to have $f(x) > 0$ for all $x \in [0,1]$? If not, $f(x_0) \le 0$ for some $x_0 \in [0,1]$.

Is it possible to have $f(x) < 0$ for all $x \in [0,1]$? If not, $f(x_1) \ge 0$ for some $x_1 \in [0,1]$.

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thanks and I can guess the function to achieve my target now. – nam Jul 27 '14 at 8:27

$$\cdot f(x) = a_0+a_1x+\dots+a_nx^n , F(x)= \frac{a_0x}{1}+\frac{a_1x^2}{2}+\dots+\frac{a_n x^{n+1}}{n+1}.$$

Comment: $F(0) = F(1) = 0 \Rightarrow \exists z \in (0,1)$ such that $F'(z) = f(z) = 0$ by Rolle's Theorem.

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