In my first attempt I tried to use IVT but I couldn't find the two values for which the function gives opposite sign. Since the question asked about the range I was quite sure it would be based on IVT.
This was all I could think of. Any hints?
Find in which interval does the real root of this cubic lies
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algebra-precalculus
polynomials
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0If you aren't looking to solve the problem mathematically, but are only looking to address the question on the test -- it may be better for you to choose values of $a_0,a_1,a_2,a_3$ that satisfy the condition, and check which of the given (disjoint) intervals work! – 2017-02-25
1 Answers
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The correct answer is $[0,1]$.
Indeed, consider the polynomial function $f$ defined by :
$$\forall x\in\mathbb{R},\,f(x)=a_0+a_1x+a_2x^2+a_3x^3$$
The assumption may be written as follows :
$$\int_0^1f(x)\,dx=0$$
If $a_0$, $a_1$, $a_2$ and $a_3$ are not all zero, then $f$ is not identically zero on $[0,1]$ but, its integral being $0$, it cannot be of constant sign. So by IVT, $f$ must cancel somewhere in $[0,1]$ (and even in $(0,1)$).
And if $a_0=a_1=a_2=a_3=0$, then every interval works !