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anemone
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Given \(\displaystyle f(x)=5tx^4+sx^3+3rx^2+qx+p\) for $f(x)\in R$. If $r+t=-p$, prove that there is a real root for $f(x)=0$ in $[-1,1]$.
anemone said:Given \(\displaystyle f(x)=5tx^4+sx^3+3rx^2+qx+p\) for $f(x)\in R$. If $r+t=-p$, prove that there is a real root for $f(x)=0$ in $[-1,1]$.
zzephod said:Consider the polynomial \(\displaystyle p(x)=tx^5+(s/4)x^4+rx^3+(q/2)x^2+px-(s/4+q/2)\).
Then \(\displaystyle p(1)=p(-1)=0\) and so \(\displaystyle p\) has an extremum in \(\displaystyle [-1,1]\), so \(\displaystyle p'(x)\) has a root in \(\displaystyle [-1,1]\) ...
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To prove that a real root exists means to show that there is at least one value within the given interval ([-1, 1] in this case) that satisfies the given equation or inequality. In other words, it is showing that the equation has a solution within the given range of numbers.
Proving the existence of a real root is important because it confirms the validity of the given equation or inequality. It also allows us to find the specific value(s) of the root, which can help in solving real-world problems and making accurate predictions.
The interval [-1, 1] represents the range of values within which we are trying to find a real root. In this case, it means that we are looking for a value that falls between -1 and 1 (including both endpoints) that satisfies the given equation or inequality.
There are various methods that can be used to prove the existence of a real root in the given interval. These include the Intermediate Value Theorem, Descartes' Rule of Signs, and the Rational Root Theorem, among others. The specific method used will depend on the given equation or inequality.
Yes, a real root can exist outside of the interval [-1, 1]. The given interval is just a specific range that has been chosen for the purpose of proving the existence of a real root. It is possible for the equation or inequality to have one or more real roots outside of this interval.