Let a0,a1,…,an be real numbers which satisfy
n+1an+nan−1+…+3a2+2a1+a0=0.
Show that the equation
anxn+an−1xn−1+…+a1x+a0=0
has at least one solution in the interval (0,1).
This is a list of Calculus proof based problems that I like. Please avoid posting complete solutions, so that others can work on it.
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#Proofs
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Consider the polynomial function : f(x)=n+1anxn+1+nan−1xn+⋯+2a1x2+a0x. Clearly f(0)=0, and f(1)=0 by the first given equality. By Rolle's theorem there is some x∈(0,1), such that f′(x)=0 which can be written as : anxn+an−1xn−1+…a1x+a0=0.
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Good work! I did the same, but with due respect "please avoid posting complete solutions, so that others can work on it"
It appears to b a problem based on ROLLE's THEOREM........ On integrating the given equation we get a new equation whose roots are 0 &1. Now the new equation is a polynomial so it is continuos as well as differentiable in (0,1) . Now apply ROLLE'S THEOREM which states that between any 2 roots of a continuous and differentiable function there must b atleast one root of its differentiable function.
Rolle's theorem for interval (0,1). :)
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