If the equation a_n x^n + a_{n-1} x^{n-1} + d | vedclass

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(C) Let $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x$.For $x = 0$,we have $f(0) = 0$.Given that $x = \alpha$ is a root of the equation $f(x) = 0$

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Smaller than $\alpha$

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(C) Let $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x$.For $x = 0$,we have $f(0) = 0$.Given that $x = \alpha$ is a root of the equation $f(x) = 0$,we have $f(\alpha) = 0$.Since $f(x)$ is a polynomial,it is continuous on $[0, \alpha]$ and differentiable on $(0, \alpha)$.By Rolle's Theorem,there exists at least one point $c \in (0, \alpha)$ such that $f'(c) = 0$.The derivative is $f'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \dots + a_1$.Thus,the equation $n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \dots + a_1 = 0$ has a root $c$ such that $0 Therefore,the root is smaller than $\alpha$.

If the equation $a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x = 0$,where $a_1 \neq 0$ and $n \geq 2$,has a positive root $x = \alpha$,then the equation $n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \dots + a_1 = 0$ has a positive root which is:

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