The roots of (x-41)^{49}+(x-49)^{41}+(x-2009) | vedclass

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(B) Let $f(x) = (x-41)^{49} + (x-49)^{41} + (x-2009)^{2009}$.To determine the number of real roots,we examine the derivative of the function $f(x)$:$f

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non-real except one positive real root

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(B) Let $f(x) = (x-41)^{49} + (x-49)^{41} + (x-2009)^{2009}$.To determine the number of real roots,we examine the derivative of the function $f(x)$:$f'(x) = 49(x-41)^{48} + 41(x-49)^{40} + 2009(x-2009)^{2008}$.Since the exponents $48$,$40$,and $2008$ are all even,each term $(x-a)^{2n}$ is non-negative for all $x \in \mathbb{R}$.Specifically,$(x-41)^{48} \ge 0$,$(x-49)^{40} \ge 0$,and $(x-2009)^{2008} \ge 0$.Since the coefficients $49$,$41$,and $2009$ are all positive,$f'(x) \ge 0$ for all $x \in \mathbb{R}$.Furthermore,$f'(x)$ is never zero for all $x$ simultaneously,as the terms vanish at different points $(x=41, 49, 2009)$.Thus,$f'(x) > 0$ for all $x \in \mathbb{R}$,which implies that $f(x)$ is a strictly increasing function.$A$ strictly increasing continuous function can cross the $x$-axis at most once.As $x 	o \infty$,$f(x) 	o \infty$,and as $x 	o -\infty$,$f(x) 	o -\infty$.By the Intermediate Value Theorem,there exists exactly one real root.To check if the root is positive,we evaluate $f(0)$:$f(0) = (-41)^{49} + (-49)^{41} + (-2009)^{2009} Since $f(0) Therefore,there is exactly one positive real root.

The roots of $(x-41)^{49}+(x-49)^{41}+(x-2009)^{2009}=0$ are

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