If alpha_1

Question

(D) Let $f(x) = (x-\alpha_1)(x-\alpha_3)(x-\alpha_5) + 3(x-\alpha_2)(x-\alpha_4)(x-\alpha_6)$.Evaluate $f(x)$ at the given points:$f(\alpha_1) = 0 + 3

Vedclass · Accepted Answer

No real root in $(-\infty, \alpha_1)$

Vedclass · Answer

(D) Let $f(x) = (x-\alpha_1)(x-\alpha_3)(x-\alpha_5) + 3(x-\alpha_2)(x-\alpha_4)(x-\alpha_6)$.Evaluate $f(x)$ at the given points:$f(\alpha_1) = 0 + 3(\alpha_1-\alpha_2)(\alpha_1-\alpha_4)(\alpha_1-\alpha_6) = 3(-)(-)(-) $f(\alpha_2) = (\alpha_2-\alpha_1)(\alpha_2-\alpha_3)(\alpha_2-\alpha_5) + 0 = (+)(-)(-) > 0$.$f(\alpha_3) = 0 + 3(\alpha_3-\alpha_2)(\alpha_3-\alpha_4)(\alpha_3-\alpha_6) = 3(+)(-)(-) > 0$.$f(\alpha_4) = (\alpha_4-\alpha_1)(\alpha_4-\alpha_3)(\alpha_4-\alpha_5) + 0 = (+)(+)(-) $f(\alpha_5) = 0 + 3(\alpha_5-\alpha_2)(\alpha_5-\alpha_4)(\alpha_5-\alpha_6) = 3(+)(+)(-) $f(\alpha_6) = (\alpha_6-\alpha_1)(\alpha_6-\alpha_3)(\alpha_6-\alpha_5) + 0 = (+)(+)(+) > 0$.Since $f(\alpha_1)  0$,there is a root in $(\alpha_1, \alpha_2)$.Since $f(\alpha_3) > 0$ and $f(\alpha_4) Since $f(\alpha_5)  0$,there is a root in $(\alpha_5, \alpha_6)$.As $x 	o -\infty$,$f(x) 	o -\infty$. Since $f(\alpha_1) < 0$,there is no guaranteed root in $(-\infty, \alpha_1)$ based on the intermediate value theorem.

If $\alpha_1 < \alpha_2 < \alpha_3 < \alpha_4 < \alpha_5 < \alpha_6$,then the equation $(x-\alpha_1)(x-\alpha_3)(x-\alpha_5) + 3(x-\alpha_2)(x-\alpha_4)(x-\alpha_6) = 0$ has :-

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