Question #04999

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Question #04999

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Answer 1 · 2016-10-15T08:46:22

$p > \frac{1}{2} (6 + \sqrt{36 + f^{2}})$ $p > 1/2 (6 + sqrt[36 + f^2])$

Explanation:

If $y > 0 \forall x$ $y >0 forall x$ then $p > 0$ $p > 0$ because $p$ $p$ is the coeficient of $x^{2}$ $x^2$ and $y (x)$ $y(x)$ cannot have $x$ $x$ crossings so $y (0) = p - 6 > 0$ $y(0) = p-6 > 0$

Anyway the $x$ $x$ crossings are given at

$x = \frac{- f \pm \sqrt{f^{2} + 24 p - 4 p^{2}}}{2 p}$ $x=(-f pm sqrt[f^2 + 24 p - 4 p^2])/(2 p)$ so if

$f^{2} + 24 p - 4 p^{2} < 0$ $f^2+24p-4p^2<0$ no feasible crossings

or solving for $p$ $p$

$p < \frac{1}{2} (6 - \sqrt{6^{2} + f^{2}})$ $p < 1/2 (6 - sqrt[6^2 + f^2])$ and $p > \frac{1}{2} (6 + \sqrt{6^{2} + f^{2}})$ $p > 1/2 (6 + sqrt[6^2 + f^2])$

Concluding

$p > 6$ $p > 6$ and $p > \frac{1}{2} (6 + \sqrt{6^{2} + f^{2}})$ $p > 1/2 (6 + sqrt[6^2 + f^2])$ or finally

$p > \frac{1}{2} (6 + \sqrt{6^{2} + f^{2}})$ $p > 1/2 (6 + sqrt[6^2 + f^2])$

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Question #04999

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