Question 1037706
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 7x^2\ +\ 8x\ +\ 2\ >\ 4x\ \forall\ x\ \in\ \mathbb{R}]


You can prove this to yourself in a couple of ways:


You can find the vertex of the parabola that graphs the quadratic which is a minimum value of the function and then show that it is greater than *[tex \Large 4x_{min}].


You can show that *[tex \Large 7x^2\ +\ 8x\ +\ 2\ =\ 4x] has complex roots, proving that the graphs of *[tex \Large 7x^2\ +\ 8x\ +\ 2] and *[tex \Large 4x] never intersect.


You can use a graphing calculator or graphing utility on your computer to show that *[tex \Large 4x] never gets bigger than the quadratic.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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