Question 397318
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It should be pretty obvious which of the two points is E.  If E is the local maximum as marked on your graph, then the coordinates of E are *[tex \Large \left(1,\frac{-25}{2}\right)].  Otherwise the coordinates of E are *[tex \Large (4, -26)]


But just to make sure, take the second derivative:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2y}{dx^2}\ =\ 6x\ - 15]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f^{(2)}(1)\ =\ -9\ <\ 0]


So *[tex \Large \left(1,\frac{-25}{2}\right)] is a local maximum, and 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f^{(2)}(4)\ =\ 9\ >\ 0]


So *[tex \Large (4, -26)] is a local minimum.


If G is on the x-axis, then G is the single real root.  If the x-coordinate of G is a rational number, then the Rational Roots Theorem says that the root must be *[tex \Large \pm1,\pm2,\pm3,\pm6,\pm9] or *[tex \Large \pm18]


The process is to use synthetic division to test each of the potential rational roots until you find one or have tested them all and concluded that there is no rational root.  In this case there is one, but I'll let you find it for yourself.  If you need a refresher on synthetic division, go to:


<a href="http://www.purplemath.com/modules/synthdiv.htm">Purple Math - Synthetic Division</a>


Make sure you read all four pages.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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