Question 640576
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Technically speaking, what you ask cannot be done with the information given.  That is becsuse *[tex \LARGE G(x)] and *[tex \LARGE g(x)] are not the same thing.  But just in case you were only being sloppy:


Part A:  Set *[tex \LARGE g(x)\ =\ 9]. That is: *[tex \LARGE 2x^2\ -\ 5\ =\ 9], then solve for *[tex \LARGE x]


Part B:  Substitute -3 for *[tex \LARGE x] in *[tex \LARGE 2x^2\ -\ 5] and do the arithmetic.


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