Question 305588
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ x^2\ -\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ g(x)\ =\ 2x\ +\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (f\circ g)(x)\ =\ f\left(g(x)\right)\ =\ \left(2x\,+\,1\right)^2\ -\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (f\circ g)(-2)\ =\ f\left(g(-2)\right)\ =\ \left(2(-2)\,+\,1\right)^2\ -\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (g\circ f)(x)\ =\ g\left(f(x)\right)\ =\ 2\left(x^2\ -\ 3\right)\ +\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (g\circ f)(3)\ =\ g\left(f(3)\right)\ =\ 2\left((3)^2\ -\ 3\right)\ +\ 1]


In either case, you get to do your own arithmetic.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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