Question 326062
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First things first.  If *[tex \Large f(x)\ =\ 5x\ -\ 2], then *[tex \Large f(a)\ =\ 5a\ -\ 2], *[tex \Large f\left(\frac{y^3z}{w^2}\right)\ =\ 5\left(\frac{y^3z}{w^2}\right)\ -\ 2].  Whatever is in the parentheses following the function designator replaces the independent variable in the function definition.


Next:  *[tex \Large (f\circ{g})(x)] means *[tex \Large f\left(g(x)\right)].  So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (f\circ{g})(x)\ =\ f\left(g(x)\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f\left(g(x)\right)\ =\ 5\left(g(x)\right)\ -\ 2]


But since *[tex \Large g(x)\ =\ x^2\ +\ x],


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5\left(g(x)\right)\ -\ 2\ =\ 5\left(x^2\ +\ x\right)\ -\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5\left(x^2\ +\ x\right)\ -\ 2\ =\ 5x^2\ +\ 5x\ -\ 2]


Note that *[tex \Large (f\circ{g})(x)\ \neq\ (g\circ{f})(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (g\circ{f})(x)\ =\ \left(5x\ -\ 2\right)^2\ +\ (5x\ -\ 2)\ =\ 25x^2\ -\ 15x\ +\ 2]


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