Question 498493
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If *[tex \Large f(x)\ =\ 4x^2\ +\ 5x\ -\ 12], then *[tex \Large f(\text{stuff} )\ =\ 4 \left(\text{stuff}\right)^2\ +\ 5\left(\text{stuff}\right)\ -\ 12]


Whatever is given between the parentheses in *[tex \Large f\left(\text{stuff}\right)] gets put in place of *[tex \Large x] in the definition of *[tex \Large f(x)].  Then you get to simplify.


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