Question 343026
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If you have a function *[tex \Large \rho] such that the value of the function at *[tex \Large x], namely *[tex \Large \rho(x)] is defined by some expression involving *[tex \Large x], then you evaluate *[tex \Large \rho(1)] by replacing *[tex \Large x] in the definition of *[tex \Large \rho(x)] with a *[tex \Large 1] and then doing the arithmetic.


Just as an example, let's define *[tex \Large \rho(x)\ =\ x^2\ +\ 2x\ - 3].  Given that definition, *[tex \Large \rho(1)\ =\ (1)^2\ + 2(1)\ - 3], whereas *[tex \Large \rho(2)\ =\ (2)^2\ + 2(2)\ - 3].  Of course, you need to do the arithmetic and simplify, but the concept is simply that you replace the independent variable by whatever is in the parentheses following the function identifier.


Therefore, if someone says, evaluate *[tex \Large \rho(a)] where *[tex \Large \rho(x)\ =\ x^2\ +\ 2x\ - 3], you replace the variable with *[tex \Large a], thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(a)\ =\ a^2\ +\ 2a\ - 3].


And the same idea to evaluate *[tex \Large \rho\left(\frac{-b}{2a}\right)] which would look like:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho\left(\frac{-b}{2a}\right)\ =\ \left(\frac{-b}{2a}\right)^2\ +\ 2\left(\frac{-b}{2a}\right)\ - 3].


So you need to write out your *[tex \Large g(x)] and then replace the independent variable, *[tex \Large x], with the definition of *[tex \Large f(x)].  And then simplify, of course.


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