Stanton D. answered • 04/23/14

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Dear Diana,

Looks like you haven't been reading your textbook, or you're confused by some of the language used? First, f:x [usually written as f(x)] is

*defined*, then the*domain*(set of input values for x) is listed, here*= the real numbers.***R**i) "Image" is the set of outputs the function makes from the input. In simple cases, such as linear algebra (what you have here), that's just the value of the function (element of the range) for that element of the domain. So plug in and solve: f(3) = 5 * 3

^{2}+ 2 = 5*9 + 2 = 47ii) I don't recognize the form f(3).f(2) ; is that a multiplication? If so, you should be able to solve as above (find f(3) and f(2), then multiply those two values).

iii) This one requires setting up the expression for the function, and solving for the specific x that satisfies the equation: So 5*x

^{2}+ 2 = 225*x

^{2}= 20x

^{2}= 4x= {-2, +2}

Incidentally, if you came up with an equation to do this, such as g(y) = sqrt((y-2)/5), you would have found g(y) = f

^{-1}(x), the inverse function of f(x). Note that I've used "y" for the result of f(x) so that you could easily see what went on here; but if you're talking about only the function g itself, not specifically about the value of g(x) for some particular value of x, then it doesn't matter what variable you put as the domain of the function, g(x) is as good as g(y). The variable is just a 'placeholder' in that case.
Diana C.

04/24/14