Sometimes, Always, or Never True?
Date: 02/12/2002 at 19:02:16 From: Jon Subject: Functions I can find no explanation for this question. Tell whether this statement is always, sometimes, or never true: f(g(x)) = g(f(x))
Date: 02/12/2002 at 23:05:39 From: Doctor Ian Subject: Re: Functions Hi Jon, Suppose f(x) = x + 1, and g(x) = x^2. Then f(g(x)) = g(x) + 1 = x^2 + 1 and g(f(x)) = (f(x))^2 = (x + 1)^2 = x^2 + 2x + 1 So in this case, f(g(x)) is clearly not equal to g(f(x)), except when x happens to be zero. So we know that it's not _always_ true. On the other hand, suppose f(x) = x + 1 and g(x) = x - 1. Then f(g(x)) = g(x) + 1 = (x - 1) + 1 = x and g(f(x)) = f(x) - 1 = (x + 1) - 1 = x So for these functions, f(g(x)) = g(f(x)), regardless of what the value of x happens to be. This is easiest to think about if you think of functions as little processes. Then the composition of two processes says: Put something into the inner function; and whatever comes out, use that as input to the outer function. The functions don't have to be mathematical. For example, if f(x) means 'paint x blue', and g(x) means 'wrap x in paper', then f(g(x)) => wrap x in paper, and paint the whole thing blue but g(f(x)) => paint x blue, and wrap the painted object in paper In the first case, the paper gets painted; in the second, the object itself gets painted. In many cases where f(g(x)) = g(f(x)), the functions are inverses of each other. That is, what one does, the other undoes, so when you compose them (in either order), you get x back as the final result. You've seen one example. Here are some others: __ f(x) = x^2, g(x) = \| x f(x) = 10^x, g(x) = log(x) f(x) = 2x, g(x) = x/2 f(x) = sin(x), g(x) = arcsin(x) But this isn't always the case, e.g., f(x) = 1/x, g(x) = -x Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
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