Find the Function (fog)(x)
Date: 07/02/2003 at 19:44:55 From: Pamela Subject: Functions Let f(x) = 2x-1 and g(x) = (x+5)/2. Find the function (fog)(x). Simplify your answer. g(x) = (x+5)/2 = 2[(x+5)/2]-1 = x+5-1 = x+4
Date: 07/03/2003 at 11:02:27 From: Doctor Mike Subject: Re: Functions Pamela, Your final answer is right. But there is one thing I should mention about the way you wrote down the steps leading to that final answer. Notice that you have the correct equation for g(x), and then right after that is an equals sign. If you think about that for a moment you will probably see what is wrong with that. You clearly understand what you are doing here, since you carry through to the correct final answer, BUT when you put the "=" right at the start of that second line, it looks as if you are saying that the first line EQUALS the second line. That is not what you meant, I'm sure. So how to fix it? Just make sure that the equations and expressions you write down really do say the things you are thinking. Here is the way I would do it. g(x)=(x+5)/2 fog(x) = f(g(x)) = f((x+5)/2) = 2[(x+5)/2]-1 = x+5-1 = x+4 Do you see the important difference? When I start out the second line with "fog(x)=" I am making it clear that I am starting out with the next step. The first step was to write down what "g" means. Then with that done I start on the main part, which is to get an equation for fog. It is very clear which part of what I'm writing has to do with g and which part has to do with fog. You might be thinking, "Well, Dr. Mike, you also wrote down a bunch of things connected by the = sign." Yes, I did, but with what I wrote down, all the things connected by = really ARE all equal to each other. The only objection I have to what you wrote down is that single = you used to connect the expression for g and the expression for fog. This is a common thing to happen when you are eager to get your thoughts written down. Just make sure what your fingers write down is the same as what your mind is thinking. Thanks for writing to Dr. Math. - Doctor Mike, The Math Forum http://mathforum.org/dr.math/
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