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Function Notation: Evaluation (page 2 of 3) Sections: Introduction & Evaluating at a number, Evaluating at a variable, Even and odd functions You can evaluate functions at variables or expressions, rather than at numbers, also.
Everywhere that your formula has an "x", you now plug in an "h": f(x) = 3x2 + 2x f( ) = 3( )2 + 2( ) f(h) = 3(h)2 + 2(h) = 3h2 + 2h
Everywhere that your formula has an "x", you now plug in an "x + h". f(x) = 3x2 + 2x f( ) = 3( )2 + 2( ) f(x + h) = 3(x + h)2 + 2(x + h) = 3(x2 + 2xh + h2) + 2x + 2h = 3x2 + 6xh + 3h2 + 2x + 2h If you're not sure how I got the stuff inside the parentheses (the set that the 3 was multiplied through), then you'll need to review polynomial multiplication.
Do not try to do this all at once. Instead, break this into pieces: first, find f(x + h); then find f(h); then do the subtraction. f(x) = 3x2 + 2x f( ) = 3( )2 + 2( ) f(x + h) = 3(x + h)2 + 2(x + h) = 3x2 + 6xh + 3h2 + 2x + 2h f(x) = 3x2 + 2x f( ) = 3( )2 + 2( ) f(h) = 3(h)2 + 2(h) = 3h2 + 2h f(x + h) – f(h) = [3x2 + 6xh + 3h2 + 2x + 2h] – [3h2 + 2h] = 3x2 + 6xh + 3h2 + 2x + 2h – 3h2 – 2h = 3x2 + 6xh + 3h2 – 3h2 + 2x + 2h – 2h = 3x2 + 6xh + 2x In particular, notice that f(x + h) – f(h) does not equal f(x + h – h). That is, you cannot "simplify" the different functions' arguments! Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved
This is actually something you will see again in calculus. I guess they're trying to "prep" you when they give you exercises like this, but it's not like anybody remembers these by the time they get to calculus, so it's really a lot of work for no real purpose. However, this type of problem is quite popular, so you should expect to need to know how to do it. The best course of action that I've found is to break this up into pieces. That is, find f(x + h) and f(x), and then subtract. Once you've simplified, only then do you do the division. It looks like this: f(x) = 3x2 + 2x f( ) = 3( )2 + 2( ) f(x + h) = 3(x + h)2 + 2(x + h) = 3x2 + 6xh + 3h2 + 2x + 2h f(x) = 3x2 + 2x f(x + h) – f(x) = [3x2 + 6xh + 3h2 + 2x + 2h] – [3x2 + 2x] = 3x2 + 6xh + 3h2 + 2x + 2h – 3x2 – 2x = 3x2 – 3x2 + 6xh + 3h2 + 2x – 2x + 2h = 6xh + 3h2 + 2h [f(x + h) – f(x)] / h = [6xh + 3h2 + 2h] / h = h[6x + 3h + 2] / h = 6x + 3h + 2 Be careful, and split the problems into small, simple steps, and you should be able to complete these successfully. << Previous Top | 1 | 2 | 3 | Return to Index Next >>
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Copyright © 2006-2008 Elizabeth Stapel | About | Terms of Use |
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