The
Purplemath Forums 
Function Notation: Evaluation (page 2 of 3) Sections: Introduction & Evaluating at a number, Evaluating at a variable, Even and odd functions As demonstrated on the previous page in the "f(@)" example, you can evaluate functions at variables or expressions, rather than merely at numbers.
Everywhere that my formula has an "x", I now plug in an "h": f(x) = 3x^{2} + 2x f( ) = 3( )^{2} + 2( ) f(h) = 3(h)^{2} + 2(h) = 3h^{2} + 2h
Everywhere that my formula has an "x", I now plug in an "x + h". f(x) = 3x^{2} + 2x f( ) = 3( )^{2} + 2( ) f(x + h) = 3(x + h)^{2} + 2(x + h) = 3(x^{2} + 2xh + h^{2}) + 2x + 2h = 3x^{2} + 6xh + 3h^{2} + 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 want to review how to simplify with parentheses and how to do polynomial multiplication.
I should not try to do this all at once. Instead, I'll break this into smaller, more manageable pieces. First, I'll find f(x + h); then I'll find f(h); only then will I do the subtraction and simplification. f( ) = 3( )^{2} + 2( ) f(x + h) = 3(x + h)^{2} + 2(x + h) = 3x^{2} + 6xh + 3h^{2} + 2x + 2h f( ) = 3( )^{2} + 2( ) f(h) = 3(h)^{2} + 2(h) = 3h^{2} + 2h = [3x^{2} + 6xh + 3h^{2} + 2x + 2h] – [3h^{2} + 2h] = 3x^{2} + 6xh + 3h^{2} + 2x + 2h – 3h^{2} – 2h = 3x^{2} + 6xh + 3h^{2} – 3h^{2} + 2x + 2h – 2h = 3x^{2} + 6xh + 2x Notice that f(x + h) – f(h) does not equal f(x + h – h). You cannot "simplify" the different functions' arguments. Copyright © Elizabeth Stapel 19992011 All Rights Reserved
This is actually something you will see again in calculus. I guess they're trying to "prep" you for upcoming courses 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.
My best course of action is to break this up into pieces. First, I'll find the expressions for each of f(x + h) and f(x), and then I'll subtract. Once I've simplified the subtraction in the numerator, only then will I do the division. My work looks like this: f( ) = 3( )^{2} + 2( ) f(x + h) = 3(x + h)^{2} + 2(x + h) = 3x^{2} + 6xh + 3h^{2} + 2x + 2h = [3x^{2} + 6xh + 3h^{2} + 2x + 2h] – [3x^{2} + 2x] = 3x^{2} + 6xh + 3h^{2} + 2x + 2h – 3x^{2} – 2x = 3x^{2} – 3x^{2} + 6xh + 3h^{2} + 2x – 2x + 2h = 6xh + 3h^{2} + 2h = h[6x + 3h + 2] / h = 6x + 3h + 2 When working on complicated exercises like the last example above, exercise caution and start by splitting the exercises into smaller, simpler steps, so that you can complete these exercises successfully. << Previous Top  1  2  3  Return to Index Next >>



Copyright © 19992012 Elizabeth Stapel  About  Terms of Use 




