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.

• Given that f(x) = 3x2 + 2x, find f(h).

Everywhere that my formula has an "x", I now plug in an "h":

f(x) = 3x2 + 2x

f(   ) = 3(  )2 + 2(  )

f(h) = 3(h)2 + 2(h) = 3h2 + 2h

• Given that f(x) = 3x2 + 2x, find f(x + h).

Everywhere that my formula has an "x", I 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 want to review how to simplify with parentheses and how to do polynomial multiplication.

• Given that f(x) = 3x2 + 2x, find f(x + h) – f(h).

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(x) = 3x2 + 2x for x + h:
• f(   ) = 3(  )2 + 2(  )

f(x + h) = 3(x + h)2 + 2(x + h)

= 3x2 + 6xh + 3h2  + 2x + 2h

• f(x) = 3x2 + 2x for h:
• f(   ) = 3(  )2 + 2(  )

f(h) = 3(h)2 + 2(h) = 3h2 + 2h

• f(x + h) – f(h) (doing the subtraction):
• = [3x2 + 6xh + 3h2  + 2x + 2h] – [3h2 + 2h]

= 3x2 + 6xh + 3h2  + 2x + 2h – 3h2   – 2h

= 3x2 + 6xh + 3h2 – 3h2 + 2x + 2h   – 2h

= 3x2 + 6xh + 2x

Notice that f(x + h) – f(h) does not equal f(x + hh). You cannot "simplify" the different functions' arguments.   Copyright © Elizabeth Stapel 1999-2011 All Rights Reserved

• Given that f(x) = 3x2 + 2x, find [f(x + h) – f(x)] / h.

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(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

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.

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 Cite this article as: Stapel, Elizabeth. "Function Notation: Evaluating at an Expression." Purplemath. Available from     http://www.purplemath.com/modules/fcnnot2.htm. Accessed [Date] [Month] 2016

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