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Function Notation: Intro / Evaluation (page 1 of 3)

Sections: Introduction & Evaluating at a number, Evaluating at a variable, Even and odd functions

You've been playing with "y =" for some time now. And you've seen that the "nice" equations are the ones that you can solve for "y =". These "y =" equations are functions. But the question you are facing at the moment is "Why do I need this function notation, and how does it work?"

Remember when you were in elementary school, and your teacher gave you worksheets containing statements like "[  ] + 2 = 4" and told you to fill in the box? But now you are given "x + 2 = 4" and told to "solve for x". Why did they switch from boxes to variables? Well, think about it: How many shapes would you have to use for things like, say, the formula for the area of a trapezoid?

A = ( ^h/₂ )(a + b)

You'd quickly run out of shapes for your boxes. And, besides, you know from experience that "A" means "area", "h" means "height", and "a" and "b" are the lengths of the top and bottom. Heaven only knows what a square box or a triangular box might stand for. So they switched from boxes to variables because, while they mean the exact same thing (a hole waiting to be filled with a value), variables are better. Variables are more flexible and can give you more information.

The same is true of "y" and "f(x)" (pronounced as "eff-of-eks"). For functions, the two notations mean the exact same thing, but "f(x)" gives you more flexibility and more information. You used to say "y = 2x + 3; solve for y when x = –1". Now you say "f(x) = 2x + 3; find f(–1)" (pronounced as "f-of-x is 2x plus three; find f-of-negative-one"). You do exactly the same thing in either case: you plug in –1 for x, multiply by 2, and then add the three, to get a value of +1.

But now you have more flexibility. Your graphing calculator will list the functions as y1, y2, etc. More standardly, we use names like f(x), g(x), h(x), s(t), etc. The point is that you can use more than one function at a time, and not mix them up, wondering "Okay, which 'y' is this, anyway?" And the notation can be explanatory: "A(r) = (pi)r²" indicates the area of a circle, while "C(r) = 2(pi)r" indicates the circumference.   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

How do we go about evaluating functions? First, remember this: While parentheses have, up until now, always indicated multiplication (look back at that circumference formula in the last paragraph above), parentheses do not indicate multiplication in function notation. "f(x)" means "plug in a value for x"; it does not mean "multiply f and x"!! If you read "f(x)" as "f times x", I can guarantee that you'll drive your instructor batty! (By the way, that is not a suggestion!)

Note: When you have function notation, the "x" in "f(x)" is called "the argument of the function", or just "the argument". So if they give you "f(2)" and ask for the "argument", the answer is just "2". Now that we have that straight, let's proceed to "evaluation": You evaluate "f(x)" just as you would evaluate "y".

• Given  f(x) = x² + 2x – 1, find  f(2).

  f(2) = (2)² +2(2) – 1
       = 4 + 4 – 1
       = 7

• Given  f(x) = x² + 2x – 1, find  f(–3).

  f(–3) = (–3)² +2(–3) – 1
       = 9 – 6 – 1
       = 2

(If you experience difficulties with negatives, try using parentheses as I just did; it helps keep track of things like whether or not the exponent is on the negative.) Remember that "x" is just a box, waiting for something to be put into it. Don't let odd-looking problems scare you:

• Evaluate f(@).

Well, f(@) = (@)² + 2(@) – 1 = @² + 2@ – 1.

Kinda pointless, I'll grant you, but you can see how the notation works.

Now YOU try!

An important type of function is called a "piecewise" function, so called because, well, it's in pieces:

As you can see, this function is split into two halves: the half that comes before x = 1, and the half that goes from x = 1 to infinity. Which half of the function you use depends on what x is. If we evaluate f(0), we must use the first half of the function, since 0 < 1. Then f(0) = 2(0)² – 1 = 0 – 1 = –1. If we evaluate f(2), then we must use the second half of the function, since 2 > 1. Then f(2) = (2) + 4 = 6. If we evaluate f(1), we must also use the second half, because the first half is only for x's that are strictly less than 1. The second half is for x's that are equal to or greater than one. Then f(1) = (1) + 4 = 5.

Now YOU try!

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