Function notation

In your elementary school years, you were given worksheets bearing statements like "[ ] + 2 = 4" and asked to fill in the box by your teacher. At this your level, your teacher would give you worksheets bearing statements like "x + 2 = 4" and asked you to "solve for x". Why did your teachers change from boxes to variables? Well, try and figure out why? How many shapes would you have to make use of for formulas like the one for the area A of a trapezoid with upper base a, lower base b, and height h?

A = ( h/2 )(a + b)

If you try to represent the above equation or something more complicated, with the use of differently-shaped boxes, you'd quickly run out of shapes. Apart from that, you know from experience that "A" stands for "area", "h" stands for "height", and "a" and "b" stand for the lengths of the parallel top and bottom sides. It would be difficult to know what a square box or a triangular box might stand for. Therefore, it make sense why they shifted from boxes to variables because, whilst the boxes and the letters mean the exact same thing (, a gap waiting to be filled with a value), variables are better. Variables are more bendable, easier to read, and can provide more information.

The same thin is true of "y" and "f(x)" (pronounced as "eff-of-eks"). For functions, the two notations mean the same thing, but "f(x)" allows 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 any of the cases: you plug in –1 for x, multiply by 2, and then add the 3, simplifying to obtain a final value of +1.

But function notation offers you greater flexibility than using just "y" for every formula. Your graphing calculator will list various functions as y1, y2, and so on. In textbooks and any time we are writing things out, we make use of names like f(x), g(x), h(x), s(t), and so on. With this notation, you can currently make use of more than one function at a time without confusing yourself or mixing up the formulas, wondering " which 'y' you are using. And the notation can be helpfully illustrative: "A(r) = (pi) r2" indicates the area of a circle, whereas "C(r) = 2(pi)r" indicates the circumference. Both functions have the same plug-in variable (the "r"), but "A" reminds you that this is the formula for "area" and "C" reminds you that this is the formula for "circumference".

How do we evaluate functions? First, bear this in mind: While parentheses have, up until now, constantly been used to stand for multiplication, the parentheses do not indicate multiplication in function notation. The expression "f(x)" means "put a value for x into a formula f "; the expression does not mean "multiply f and x"! Don't think that "f(x)" stands for "f times x".

In function notation, the "x" in "f(x)" is known as "the argument of the function", or just "the argument". So if you are given "f(2)" and asked for the "argument", the answer is simply "2".

Let's go over to "evaluation": You evaluate "f(x)" exactly 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 to make use of parentheses as it is done above; it aids keep track of things like whether or not the exponent is on the "minus" sign.) Bear in mind that "x" is just a box, waiting for something to be put into it. Don't allow apparently-awkward problems scare you:

Evaluate f (@).

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

The above example is just to show you clearly how the notation works.

A significant type of function is referred to as a "piecewise" function. It is so called because, it's in pieces:

As you can see, this function is divided 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 make use of would depend on what the value of x is. If we want to evaluate f (0), we must, because 0 < 1, use the first half of the function. Then f (0) = 2(0) 2 – 1 = 0 – 1 = –1. If we want to evaluate f (2), then we ought to, since 2 > 1, make use of the second half of the function. Then f (2) = (2) + 4 = 6. If we want to solve for f (1), we ought to as well make use of the second half, due to the fact that first half is only for x's that are strictly less than 1. The second half is for x's that are greater than or equal to1. Then f(1) = (1) + 4 = 5

As illustrated in above example with the "f(@)" example, you can calculate functions at variables or expressions, instead of just 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 don’t know the stuff inside the parentheses was obtained, (the set that the 3 was multiplied throughout), then you would go ahead 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).

You ought not to do this all at the same time, rather you would break it into smaller, more convenient pieces. First, you will find f(x + h); then you will find f(h); only then will you 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

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

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

This is in fact something you will come across again in calculus. Exercises like this is trying to prepare you for upcoming courses, but it doesn’t seem like remembers these when they get to calculus, so it's really a huge work for no authentic purpose. Nevertheless, this type of question is rather popular, so you ought to know how to do it.

The best way to go about it is to break this up into pieces. First, try to figure out the expressions for each of f(x + h) and f(x), and then move on and do subtraction. Immediately, You've simplified the subtraction in the numerator, it is only then that you will do the division. Your work would appear 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 complex exercises like the last example above, be cautious and begin by splitting the exercises into smaller, simpler steps, so that you can finish the exercises effectively.

Function Notation: even or odd

You may be asked to "establish algebraically" whether a function is even or odd. To work this out, you take the function and put –x in for x, and then simplify. If you end up with the exact same function that you began with (that is, if f(–x) = f(x), so all of the signs are the same), then the function is even. If you end up with the exact opposite of what you began with (that is, if f(–x) = –f(x), which means that you got all of the "plus" signs replaced by "minus" signs, and vice versa), then the function is odd. In all other cases, the function is "neither even nor odd".

Determine algebraically whether f(x) = –3x2 + 4 is even, odd, or neither.

If I graph this, I will see that this is "symmetric about the y-axis"; in other words, whatever the graph is doing on one side of the y-axis is mirrored on the other side:

This mirroring about the axis is a feature of even functions.

Also observe that all the exponents are even (the exponent on the constant term being zero: 4x0 = 4 ×1 = 4).

But the question asks you to make the determination algebraically, which means that you require to do it with algebra, not with graphs.

So you'll plug –x in for x, and simplify:

f(–x) = –3(–x)2 + 4

= –3(x2) + 4

 = –3x2 + 4

The final expression is the same thing you'd began with, and this implies that f(x) is even.

Determine algebraically whether f(x) = 2x3 – 4x is even, odd, or neither.

If you plot this in a graph, you would see that it is "symmetric about the origin"; that is, if I began at a point on the graph on one side of the y-axis, and draw a line from that point through the origin and extending the same length on the other side of the y-axis, I will get to another point on the graph.

This symmetry is a feature of odd functions.

Observe as well that all the exponents are odd (since the second term is 4x = 4x1). This is a helpful clue. I ought to expect this function to be odd.

But the question asks me to make the determination algebraically, therefore, I'll plug –x in for x, and simplify:

f(–x) = 2(–x)3 – 4(–x)

= 2(–x3) + 4x

= –2x3 + 4x

The final expression is the exact opposite of what I began with, by which I mean that the sign on each term has been changed to its opposite, just as if I'd multiplied through by –1:

–f(x) = –1[f(x)]

= –[2x3 – 4x]

 = –2x3 + 4x

This means that f(x) is odd.

Determine algebraically whether f(x) = 2x3 – 3x2 – 4x + 4 is even, odd, or neither.

This function is the sum of the earlier two functions.

Observe that its graph does not have the symmetry of any of the previous ones, nor are all its exponents either even or odd.

You would anticipate this function to be neither even nor odd.

You’ll put –x in for x, and simplify:

f(–x) = 2(–x)3 – 3(–x)2 – 4(–x) + 4

 = 2(–x3) – 3(x2) + 4x + 4

 = –2x3 –3x2 + 4x + 4

This is neither the same thing you began with (which is, 2x3 – 3x2 – 4x + 4) nor the exact opposite of what you began with (which is, –2x3 + 3x2 + 4x – 4).

This means that f(x) is neither even nor odd.

You may find it useful, when answering this "even or odd" type of question, to write down f(x) and –f(x) clearly, and then evaluate them to anything you get for f(–x). This can assist you make a confident determination of the exact answer. View more on psalmfresh.blogspot.com for more educating and tricks article Thank you