Untitled Document

From: Subject: Untitled Document Date: Fri, 14 Jan 2011 13:01:02 -0600 MIME-Version: 1.0 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Content-Location: http://www.alamo.edu/sac/math/faculty/lgreene/SACwebsite/Algebra%20Notes/algebrafns.htm X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.5994 Untitled Document

Week 7: 0303 INTERNET

Topic: The=20 Algebra of Functions

Section 8.1
This section presents the topic of = functions which=20 was also partially covered at the end of chapter 3.

When you have two functions: f(x) and g(x) = you can=20 perform algebraic operations with them. Getting the algebraic expression = is the=20 easy part:
(f + g)(x) =3D f(x) + g(x)
(f - g)(x) =3D f(x) = - =20 g(x)
(fg)(x) =3D f(x) * g(x)
(f/g)(x) =3D = f(x)/g(x)
Just plug in=20 the functions and simplify as they do in examples 1 & = 2.

In=20 example 1, they perform the arithmetic operation, then they plug in the = value=20 given for x. In example 2, they perform the arithmetic operation but do = not give=20 a value for x that we can plug in.

Example 1a) f(x) =3D x² + 1, g(x) = =3D 3x +=20 5
The general function is: (write the f, write a plus sign, then = write the g,=20 then simplify)
(f + g)(x) =3D (x² + 1) + (3x + 5) =3D = x² +=20 3x + 6
The example says: (f + g)(1) which means they want you to plug = in the=20 number (1) in the above expression wherever there's an x.
so, (f + = g)(1)=3D=20 (1)² + 3(1) + 6 =3D 10

DIFFERENCE QUOTIENT: This formula is written in = the center=20 of page 482. Example 3 demonstrates it's use.

Procedure: Notice that there is a = given=20 function f(x). They first plug (x + h) into this given function, then = they plug=20 in (x) and take these two answers and subtract them. Then the bottom = value is=20 always the letter "h". This equation is used to find the slope of = a secant=20 line. So it's a variation of the slope equation = m=3D(y2-y1)/(x2-x1). But=20 here we don't know the numbers that need to be plugged in so instead = they plug=20 in variables. Depending on what was given as the function, you may = get a=20 single number as an answer or you might get a polynomial as in example = 3b. =20 Example 3a is only figuring out the top part of the equation.

p. 483 COMPOSITION OF FUNCTIONS
(This was also = explained in the=20 chapter 3 notes, here's the same explanation I gave there)
Now they = give you=20 two functions f(x) and g(x). They want you to plug one into the = other in=20 this way:
    f[g(x)] =3D = replace the x's=20 in the f(x) equation with the whole function = g(x)
   =20 g[f(x)] =3D plug the f function into the g function (replace the = x's)

Example 4: a) Plug the 2 into the g function, then plug = that=20 answer into the f function.
c) plug -3 into the f function then plug = that=20 answer into the g function.

Ex 5: Is doing the same thing but they are not giving = you an "x"=20 to plug in.

Ex 6: Domain of composite functions: Find the domain of = the=20 f function and then the g function. Then when you have = done the=20 composition, find the domain of the new function.

In example 6: Since f(x) =3D 1/x, it's domain is = restricted=20 because x cannot be equal to zero (you can't have zero on the bottom of = a=20 fraction). g(x) =3D square root of (3 - x) restricts it's domain = because x=20 cannot be smaller than 3 (if it is smaller than 3, then 3 - 4, for = example,=20 gives you a -1. This produces an imaginary number). So the domain of = f is=20 {all real numbers x | x (is not equal to) 0} and the domain of g = is=20 (-infinity, 3].

In the composition function: after the f is = plugged into=20 the g and it is simplified, you have to set the inside of the = square root=20 < or equal to 0 and see which values produce a negative square root. = Also,=20 the square root (bottom of p.484) is on the bottom of a fraction. So you = have to=20 do two tests, (3-x) must be greater than or equal to zero and sqrt(3-x) = must not=20 be equal to zero. The first test produces x (less than or equal to = 3), and=20 the second test produces x not equal to 3.

Next, you have to combine the information from all three = domains:=20 f can't equal zero, g can't be bigger than 3. And the = composition=20 must be less than or equal to three, but not equal to 3. So=20 all-together,
the domain of f o g is all real values of x such = that=20 (-infinity, 0)U(0,3). (All real numbers up to but not including 3 and = also=20 not including zero)

A similar method is done in part b of example 6. = But be=20 aware that since the final composition is an inequality, (3x-1)/1 = >(or equal=20 to)0, you will have to draw a graph and test the signs of each region to = see=20 which areas are positive. (As in section 7.5)

Example 7 is well explained in the text.

Graphing Functions: f(x) =3D a(x - h) + k = is the=20 basic form of all the functions.

The vertex is (h,k) and the altitude is the = a.=20 Notice that the h sits inside the parentheses and the k = sits by=20 itself outside the parentheses and the a is always to the left of = the=20 (x - h) group.

f(x)=3D 2(x - 1) +=20 3 the a =3D = 2, h =3D 1, k=3D = 3

If there are no parentheses in the function that means = the=20 h had the value zero, so they didn't bother to write (x-0), they = just=20 wrote the x.

f(x)=3D = 2x=20 + 1, = or [2(x-0)+1] =20 the=20 a =3D 2, h =3D 0, k=3D = 1

If the k = =3D 0 then it=20 is just missing altogether.

f(x)=3D 2(x = + 1),[or 2(x+1)+0] the a =3D=20 2, h =3D -1, k = =3D=20 0

NOTE: The formula has a minus sign in front of the = h so it=20 changes the sign of it (i.e. it always looks like the h has the = wrong=20 sign). You want to use the original sign of h when graphing the=20 vertex.
Ex. in f(x)=3D (x - = 1), the=20 h is a +1,
in = f(x)=3D (x + 3), the h is a (-3) [i.e. (x - = (-3)].

The a is the altitude (or amplitude) which = determines=20 whether the figure opens "up" or "down"
The = a also=20 controls the "stretch or shrink" .
h is the horizontal shift = (picture=20 shifts right or left)
k is the vertical shift (picture shifts = up or=20 down)
The powers on the group (x - h) determines the shape of the = picture

(x-h) is a line
(x-h)^2 is a parabola
(x-h)^3 = is a=20 cube
square-root of (x-h) is a square root
1/(x-h) is a rational=20 function

Note: The book always draws the figure at (0,0) then = shifts it.=20 You can do the problems this way or you can draw the graph starting from = the=20 final location of the vertex, after it has been shifted. I will count = either one=20 correct.
Your textbook concentrates on parabolas and just = touches on=20 the other figures. The exam will test mainly parabolas, but be = aware that=20 in the next course you will be expected to know how to draw all the = other=20 figures as well. The graphing procedures are the same for the other = figures, you=20 are just drawing an altered parabola. (See my "graphing parabolas" notes = for=20 details).

The vertex can be found in the function (the = (h,k)). The=20 rest of the points can be found by choosing 2 points to the right and = left of=20 the vertex and plugging those values into the original function. This = will give=20 you a total of 5 points on the graph. If the figure is too big to = fit on=20 your graph, simply label those points that are off the edges so that I = will know=20 that you knew where they should have gone. You do not need to = label the=20 points that fit on the graph.

Using symmetry: If you will notice, each parabola = has the=20 same y-values on the left and right sides of the figure. You can = cut down=20 on the amount of calculations you need to do by plugging in only those = two=20 values on the right and then graphing the "mirror image" (symmetry) on = the other=20 side of the vertex.