15.4 - Rational Functions: Discontinuities, Limits, Partial Fractions

Hey everybody!

to update from Anna's post: two and a half weeks until summer! yay, the greatest moment ever!

down to business...when you divide a polynomial by a linear factor, the result usually resembles a quadratic function:

f(x) = (x^3 - 5x^2 + 8x - 6)/(x-3)

when you use synthetic substitution:
3} 1 -5 8 -6
3 -6 6
1 -2 2 0 is the answer,
so the net equation is x^2 - 2x + 2. there is a discontinuity at x = 3. this means that the answer is undefined because when you plug 3 in for x, the denominator is equal to 0. you can use this later for finding limits.The dot above the 3 on the x axis is the representation of the discontinuity.

LIMITS! If you want to find the y value that the discontinuity should would be, plug the equation into your calculator. then go to TBLSET, change TBLstart to a few decimal places before the desired number, and change the scale to 0.1, or some very small number. then go to TABLE and find the y values for 2.99 and 3.01. it should be obvious what the missing number is. in this case it is 5.

lim f(x) = 5

you can also find this algebraically by plugging 3 back into the net equation:
x^2 - 2x + 2
3^2 - 2(3) + 2
9-6+2 = 5

You can remove the discontinuity algebraically by simplifying the following:

(x-3)(x^2-2x +2)/(x-3) you can cancel the two (x-3)'s.

there are some equations that have more complex discontinuities that are not removable:

with the equation g(x) = (x^3 - 5x^2 + 8x - 5)/(x-3) there is an asymptote at x=3.
you can use synthetic substitution to find the factored form:

3} 1 -5 8 -5
3 -6 6
1 -2 2 1 is the answer, so it has a remainder. the net equation is therefore x^2-2x+2+1/(x-3) and the graph...


You can't factor this type of problem so it has an infinite or irremovable discontinuity.
the limit of g:
lim g(x) = does not exist or, in some cases, is infinity

PARTIAL FRACTIONS: polynomial/polynomial = partial fraction

(9x-7)/(x^2-x-6) = ....

you need to factor the denominator to (x-3)(x+2). you can represent the numerator with variables, A and B.

A/(x-3) + B/(x+2) --> give them equal denominators by multiplying the first value by (x+2) and the second by (x-3).

A(x+2)/(x-3)(x+2) + B(x-3)/(x-3)(x+2) --> distribute the A and B values.

(Ax +2A +Bx+ 2B)/(x-3)(x+2) = (9x-7)/(x^2-x-6)

you know there are 9x's and that you have A + B x's, so A+B=9
you also know that... 2A-3B = -7,
so you can use matrices or substitution to solve for A and B. they equal 4 and 5, respectively.

a shorter way to do this is by making the factors in the denominator equal zero...like this:
(9x-7)/(x-3)(x+2)
plug in 3 for x... (27-7)/(3+2) --> 20/5 --> 4
plug in -2 for x... (-18-7)/(-2-3) --> -25/-5 --> 5
plug those values back in to get the partial fraction:

4/(x-3) + 5/(x+2) and that is the answer.

Examples!

What is the partial fraction for (4x-14)/(x^2 + 2x - 8)

first, factor the denominator: (x-2)(x+4).

then, set up two fractions: A/(x-2) + B/(x+4).
make equal denominators: A(x+4)/(x-2)(x+4) + B(x-2)/(x-2)(x+4)
distribute: (Ax + 4A +Bx - 2B) /(x-2)(x+4) = 4x-14
A + B = 4 and 4A - 2B = -14
solve with matrices or substitution.
A = 1, B = 5

or...

this way is better for speed.
(4x-14)/(x-2)(x+4)
plug in 2 for x to find A value, then plug in -4 to find B.
A = -6/6 --> -1
B = -30/-6 --> 5

the partial fraction is 1/(x-2) + 5/(x+4)

LINKS:

here is some extra help for...
partial fractions: http://www.sosmath.com/algebra/pfrac/pfrac.html
discontinuities: http://www.tpub.com/math2/39.htm
limits: http://en.wikipedia.org/wiki/Limit_(mathematics)

Reminder: Mariclare is next

this is good fun: it's german AND stop motion! two of my fave things!
http://youtube.com/watch?v=4RST7CYjfjo
i don't know what he's saying though. i could be exposing you to crude language without knowing it... :(

i hope all of this helped!