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[ap-calculus] Re: TI83 question - undefined fuction value
Posted:
Dec 11, 2005 3:46 AM
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In a message dated 12/10/05 5:21:54 AM, tguay@cinci.rr.com writes:
<< In the rational function y = (6x^4-17x^3-25x^2+19x-3
)/(x^3-2x^2-7x+2), the factor (x^2-4x+1) appears in the numerator and
denominator. So I believe there are two removable discontinuities or
holes in the graph: at x = 2 + sqrt(3) and x = 2 - sqrt(3).
However, the TI83 plus calculator gives a confusing picture of this
situation. Using trace, the y value for the x value of 2 - sqrt(3) is
undefined as we expect. But the y value for the x value of 2 + sqrt(3)
is 9 which is nowhere near the rest of the graph of the function (looks
like a jump discontinuity instead of an undefined situation).
What I'd like to know is how the calculator is coming up with this >>
Terry,
The calculator is coming up with an erroneous answer because it does not find
the denominator of your function to be exactly zero for x = 2 + sqrt 3. It
does find the denominator of your function to be exactly zero for x = 2 -
sqrt 3, and therefore correctly identifies a discontinuity at x = 2 - sqrt 3.
For x = 2 + sqrt 3,
your denominator is 1 x 10^(-12) and your numerator is 9 X 10^(-12). That
is why you are getting an answer of "9." The only way I could get your
graph to display correctly at x = 2 + sqrt 3 was to use the 'round' function (
math menu, num menu, round) on the denominator of your function as it is entered
in the Y= menu. Doing this will force the denominator to be exactly zero at
x = 2 + sqrt 3. On the home screen, I was able to enter your function and
get the discontinuity at x = 2 + sqrt 3 by going to the mode menu and setting
the 'float' entry to a fixed number of decimal places ( 9 will work).
Apparently, changing the 'float' entry does not affect the number of decimal
places being carried on the graph. The ideal solution would be to change the
number of decimal places being carried during the graphing operation. I don't
know if this can be done.
Maybe somebody can give us all a way do deal with a 'zero' that is not an
exact 'zero'
when an exact 'zero' is needed.
Bob Coger
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