Re: TI-M: trig and continuity
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Re: TI-M: trig and continuity
In a message dated 11/6/00 2:04:42 PM Mountain Standard Time,
skinnej16@juno.com writes:
How can i find out where a function is discontinuous? Lets take for
example:
f(x)=sin(x^2-2)?
When i put it into my calculator using the zeros( function
zeros(sin(x^2-2),x) the answer the calculator gives me is
{when(@n2greater than or equal to the square root of (@n2*pi+2) and a
whole bunch of other junk, am i doing it wrong? because the answer should
be none. Help, also if anybody has any programs for this and trigometric
limits can you tell me where to get them? Thanks a lot
.
As I remember, I think there's like 4 types of discontinuities:
1) Jump discontinuity, where the function "jumps" from one y value as you
move along the x-axis without taking the values of any of the intermediate y
values. Some piecewise functions can do this, but most "normal" functions
don't.
2) Removable discontinuity, where there's just a single point that's out of
place. Since it's just a single point, it's "removable" in that you can
"fix" the function up easily. For example, f(x) = sin(x)/x is discontinuous
at x = 0. However, lim sin(x)/x as x -> 0+ = lim sin(x)/x as x -> 0- = 1.
3) Infinite discontinuity, where lim f(x) as x -> c (from either side) = +/-
infinity (ie, and asymptote).
4) Oscillating discontinuity, such as with f(x) = sin(1/x) at x = 0.
f(x) = sin(x^2 - 2) is continuous throughout its domain (which is all reals),
since x^2 - 2 is continuous throughout its domain (all reals) and sin(x) is
continuous throughout its domain (also all reals).
If f(x) = sin(x^2 - 2) = 0, then x^2 - 2 = arcsin(0) = n*pi (for any integer
n). So the TI-89's solver should spit out something like "x = sqrt(@n1*pi +
2) or x = -sqrt(@n1*pi + 2)". I'm hoping all this I've said is correct...
JayEll