anonymous
  • anonymous
how do you find out if something is continuous? for example, the first question i am given on this is: f(x) = 2x+x^2/3 on [-1,1]. i know that the answer is that it is continuous but i don't understand how to get to that point.
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
the lim from the left hand=lim from right hand = value at the poit we need to find out the f is continuous
anonymous
  • anonymous
how do you find that...sorry, i don't understand cont. at all...
anonymous
  • anonymous
If f'(c) exist, then f is continous at c

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anonymous
  • anonymous
if f` = \[(-\cos (1/x ^{2})/x^{2}\] then f`c does not exist at x =0. Hence there has to be a discontinuity there. Does this make any sense to you?
anonymous
  • anonymous
kind of, thanks for trying to explain it! i really appreciate it hahahah this is just over my head LOL (: imma try going to my ta.
anonymous
  • anonymous
Take the derivative of the function. If the derivative exist, the function is continous at that point.
anonymous
  • anonymous
This is not actually true. If the function is discontinuous at a point it will not be differentiable at that point, but the reverse is not always true. As an example check out : http://en.wikipedia.org/wiki/Weierstrass_function Which has the neat property that it has no discontinuities but also has no intervals over which it is differentiable.
anonymous
  • anonymous
There are lots of formal bits about continuity, but some trends do show up. Discontinuities usually happen because you have to divide by 0 in some spot. If you ever do, then it is discontinuous at that point. eg . 1 /( x-1) = 1/0 at x=1; therefore we have a discontinuity at x=1. You can check for this by looking at all denominators and setting them equal to zero. If any are equal to zero then you have a discontinuous functions. The other main place where discontinuities show up is if there is ever a spot where a function evaluates to infinite. A fine example of this the tan function. tan(pi/2) = infinite. This means if you get this in any function it is discontinuous. It also helps to learn features about certain kinds of functions. It is exceptionally valuable to know that sin(x) = 0, when x=n(pi), where n = all integers (ie ...-2,-1,0,1,2...) . If you want to be very good at this, then find out any important points (ie where the function = 0, infinite, etc...) The internet is a great tool for this, and so are most pre-calculus textbooks. y= cos(x) y=sin(x) y=tan(x) y=e^x y= cosh(x) y=sinh(x) y= tanh(x) Polynomials in general. I know this isn't the most rigorous test, but it a good basis to pick up on major discontinuities.
anonymous
  • anonymous
The inquiry was about a "well behaved" function! The inquiry was not about a function that literally has a "kink" at every point.

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