How do you prove a function to be continuous on an interval?

Question

My math books just lists the definition of functions which are continuous intervals (i.e. they are continous on every point of that interval etc...), but a question asks you to find the intervals over which a function is continous (e.g, f(x) = x + sqrt(1 - x^2) ) and I couldn't find a method for that anywhere? Isn't there a particular technique for doing that, like there is for proving a function is continuous on a point?

Answer 1

http://en.wikipedia.org/wiki/Continuous_function
http://mathworld.wolfram.com/ContinuousFunction.html

Answer 2

One way to check for continuity is to find the limit of the function at a discontinuity. If the limit is the same (right or left), then the function is indeed continuous.

In the example

f(x) = x + sqrt(1-x^2)

The equation is defined for (1-x^2) greater or equal to zero (sqrt of a negative number is a complex quantity).

In this case, x<=1. So for the interval (-inf to 1) the function is defined and is continuous. For x>1, the function is not defined.

Hope this explanation helps.

Answer 3

Many types of functions, including polynomials, rational functions, trigonometric functions, or exponential functions (and other types; your text should have the complete list) are known to be continuous at every point at which they are defined, so finding points of discontinuity is equivalent to finding points at which the function is not defined, for these functions. Once you find the discontinuities, you know the intervals on which the function is continuous.

Many calculus texts feature problems in which a function is defined in pieces, such as f(x) = x+2 for x < 0, and x^2 - 1 for x >= 0. Then, the function may be discontinuous between pieces. For example, my example is discontinuous at x=0, but it's equal to a polynomial everywhere else, so it's continuous on (-infinity,0) and (0,infinity).

Answer 4

If you look for places where the function becomes undefined (division by zero), you will find infinite limits. Since the function is not differentialble at this point, it will not be continuous.

In short, look at the function as it approaches extrema. If there is an approach to infinity or a change in sign, the function is not continuous at that point.

Answer 5

ordinary, ill permit u do the concern by using ur self, yet ill tell u the technique shall we are saying u prefer to renowned if the the equation is continous at 0 so what we do is cut back (as x techniques 0 from useful direction) f(x) get the respond then: cut back(as x techniques 0 from detrimental direction) f(x) get the respond if the two solutions adventure, then the equation is continous at 0

Answer 6

f(x)=x+sqr(1-x^2)
1-x^2>0
(1-x)(1+x)>0
-1<=x<=1
observe the limits of the function from both 1 and -1.