Homework Help: Basic definition of discontinuity

f(x)=3/(x²+x-6)

A simple definition would be that a point of discontinuity on the graph of f(x) occurs at x=a such that f(a) is undefined.

So where would your function be undefined for?

 

Say your function was

f(x)=1/(x-a)

Wouldn't you agree that [itex]\lim_{x \rightarrow a} = \infty[/itex] and so f(a) is undefined?

 

your f(x) is 3/(x-2)(x+3), as x→2, f(x)→∞, thus f(2) is undefined, similarly for f(-3)

 

but if x=2, don't you have one?

[tex]f(x)=\frac{3}{(x-2)(x+3)}[/tex]

 

cancel out with what?

f(x)=1/(x+3) is discontinuous at x=-3.

 

A function is discontinuous at a point c if the limit as x->c of f(x) does not equal f(c).

In this case, consider your c=2. f(c) is not defined. The limit of f(x) as x->2 from the positive side is +infinity. The limit of f(x) as x->2 from the negative side is -infinity.

The first way you can know this function is discontinuous at 2 is that it is not defined at 2. Then no matter what the limit of f(x) is as x->2, it is not equal to f(2) (even if the limit does not exist).

The second way you can know this function is discontinuous at 2 is that the limit of f(x) as x->2 from the positive side does not equal the limit of f(x) as x->2 from the negative side. Therefore the limit does not exist. If the limit does not exist, then it does not equal f(2) (even if f(2) is not defined).

Remember to keep in mind that a function is discontinuous at a point c if the limit as x->c of f(x) does not equal f(c). f(c) could exist, it just happens to not exist in this example.

 

Actually the limit of f(x) as x->a from the left is negative infinity, while the limit of f(x) as x->a from the right is positive infinity.

Therefore the limit of f(x) as x->a doesn't exist.

I think you mean x = -3.

 

Not necessarily, a rational function just means that it can be written as the ratio of two polynomials. But if, for example, the "denominator polynomial" is a constant, then the function is continuous over its domain.

 

If you choose your numerator to be a first order polynomial, then yes. But what about f(x) = [(x+1)^2]/3

 

That function would be discontinuous at x=2 and x=-2