Finding Discontinuity Algebraically

[Aresius]

I have a test tomorrow and this is a subject we only briefly touched on. I can find points of discontinuity graphically very easily, but I have no idea how to find them algebraically using just the equation.

I know that when the denominator = 0 and in most piecewise functions there is discontinuity. I suppose i'll put up a couple of examples.

$$\frac {x} {2x^2+x}$$

I canceled the x on this one which made the graph eqn look like a 1/x graph, which I know is not continuous at x=0, but it is also not continuous at x=-1/2 and I don't know why.

$$\frac {3} {x} + \frac{x-1} {x^2-1}$$

I assume that I can use one of the theorems which states that if both f(x) and g(x) are continuous then f(x)+g(x) is continuous? I'm stumped on how to find the discontinuities.

$$\frac {x^2+6x+9} {|x|+3}$$

I assume the answer is continuous everywhere here, since you cannot have a 0 on the denominator.

Also consequently, what is the standard 'working' that you should show for these problems in your answer?

 

[quantumdude]

$$\frac {x} {2x^2+x}$$

I canceled the x on this one which made the graph eqn look like a 1/x graph, which I know is not continuous at x=0, but it is also not continuous at x=-1/2 and I don't know why.

Well, when you canceled the x, what did you have left over in the denominator? If you did it correctly, then it should have been a 1st degree polynomial whose zero is x=-1/2.

$$\frac {3} {x} + \frac{x-1} {x^2-1}$$

I assume that I can use one of the theorems which states that if both f(x) and g(x) are continuous then f(x)+g(x) is continuous?

Yes.

I'm stumped on how to find the discontinuities.

You said it yourself earlier in this very post: When you get a zero in the denominator of a rational function, then you have a discontinuity. So where are the zeros of your two rational functions?

$$\frac {x^2+6x+9} {|x|+3}$$

I assume the answer is continuous everywhere here, since you cannot have a 0 on the denominator.

You assume rightly. Your teacher might want more than that though (I would).

Also consequently, what is the standard 'working' that you should show for these problems in your answer?

In all of these examples, you should set the denominator equal to zero and solve the resulting equation. In the first two examples that should explicitly involve factoring. In the last one it should involve isolating the absolute value, and then noting that |x|=(some negative number) has no solution.