Caculus help please?

Question

6a) Check for discontinuity for this function

y = 2x/x-3 or (2x) (Divided by) (x-3)

i dont remember learning this in class so maybe someone can help me by showing me the steps, thanks a bunch

Answer 1

A rational function is always discontinuous when the denominator equals 0. In this case, discontinuity at x=3

Answer 2

When x = 3, the denominator of the function is
x - 3
= 3 - 3
= 0

When you divide by zero, a function blows up (to plus or minus infinity, depending on the direction from which you approach). You would say, then, that the function has a discontinuity at x = 3.

Answer 3

Discontinuity is where the bottom =0. So, it's a matter of setting x-3=0, then solving for x. x=3, so that is where you will have discontinuity. In cases where you have a bottom that is a more complex equation (like 3x^2-7x), then you will have multiple discontinuities. That's really all there is to it.

Answer 4

When x = 3, the quantity x - 3 equals zero which would have the function dividing by zero at x = 3. It is, therefore, discontinuous. (At x = 3.)

Answer 5

i believe it is just discontinuous where the denominator is zero so set it to 0 and solve for it

x-3 = 0
Therefore the function is discontinuous at x=3

Answer 6

Thinking algebraically, where do we run into trouble with this function? Think: domain restrictions... What can x "not" be?

Looking at the graph, you will see asymptotes at that same point.

If a function is discontinuous at some x, it is not differentiable there either.

Answer 7

x= 3 cannot happen, since this would make the fraction have a 0 in the denominator