Explain why f(x)= 1/(x-x3)^3 is not continuous at x = 3.

Option b.

Step-by-step explanation:

 1. You have the function given in the problem above.

2. You must keep on mind that. by definition, the division by zero does not exist.

3. The value x=3 makes the denominator of the function f(x) equal to zero. Therefore you can conclude that the function shown in the problem is not defined at x=3.

The answer is the option b.

this may not be at all, but i use the same website you're using for school, and if i'm stuck doing one of these i try to look for it on the section.

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation: