Question

What are the removable discontinuities of the following function? f (x) = StartFraction x squared minus 36 Over x cubed minus 36 x EndFraction

Answer 1

Answer:

the removable discontinuities are at x = 6 and at x = -6

Step-by-step explanation:

Notice that the function has common binomial factor in numerator and denominator:

$f(x)=\frac{x^2-36}{x^3-36x} =\frac{(x-6)\.(x+6)}{x\,(x-6)\.(x+6)}$

therefore, the removable discontinuities are those at x= 6 and x = -6 that correspond to zeros common in numerator and denominator, and therefore those associated with the (x + 6) factor, with the (x - 6) factor.

There is a non-removable discontinuity at x = 0.

The discontinuities can be removed by re-assigning the value of f(x) at x=6 (as 1/6), and at x = -6 (as -1/6)

Answer 2

Answer:

the removal discontinuity of the following function at x=-6 or x=6.

Step-by-step explanation:

factoring  f(x)  = (x - 6)(x + 6)

                         ----------------

                          x(x - 6)^x + 6)