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)
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:
[tex]f(x)=\frac{x^2-36}{x^3-36x} =\frac{(x-6)\.(x+6)}{x\,(x-6)\.(x+6)}[/tex]
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)
