Discuss the Discontinuity of
f(x) = x+7/x^2-49
non-removable and removable
Discuss the Discontinuity of
f(x) = x+7/x^2-49
non-removable and removable
Hi!
Happy Easter!
There is classification of discontinuities (you have at x=0 the Type 3)
+++ IHM
SEE OTHER ANSWER. IT IS MORE COMPLETE.
A discontinuity occurs when the function does not have an answer at a certain point. Usually this is caused by one of two things: division by zero or taking the square root of a negative number.
Here, f(7) creates a discontinuity, because the denominator becomes (7)^2 - 49 = 0, and division by zero is not possible.
Factoring the denominator to read (x+7)/[(x+7)(x-7)] lets us simplify the equation to be 1/(x-7), but unfortunately this does NOT remove the discontinuity. If the numerator had read (x-7) as opposed to (x+7), we would have been able to end up with 1/(x+7), and we would no longer have the discontinuity.
Visualising this discontinuity is easy. Simply substitute in numbers that gradually approach 7, first 'from the left' (as in, going from 6 to 7), then 'from the right' (going from 8 to 7). From the left, the numerator is positive, but the denominator is negative, as (6.99999)^2 < 49. You are dividing by ever smaller negative numbers, thus, f(x) approaches
NEGATIVE infinity. From the right, (7.0000001)^2 > 49, so it is always positive as it approaches zero, thus, f(x) approaches
positive infinity.
When this happens, we call it a vertical asymptote, and it occurs at f(7)