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# Removable and Nonremovable discontinuity?

1. Give an example of a function f(x) that is continuous for all values of x except x=2, where it has a removable discontinuity. Explain how you know that f is discontinuous at x=2, and how you know the discontinuity is removable.

2.Nonremovable discontinuity. Give an example of a function g(x) that is continuous for all values of x except x=-1, where it has a nonremovable discontinuity. Explain how you know that g is discontinuous there and why the discontinuity is not removable.

3. If funcitons f(x) and g(x) are continuous for 0≤x≤1, could f(x)/g(x) possibly be discontinuous at a point of [0,1]? Give reasons for your answer.

### 4 Answers

- Old TeacherLv 78 years agoFavorite Answer
Look at a rational function, and factor numerator and denominator; if a factor "cancels" then it represents a hole in the graph, which is removeable discontinuity. You can "fix" the hole by filling it in.

If a factor in the denominator does not cancel, then it represents a vertical asymptote, which is nonremoveable, since you cannot fix the space caused by the asymptote.

1. Give an example of a function f(x) that is continuous for all values of x except x=2, where it has a removable discontinuity. Explain how you know that f is discontinuous at x=2, and how you know the discontinuity is removable.

f(x)= [(x+1)(x-2)] /(x-2)

It is discontinuous at x= 2, because you would have zero in the denominator.

This would simplify to f(x)=( x+1), but since 2 is not in the original domain, there would be a hole at (2,3)

2.Nonremovable discontinuity. Give an example of a function g(x) that is continuous for all values of x except x=-1, where it has a nonremovable discontinuity. Explain how you know that g is discontinuous there and why the discontinuity is not removable.

g(x)= x/(x+1)

Since g(-1)= -1/0 , this is undefined, and represents a vertical asymptote at x= -1

A vertical asymptote is a nonremoveable discontinuity.

3. If funcitons f(x) and g(x) are continuous for 0≤x≤1, could f(x)/g(x) possibly be discontinuous at a point of [0,1]? Give reasons for your answer.

If g(x) = 0 when x is in [0,1],then f(x)/ g(x) would be undefined at that point, even though f(x) is defined and g(x) is defined.

Ex: f(x)=4x-3 and g(x) = x-1/2

Then f(1/2)= -1 and g(1/2)= 0

So f(x)/g(x) would have a vertical asymptote at x= 1/2

Hoping this helps!

- BrianLv 78 years ago
1.) Let f(x) = (x - 2) / (x^3 - 2x^2 + 2x - 4). Now the denominator factors to

x^2*(x - 2) + 2*(x - 2) = (x^2 + 2)(x - 2), which would suggest that f(x) is

discontinuous at x = 2 since this is where the denominator is 0, but since we can

cancel out (x - 2) from numerator and denominator with the condition that x does

not equal 2 we end up with f(x) = 1 / (x^2 + 2) for x not equal to 2, where we have

'removed' the discontinuity and replaced it with an open dot at (2, 1/6). Finally,

since (x^2 + 2) > 0 for all x we can conclude that f(x) is continuous for all values

of x except x = 2, where it has a removable discontinuity.

2.) The simplest example would be g(x) = 1 / (x + 1). The denominator is 0 for

x = -1 and cannot be removed. g(x) is continuous for all other x. Note that

g(x) -> infinity as x -> -1 from the right and g(x) -> negative infinity as x -> -1

from the left, so the two-sided limit as x -> -1 does not exist.

3.) Yes, it could. For example, let f(x) = x^2 and g(x) = x - (1/2). Then f(x) and g(x)

are both continuous on [0,1] but f(x)/g(x) = x^2 / (x - (1/2)) has a non-removable

discontinuity at x = 1/2 where g(x) = 0.

- lezakLv 44 years ago
1.) Let f(x) = (x - 2) / (x^3 - 2x^2 + 2x - 4). Now the denominator causes to x^2*(x - 2) + 2*(x - 2) = (x^2 + 2)(x - 2), which would endorse that f(x) is discontinuous at x = 2 on the grounds that this is the place the denominator is zero, but considering the fact that we can cancel out (x - 2) from numerator and denominator with the that x does no longer equal 2 we end up with f(x) = 1 / (x^2 + 2) for x not equal to 2, where we have now 'removed' the discontinuity and changed it with an open dot at (2, 1/6). Eventually, given that (x^2 + 2) > zero for all x we are able to conclude that f(x) is continuous for all values of x besides x = 2, the place it has a detachable discontinuity. 2.) The simplest illustration would be g(x) = 1 / (x + 1). The denominator is zero for x = -1 and cannot be removed. G(x) is steady for all different x. Word that g(x) -> infinity as x -> -1 from the proper and g(x) -> poor infinity as x -> -1 from the left, so the two-sided limit as x -> -1 does now not exist. 3.) sure, it might. For example, let f(x) = x^2 and g(x) = x - (half). Then f(x) and g(x) are both steady on [0,1] however f(x)/g(x) = x^2 / (x - (1/2)) has a non-detachable discontinuity at x = half where g(x) = 0.