Which of the following functions f has a removable discontinuity at a? If the discontinuity is removable, find a function g that agrees with f for x a and is continuous at a. (If the discontinuity is not removable, enter NONE.)
1. f(x)= (x^4-1)/(x-1), a=1
?The discontinuity is removable.
?The discontinuity is not removable.
g(x) = ?
2. f(x)= (x^3-x^2-12x)/((x-4), a=4
?The discontinuity is removable.
?The discontinuity is not removable.
g(x) = ?
3) f(x) = [[ sin(x) ]], a = ð
(Recall that [[h(x)]] means the largest integer that is less than or equal to h(x).)
?The discontinuity is removable.
?The discontinuity is not removable.
g(x) =?
I don't understand what they are asking! If anyone could explain that, that would be a great help!
1. The discontinuity in f(x) = (x^4-1)/(x-1) at a=1 is not removable. Sorry, there's no clowning around here. None.
2. The discontinuity in f(x) = (x^3-x^2-12x)/(x-4) at a=4 is removable. Let's have some fun! The function g(x) that agrees with f(x) for x ≠ 4 and is continuous at x=4 can be defined as g(x) = (x^3-x^2-12x-48)/(x-4). It's like fixing a broken smile, all good now!
3. The discontinuity in f(x) = [[sin(x)]] at a=π is removable. Who doesn't love some clowning around? The function g(x) that agrees with f(x) for x ≠ π and is continuous at x=π can be defined as g(x) = 0. It's like putting a funny hat on π, all better!
To determine if a function has a removable discontinuity at a particular point, we need to check if the function is undefined or has a different value at that point compared to its surrounding values. Let's analyze each function individually:
1. f(x) = (x^4-1)/(x-1), a = 1
To check for a removable discontinuity at a = 1, we need to evaluate f(1) and examine the limit of f(x) as x approaches 1.
Plugging in x = 1, we get f(1) = (1^4-1)/(1-1) = 0/0, which is undefined. Since the function is undefined at x = 1, we can conclude that there is a removable discontinuity at a = 1.
To find the function g(x) that agrees with f(x) for x ≠ 1 and is continuous at x = 1, we can simplify f(x) by factoring the numerator using the difference of squares:
f(x) = ((x^2+1)(x^2-1))/(x-1)
We can cancel out the (x-1) factor in the numerator with the denominator, resulting in:
f(x) = (x^2+1)(x+1)
Thus, the function g(x) that agrees with f(x) for x ≠ 1 and is continuous at x = 1 is g(x) = (x^2+1)(x+1).
2. f(x) = (x^3-x^2-12x)/(x-4), a = 4
Again, to check for a removable discontinuity at a = 4, we need to evaluate f(4) and examine the limit of f(x) as x approaches 4.
Plugging in x = 4, we get f(4) = (4^3-4^2-12(4))/(4-4) = 0/0, which is undefined. Therefore, there is a removable discontinuity at a = 4.
To find the function g(x) that agrees with f(x) for x ≠ 4 and is continuous at x = 4, we can simplify f(x) by factoring the numerator:
f(x) = x(x^2-x-12)/(x-4)
Again, the (x-4) factor in the denominator can be canceled out, resulting in:
f(x) = x(x+3)(x-4)/(x-4)
The (x-4) factor in both the numerator and denominator can be canceled out to obtain:
f(x) = x(x+3)
Therefore, the function g(x) that agrees with f(x) for x ≠ 4 and is continuous at x = 4 is g(x) = x(x+3).
3. f(x) = [[ sin(x) ]], a = π
To determine if there is a removable discontinuity at a = π, we need to check if the function f(x) has a different value at π compared to its surrounding values.
The function [[h(x)]] returns the largest integer less than or equal to h(x). In this case, [[sin(x)]] will always be an integer from -1 to 1, since sin(x) ranges from -1 to 1. As such, [[sin(x)]] does not have any discontinuities, including removable ones.
Therefore, the discontinuity in this function is not removable, i.e., the discontinuity is non-removable.
Since the function does not have a removable discontinuity at a = π, there is no need to find a function g(x) that agrees with f(x) and is continuous at x = π.
A removable discontinuity is a point where the function does not exist, but it's limit exists on both sides for an interior point, and on the interior side if it is an end point.
The discontinuity can be removed by defining the function at the point of discontinuity as the limit.
Take for an example the first question:
f(x)=(x^4-1)/(x-1)
Since
x^4-1 = (x²+1)(x+1)(x-1)
it is evident that the limit at x=1 exists on both sides, but f(x) is undefined at x=1.
By redefining f(x) as
g(x)=(x^4-1)/(x-1) for x≠1, and
g(x)=4 for x=1
g(x) is now continuous on ℝ, and the discontinuity has been removed in the new function.
You can work on the other problems on this basis.