A function f(x) is said to have a removable discontinuity at x=a if:
1. f is either not defined or not continuous at x=a.
2. f(a) could either be defined or redefined so that the new function IS continuous at x=a.
Let f(x)=2x^2+3x–14/x–2
Show that f(x) has a removable discontinuity at x=2 and determine what value for f(2) would make f(x) continuous at x=2.
f(2) = ?
To determine if a function has a removable discontinuity at a certain point, we can check if it is undefined or not continuous at that point. In this case, we are asked to find if the function f(x) = (2x^2 + 3x - 14) / (x - 2) has a removable discontinuity at x = 2.
To find out if the function is undefined at x = 2, we can substitute x = 2 into the expression and see if it results in an undefined value. Plugging in x = 2 into the function:
f(2) = (2(2)^2 + 3(2) - 14) / (2 - 2) = (8 + 6 - 14) / 0 = 0 / 0
Since we have a division by zero, the function is undefined at x = 2.
Now, to determine if the function is not continuous at x = 2, we can check the limit of the function as x approaches 2 from both the left and the right to see if they are equal. If the limits are not equal, then the function is not continuous at that point.
First, let's find the left-hand limit:
lim(x->2-) (2x^2 + 3x - 14) / (x - 2)
To evaluate this limit, we can substitute x = 2 - h, where h is a small positive number approaching zero:
lim(h->0-) (2(2 - h)^2 + 3(2 - h) - 14) / ((2 - h) - 2)
= lim(h->0-) (2(4 - 4h + h^2) + 6 - 3h - 14) / (-h)
= lim(h->0-) (8 - 8h + 2h^2 + 6 - 3h - 14) / (-h)
= lim(h->0-) (2h^2 - 11h)
Next, let's find the right-hand limit:
lim(x->2+) (2x^2 + 3x - 14) / (x - 2)
To evaluate this limit, we can substitute x = 2 + h, where h is a small positive number approaching zero:
lim(h->0+) (2(2 + h)^2 + 3(2 + h) - 14) / ((2 + h) - 2)
= lim(h->0+) (2(4 + 4h + h^2) + 6 + 3h - 14) / (h)
= lim(h->0+) (8 + 8h + 2h^2 + 6 + 3h - 14) / (h)
= lim(h->0+) (2h^2 + 11h)
Now, let's simplify and evaluate the limits:
lim(h->0-) (2h^2 - 11h) = 0
lim(h->0+) (2h^2 + 11h) = 0
Since the left-hand and right-hand limits are both equal to zero, the function is continuous at x = 2.
Since the function is undefined at x = 2 but has a continuous limit, it has a removable discontinuity at x = 2. To remove the discontinuity, we can redefine the function at x = 2 by evaluating the limit of the function as x approaches 2:
lim(x->2) (2x^2 + 3x - 14) / (x - 2)
To evaluate this limit, we can use algebraic manipulation or L'Hôpital's Rule. However, in this case, it is easier to see that if we factor the numerator as (2x - 7)(x + 2), we can cancel out the common factors of (x - 2) in both the numerator and the denominator:
lim(x->2) (2x - 7)(x + 2) / (x - 2)
= lim(x->2) 2x - 7
= 2(2) - 7
= 4 - 7
= -3
Therefore, f(2) = -3 would make f(x) continuous at x = 2.