Let f(x) = [(2x2+4 x −70)/(x−5)]
Show that f(x) has a removable discontinuity at x=5 and determine what value for f(5) would make f(x) continuous at x=5
Must define f(5)=
I've been stuck on this problem for some time how would i do this one? Ty!
First we observe that the (x-5) in the denominator creates a vertical asymptote at x=5, hence f(x) is undefined at x=5.
Next, we factorize the numerator:
f(x) = [(2x2+4 x −70)/(x−5)]
=(2x+14)(x-5)/(x-5)
We can see that the (x-5) cancel out in both the numerator and denominator, but we are not allowed to cancel, otherwise the properties of the function will be changed.
Does f(x) have a removable discontinuity (at x=5)? We will find out by finding the limits of f(x) at x=5+ and x=5-. Post if you are not sure how this can be done.
If they are both equal to the same value (=24 for both), then the discontinuity can be removed by defining explicitly f(5)=24.
Figured it out thanks a lot!
You're welcome!
To show that f(x) has a removable discontinuity at x = 5, we need to verify two conditions:
1. The function is undefined at x = 5.
2. The limit of f(x) as x approaches 5 exists.
First, let's determine if the function is undefined at x = 5. We can do this by substituting x = 5 into f(x). However, we notice that substituting x = 5 directly will result in division by zero, which is undefined. So instead, we factorize the numerator:
f(x) = (2x^2 + 4x - 70)/(x - 5)
The numerator can be factored as (2(x + 7)(x - 5))/(x - 5). Now, we can cancel out the common factor (x - 5) in the numerator and denominator:
f(x) = 2(x + 7)
Now that the function is defined at x = 5, the first condition for a removable discontinuity is satisfied.
Next, we need to determine the value of f(5) that would make f(x) continuous at x = 5. For this, we substitute x = 5 into the simplified function f(x):
f(5) = 2(5 + 7)
= 2(12)
= 24
Hence, if we define f(5) = 24, the function f(x) would become continuous at x = 5.
To summarize:
- f(x) has a removable discontinuity at x = 5 because the function is initially undefined at x = 5, but becomes defined after canceling out the common factor (x - 5).
- To make f(x) continuous at x = 5, we need to define f(5) = 24.