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.
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!
3 answers
Figured it out thanks a lot!
You're welcome!
2 answers