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 an answer does not exist, enter DNE.)
f(x)= (x^4  1) / (x  1), a = 1
for g(x) I got (x+1) / (x^2 + 1) but it says the answer is wrong.
f(x)= (x^4  1) / (x  1), a = 1
for g(x) I got (x+1) / (x^2 + 1) but it says the answer is wrong.

Aremovable discontinuity is a hole, that cancels out. Then you fill it in to remove the discontinuity.
I think you meant to have multiplication where you have the division.
f(x)= (x^41)/(x1)= (x^21)(x^2+1)/(x1)= (x+1)(x1)(x^2+1)/(x1)
Now the (x1) cancels out
So g(x) = (x+1)(x^2+1)= x^3+x^2+x+1, where g(1)= 4
This means f(x) and g(x) are the same except at a=1, where f(x) is undefined, and would have a hole at (1,4)
Hoping this helps!
I think you meant to have multiplication where you have the division.
f(x)= (x^41)/(x1)= (x^21)(x^2+1)/(x1)= (x+1)(x1)(x^2+1)/(x1)
Now the (x1) cancels out
So g(x) = (x+1)(x^2+1)= x^3+x^2+x+1, where g(1)= 4
This means f(x) and g(x) are the same except at a=1, where f(x) is undefined, and would have a hole at (1,4)
Hoping this helps!