Where is the removable discontinuity of f(x)= x+5/x^2+3x-10

x= -5
x= -2
x= 2
x= 5

The removable discontinuity of f(x) is at x=-5.

Step-by-step explanation:

A rational function will have the discontinuity at x=a if q(a)=0. The discontinuity is removable when p(a)=0.

Given the function:

The discontinuity of the function f(x) occurs at the zeros of the denominators   :

=0

=0

⇒ x=2 ,-5.

When x=2 we have the numerator x+5 is

x+5=2+5=7≠0

when x=-5, then the numerator is

x+5=-5+5=0.

Thus, the function f(x) has a removable discontinuity at x=-5.

you just switch the numbers basically so it would be 3/17