Respuesta :

Arnold11
Rational functions are discontinuous where the denominator is equal to zero. If f(x)=(x+5)/(x^2 + 3x -10), we need to find where x^2 + 3x -10 = 0. 
Lets factor the denominator: (x+5)(x-2). So the denominator is 0 in x=-5 and x=2. Nominator is also 0 in x=-5. So we have a hole in x=-5 because the numerator and the denominator are both zero at x=-5 and a vertical asymptote in x=2. The hole is removable discontinuity so the correct answer is x=-5. 

Answer:

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

Step-by-step explanation:

A rational function [tex]f(x)=\frac{p(x)}{q(x)}[/tex] will have the discontinuity at x=a if q(a)=0. The discontinuity is removable when p(a)=0.

Given the function: [tex]f(x)=\frac{x+5}{x^2+3x-10}[/tex]

The discontinuity of the function f(x) occurs at the zeros of the denominators   [tex]x^2+3x-10[/tex]:

[tex]x^2+3x-10[/tex]=0

[tex](x+5)(x-2)[/tex]=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.


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