Question

where is the removable discontinuity of f(x) = (x + 5)/(x²+3x - 10) located? x = - 5 x = 2 x = - 2 x = 5

Explanation:

Step1: Factor the numerator

Factor $x^{2}+3x - 10$ as $(x + 5)(x-2)$. So $f(x)=\frac{(x + 5)(x - 2)}{x + 5}$.

Step2: Simplify the function

Cancel out the common factor $(x + 5)$ (for $x
eq - 5$), getting $f(x)=x - 2$ for $x
eq - 5$. A removable discontinuity occurs when a factor can be canceled out in a rational - function. The original function $f(x)$ is undefined at $x=-5$, but the limit as $x\to - 5$ exists.

Answer:

$x=-5$