Question

where is the removable discontinuity of $f(x) = \frac{x + 5}{x^2 + 3x - 10}$ located?
$x = 2$
$x = -5$
$x = 5$
$x = -2$

Explanation:

Step1: Factor the denominator

The denominator is \(x^2 + 3x - 10\). We factor it as \((x + 5)(x - 2)\) since \(5\times(-2)=-10\) and \(5+(-2)=3\). So the function becomes \(f(x)=\frac{x + 5}{(x + 5)(x - 2)}\).

Step2: Identify removable discontinuity

A removable discontinuity occurs where the numerator and denominator have a common factor. Here, the common factor is \(x + 5\), which means we can cancel it (for \(x
eq - 5\)). The value that makes the common factor zero is \(x=-5\), so the removable discontinuity is at \(x = - 5\).

Answer:

\(x=-5\) (the option with \(x = - 5\))