Question

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

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 (which can be canceled). Here, the common factor is \(x + 5\), so we set \(x+5 = 0\), which gives \(x=-5\). After canceling \(x + 5\) (for \(x
eq - 5\)), the function simplifies to \(\frac{1}{x - 2}\), so the removable discontinuity is at \(x=-5\).

Answer:

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