Question

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

Explanation:

Step1: Factor the denominator

Factor \(x^{2}+3x - 10\). Using the formula \(ax^{2}+bx + c=a(x - x_1)(x - x_2)\) where \(x_{1,2}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(a = 1\), \(b=3\), \(c=-10\).
\(x=\frac{-3\pm\sqrt{9+40}}{2}=\frac{-3\pm7}{2}\), so \(x_1 = 2\), \(x_2=-5\) and \(x^{2}+3x - 10=(x + 5)(x - 2)\).
The function becomes \(f(x)=\frac{x + 5}{(x + 5)(x - 2)}\).

Step2: Simplify the function (for \(x

eq - 5\))
Cancel out the common factor \((x + 5)\) (when \(x
eq-5\)), and we get \(f(x)=\frac{1}{x - 2}\) (with a hole at \(x=-5\) since the original function is undefined at \(x=-5\) (because the denominator of the original function \(x^{2}+3x - 10=(x + 5)(x - 2)\) is zero at \(x=-5\)) and the factor \((x + 5)\) can be canceled).

Answer:

\(x=-5\)