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

Factor \(x^2 + 3x - 10\). We need two numbers that multiply to \(-10\) and add to \(3\). Those numbers are \(5\) and \(-2\), so \(x^2 + 3x - 10=(x + 5)(x - 2)\).

Step2: Simplify the function

The function \(f(x)=\frac{x + 5}{(x + 5)(x - 2)}\) can be simplified (for \(x
eq - 5\)) to \(f(x)=\frac{1}{x - 2}\) by canceling the common factor \((x + 5)\).

Step3: Identify the removable discontinuity

A removable discontinuity occurs where the function is undefined (due to a common factor in the numerator and denominator) but can be made continuous by redefining the function at that point. Here, the common factor is \((x + 5)\), so we set \(x+5 = 0\), which gives \(x=-5\). At \(x = 2\), the denominator is zero but there is no common factor, so it's a non - removable (infinite) discontinuity.

Answer:

\(x=-5\)