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

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

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

Answer

Explanation:

Step1: Factor the denominator

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

Step2: Identify removable discontinuity

A removable discontinuity occurs when a factor in the numerator and denominator cancels out. Here, the factor $(x + 5)$ cancels (for $x\neq - 5$). The function is undefined at $x=-5$ due to the original denominator being zero, but the $(x + 5)$ terms simplify. So the removable discontinuity is at $x=-5$.

Answer:

$x=-5$